Integrals and Series

Lobachevsky-type logarithmic integrals via partial-fraction series

2026-10-24T10:12:00+05:30

Whenever an even, \(\pi\)-periodic function \(f(t)\) is integrated against \(\frac{1}{t^2}\) over the entire real line, splitting \(\mathbb{R}\) into intervals of length \(\pi\) and summing the translates \(\frac{1}{(t+n\pi)^2}\) folds the integral onto \(\left[0, \frac{\pi}{2}\right]\). We will use this Lobachevsky periodization method to prove that for \(|a| < 1\) and \(b > 0\),

\[ \int_{0}^{\infty} \ln \left(\frac{1+a \sin^{2} (bx)}{1-a \sin^{2} (bx)} \right) \frac{1}{x^{2}}\,dx = \pi b \left( \sqrt{1+a} - \sqrt{1-a} \right), \]

and then record three related consequences.

Folding the real line onto \(\left[0, \frac{\pi}{2}\right]\)

We make use of Euler's partial-fraction expansion for the squared cosecant:

\[ \sum_{n=-\infty}^\infty \frac{1}{(t+n\pi)^2} = \frac{1}{\sin^2 (t)}. \]

Let \(I\) denote the target integral. Using the evenness of the integrand and substituting \(t = bx\), we decompose \(\mathbb{R}\) into the intervals \(\left[n\pi - \frac{\pi}{2}, n\pi + \frac{\pi}{2}\right]\) for \(n \in \mathbb{Z}\):

\[ \begin{aligned} I &= \frac{1}{2}\int_{-\infty}^\infty \ln \left(\frac{1+a \sin^{2} (bx)}{1-a \sin^{2} (bx)} \right)\frac{1}{x^2}\,dx \\ &= \frac{b}{2}\int_{-\infty}^\infty \ln \left(\frac{1+a \sin^{2} (t)}{1-a \sin^{2} (t)} \right)\frac{1}{t^2}\,dt \quad (t=bx) \\ &= \frac{b}{2}\sum_{n=-\infty}^\infty \int_{n\pi - \frac{\pi}{2}}^{n\pi + \frac{\pi}{2}} \ln \left(\frac{1+a \sin^{2} (t)}{1-a \sin^{2} (t)} \right)\frac{1}{t^2}\,dt \\ &= \frac{b}{2}\sum_{n=-\infty}^\infty \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln \left(\frac{1+a \sin^{2} (t)}{1-a \sin^{2} (t)} \right)\frac{1}{(t+n\pi)^2}\,dt \\ &= \frac{b}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln \left(\frac{1+a \sin^{2} (t)}{1-a \sin^{2} (t)} \right)\left( \sum_{n=-\infty}^\infty \frac{1}{(t+n\pi)^2} \right) dt \\ &= \frac{b}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln \left(\frac{1+a \sin^{2} (t)}{1-a \sin^{2} (t)} \right)\frac{1}{\sin^2(t)}\,dt \\ &= b \int_0^{\frac{\pi}{2}} \ln \left(\frac{1+a \sin^{2} (t)}{1-a \sin^{2} (t)} \right)\frac{1}{\sin^2(t)}\,dt. \end{aligned} \]

Expanding in a power series in \(a\)

For \(|a| < 1\), expand the logarithm using \(\ln\left(\frac{1+u}{1-u}\right) = 2\sum_{k=0}^\infty \frac{u^{2k+1}}{2k+1}\) with \(u = a\sin^2 t\), and apply Wallis's formula \(\int_0^{\pi/2} \sin^{2m}(t)\,dt = \frac{\pi}{2}\frac{1}{2^{2m}}\binom{2m}{m}\) with \(m = 2k\):

\[ \begin{aligned} I &= 2b \int_0^{\frac{\pi}{2}}\left(\sum_{k=0}^\infty \frac{a^{2k+1}\sin^{4k+2}(t)}{2k+1} \right)\frac{1}{\sin^2(t)}\,dt \\ &= 2b\sum_{k=0}^\infty \frac{a^{2k+1}}{2k+1} \int_0^{\frac{\pi}{2}} \sin^{4k}(t)\,dt \\ &= \pi b\sum_{k=0}^\infty \frac{a^{2k+1}}{(2k+1) 2^{4k}} \binom{4k}{2k}. \end{aligned} \]

To sum this series in closed form, recall the binomial expansion \((1-u)^{-1/2} = \sum_{n=0}^\infty \frac{1}{2^{2n}}\binom{2n}{n}u^n\). Taking the even part in \(u\) gives

\[ \frac{(1-u)^{-1/2} + (1+u)^{-1/2}}{2} = \sum_{k=0}^\infty \frac{u^{2k}}{2^{4k}}\binom{4k}{2k}, \]

and integrating both sides from \(u = 0\) to \(u = a\) produces

\[ \sum_{k=0}^\infty \frac{a^{2k+1}}{(2k+1) 2^{4k}} \binom{4k}{2k} = \int_0^a \frac{(1-u)^{-1/2} + (1+u)^{-1/2}}{2}\,du = \sqrt{1+a} - \sqrt{1-a}. \]

Therefore,

\[ I = \pi b \left( \sqrt{1+a}-\sqrt{1-a}\right). \]

Three companion identities

A first-order Lobachevsky integral: Using the partial-fraction expansion \(\sum_{n=-\infty}^\infty \frac{(-1)^n}{t+n\pi} = \frac{1}{\sin t}\) in the same manner shows that for \(|a| < 1\) and \(b > 0\),

\[ \int_{0}^{\infty} \ln \left( \frac{1+a \sin(bx)}{1-a \sin(bx)} \right) \frac{dx}{x} = \pi \arcsin(a). \]

A dilogarithm integral: Dividing the identity above (with \(b=1\)) by \(a\) and integrating both sides from \(a=0\) to \(a=\frac{1}{\sqrt{2}}\) using \(\int_0^z \ln\left(\frac{1+u}{1-u}\right)\frac{du}{u} = \operatorname{Li}_2(-z) - \operatorname{Li}_2(z)\) yields

\[ \int_0^\infty \frac{\operatorname{Li}_2\left(\frac{\sin x}{\sqrt{2}}\right)-\operatorname{Li}_2\left(-\frac{\sin x}{\sqrt{2}}\right)}{x}\,dx = \pi\int_0^{\frac{1}{\sqrt{2}}}\frac{\arcsin(a)}{a}\,da = \frac{\pi G}{2}+\frac{\pi^2 \ln(2)}{8}, \]

where \(G\) is Catalan's constant (since substituting \(a = \sin\theta\) and integrating by parts reduces \(\int_0^{1/\sqrt{2}}\frac{\arcsin a}{a}\,da\) to \(\frac{\pi}{8}\ln 2 + \frac{G}{2}\)).

Reducing an improper squared-arctangent integral: Applying the same \(\sum_{n=-\infty}^\infty \frac{1}{(x+n\pi)^2} = \frac{1}{\sin^2 x}\) reduction to \(f(x) = \arctan^2(\sin^2 x)\) shows that \(\int_0^\infty \frac{\arctan^2(\sin^2 x)}{x^2}\,dx = \int_0^{\pi/2} \frac{\arctan^2(\sin^2 x)}{\sin^2 x}\,dx\). By our earlier evaluation of that trigonometric integral, we immediately obtain

\[ \int_0^\infty \frac{\arctan^2 (\sin^2 x)}{x^2}\,dx = \pi \log\left(\frac{2+\sqrt{2}}{4}\right)\sqrt{\frac{\sqrt{2}+1}{2}}+\frac{\pi^2}{4} \sqrt{\frac{\sqrt{2}-1}{2}}. \]

A difference of arctangent integrals and dilogarithm identities

2026-10-24T10:11:00+05:30

In this post we use Poisson-kernel Fourier series and two-variable dilogarithm identities to prove the identity

\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \arctan\left(\sqrt{3}\tan\theta\right)\,d\theta - \int_{0}^{\frac{\pi}{6}} \arctan\left(\sqrt{3}\tan\theta\right)\,d\theta = \frac{\pi^2}{12}, \]

and then derive an additional dilogarithm relation at \(\rho = 2-\sqrt{3}\) from Hill's two-variable functional equation.

Expanding a parameterized integral in a Fourier series

Let \(I\) denote the expression on the left-hand side. The substitution \(\sqrt{3}\tan \theta = \tan \varphi\) (so \(\sqrt{3}\sec^2\theta\,d\theta = \sec^2\varphi\,d\varphi\), which simplifies to \(d\theta = \frac{\sqrt{3}}{2}\frac{d\varphi}{1+\frac{1}{2}\cos(2\varphi)}\)) transforms \(I\) into

\[ I = \frac{\sqrt{3}}{2}\left( \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\varphi}{1+\frac{1}{2}\cos(2\varphi)}\,d\varphi - \int_{0}^{\frac{\pi}{4}}\frac{\varphi}{1+\frac{1}{2}\cos(2\varphi)}\,d\varphi \right). \]

For \(0 \leq \phi \leq \frac{\pi}{2}\), consider the generalized integral

\[ I(\phi) = \frac{\sqrt{3}}{2}\int_{0}^{\phi}\frac{\varphi}{1+\frac{1}{2}\cos(2\varphi)}\,d\varphi. \]

Using the Poisson kernel identity

\[ 1+2\sum_{k=1}^\infty x^k \cos(k\theta) = \frac{1-x^2}{1-2x\cos(\theta) + x^2}, \qquad |x|<1, \]

with \(x = -\rho\), where \(\rho = 2-\sqrt{3} = \tan\left(\frac{\pi}{12}\right)\) satisfies \(\frac{2\rho}{1+\rho^2} = \frac{1}{2}\) and \(\frac{1-\rho^2}{1+\rho^2} = \frac{\sqrt{3}}{2}\), we deduce that

\[ \frac{1}{1+\frac{1}{2}\cos(2\varphi)} = \frac{2}{\sqrt{3}}\left(1+2\sum_{k=1}^\infty (-1)^k \rho^k \cos(2k\varphi) \right). \]

Integrating term by term on \([0, \phi]\) using integration by parts gives

\[ \begin{aligned} I(\phi) &= \int_0^\phi \varphi \left( 1+2\sum_{k=1}^\infty (-1)^k \rho^k \cos(2k\varphi) \right)d\varphi \\ &= \frac{\phi^2}{2} + 2\sum_{k=1}^\infty (-1)^k \rho^k \int_0^\phi \varphi \cos(2k\varphi)\,d\varphi \\ &= \frac{\phi^2}{2} + 2\sum_{k=1}^\infty (-1)^k \rho^k \left[\frac{\phi \sin(2k\phi)}{2k}-\frac{1-\cos(2k\phi)}{4k^2} \right] \\ &= \frac{\phi^2}{2} - \phi \sum_{k=1}^\infty (-1)^{k+1}\frac{\rho^k\sin(2k\phi)}{k} + \frac{1}{2}\sum_{k=1}^\infty (-1)^{k+1} \frac{1-\cos(2k\phi)}{k^2}\rho^k \\ &= \frac{\phi^2}{2} - \phi \arctan\left(\frac{\rho \sin(2\phi)}{1+\rho\cos(2\phi)} \right) + \frac{1}{2}\sum_{k=1}^\infty (-1)^{k+1} \frac{1-\cos(2k\phi)}{k^2}\rho^k, \end{aligned} \tag{1} \]

where the last step uses the Fourier sine series

\[ \sum_{k=1}^\infty (-1)^{k+1}\frac{x^k}{k}\sin(k\theta) = \arctan\left(\frac{x \sin(\theta)}{1+x\cos(\theta)}\right), \qquad |x| \leq 1. \]

Evaluating \(I(\phi)\) at \(\frac{\pi}{2}\), \(\frac{\pi}{3}\), and \(\frac{\pi}{4}\)

Recall the dilogarithm \(\operatorname{Li}_2(z) = \sum_{k=1}^\infty \frac{z^k}{k^2}\) and Legendre's chi function

\[ \chi_2(z) = \sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)^2} = \frac{1}{2}\bigl(\operatorname{Li}_2(z) - \operatorname{Li}_2(-z)\bigr) = \operatorname{Li}_2(z) - \frac{1}{4}\operatorname{Li}_2(z^2), \]

where the last equality follows from the duplication identity \(\operatorname{Li}_2(z) + \operatorname{Li}_2(-z) = \frac{1}{2}\operatorname{Li}_2(z^2)\). Using \(\rho = 2-\sqrt{3} = \tan\left(\frac{\pi}{12}\right)\), we evaluate equation (1) at our three angles:

At \(\phi = \frac{\pi}{2}\): \(\sin(\pi) = 0\) and \(1-\cos(k\pi)\) is \(2\) for odd \(k\) and \(0\) for even \(k\), so

\[ I\left(\frac{\pi}{2}\right) = \frac{\pi^2}{8}+\chi_2(\rho). \tag{2} \]

At \(\phi = \frac{\pi}{3}\): \(\frac{\rho\sin(2\pi/3)}{1+\rho\cos(2\pi/3)} = \frac{(2-\sqrt{3})\frac{\sqrt{3}}{2}}{1 - \frac{2-\sqrt{3}}{2}} = 2-\sqrt{3} = \tan\left(\frac{\pi}{12}\right)\), so the arctangent equals \(\frac{\pi}{12}\) and \(\frac{\phi^2}{2} - \phi\frac{\pi}{12} = \frac{\pi^2}{36}\). For the cosine sum, \(1-\cos\left(\frac{2k\pi}{3}\right)\) equals \(\frac{3}{2}\) when \(3 \nmid k\) and \(0\) when \(3 \mid k\), giving

\[ I\left(\frac{\pi}{3}\right) = \frac{\pi^2}{36}+\frac{1}{12}\operatorname{Li}_2\left(-\rho^3\right) -\frac{3}{4}\operatorname{Li}_2(-\rho). \tag{3} \]

At \(\phi = \frac{\pi}{4}\): \(\frac{\rho\sin(\pi/2)}{1+\rho\cos(\pi/2)} = \rho = \tan\left(\frac{\pi}{12}\right)\), so \(\frac{\phi^2}{2} - \phi\frac{\pi}{12} = \frac{\pi^2}{96}\), while separating odd and even \(k\) in \(1-\cos\left(\frac{k\pi}{2}\right)\) gives

\[ I\left(\frac{\pi}{4}\right) = \frac{\pi^2}{96}+\frac{1}{2}\chi_2(\rho)-\frac{1}{4}\chi_2\left(\rho^2\right). \tag{4} \]

Expanding \(\chi_2\) and \(\operatorname{Li}_2(-z)\) in terms of \(\operatorname{Li}_2\) at positive powers of \(\rho\) via \(\operatorname{Li}_2(-z) = \frac{1}{2}\operatorname{Li}_2(z^2) - \operatorname{Li}_2(z)\), we obtain

\[ \begin{aligned} I &= I\left(\frac{\pi}{2}\right)-I\left(\frac{\pi}{3}\right)-I\left(\frac{\pi}{4}\right) \\ &= \frac{25\pi^2}{288} + \left[-\frac{1}{4}\operatorname{Li}_2(\rho)+\frac{1}{2}\operatorname{Li}_2(\rho^2)+\frac{1}{12}\operatorname{Li}_2(\rho^3)-\frac{1}{16}\operatorname{Li}_2(\rho^4)-\frac{1}{24}\operatorname{Li}_2(\rho^6) \right]. \end{aligned} \tag{5} \]

Completing the proof with dilogarithm identities

To show that \(I = \frac{\pi^2}{12}\), it remains to prove that

\[ -\frac{1}{4}\operatorname{Li}_2(\rho)+\frac{1}{2}\operatorname{Li}_2(\rho^2)+\frac{1}{12}\operatorname{Li}_2(\rho^3)-\frac{1}{16}\operatorname{Li}_2(\rho^4)-\frac{1}{24}\operatorname{Li}_2(\rho^6) = -\frac{\pi^2}{288}. \tag{6} \]

Let \(\operatorname{Li}_2(x,\theta) = \operatorname{Re}\operatorname{Li}_2(xe^{i\theta})\). We use two classic identities from Lewin's Polylogarithms and Associated Functions:

\[ \begin{aligned} \operatorname{Li}_2 \left(\tan a, \frac{\pi}{2}-2a \right) &= a^2 +\frac{3}{4}\operatorname{Li}_2(\tan^2 a) -\frac{1}{8}\operatorname{Li}_2(\tan^4 a), \\ \operatorname{Li}_2\left(x,\frac{\pi}{3}\right) &= \frac{1}{6} \operatorname{Li}_2(-x^3) -\frac{1}{2}\operatorname{Li}_2(-x). \end{aligned} \]

Setting \(a = \frac{\pi}{12}\) and \(x = \tan\left(\frac{\pi}{12}\right) = \rho\) makes the left-hand sides of both identities equal to \(\operatorname{Li}_2\left(\rho, \frac{\pi}{3}\right)\). Equating their right-hand sides and expanding \(\operatorname{Li}_2(-\rho)\) and \(\operatorname{Li}_2(-\rho^3)\) via the duplication formula gives

\[ \begin{aligned} \frac{1}{6}\operatorname{Li}_2(-\rho^3)-\frac{1}{2}\operatorname{Li}_2(-\rho) &= \frac{\pi^2}{144} + \frac{3}{4}\operatorname{Li}_2(\rho^2)-\frac{1}{8}\operatorname{Li}_2(\rho^4) \\ \implies \frac{1}{12}\operatorname{Li}_2(\rho^6)-\frac{1}{6}\operatorname{Li}_2(\rho^3) -\frac{1}{4}\operatorname{Li}_2(\rho^2) +\frac{1}{2} \operatorname{Li}_2(\rho) &= \frac{\pi^2}{144} + \frac{3}{4}\operatorname{Li}_2(\rho^2)-\frac{1}{8}\operatorname{Li}_2(\rho^4). \end{aligned} \tag{7} \]

Dividing equation (7) by \(-2\) and rearranging terms immediately produces equation (6), completing the proof that \(I = \frac{\pi^2}{12}\).

Bonus: An identity from Hill's two-variable relation

Another attractive dilogarithm identity at \(\rho = 2-\sqrt{3}\) follows from Hill's two-variable relation:

\[ \operatorname{Li}_2(xy) = \operatorname{Li}_2(x) + \operatorname{Li}_2(y) + \operatorname{Li}_2\left(-x\left(\frac{1-y}{1-x}\right) \right)+\operatorname{Li}_2\left(-y\left(\frac{1-x}{1-y}\right)\right)+\frac{1}{2}\log^2\left(\frac{1-x}{1-y}\right). \]

Substitute \(x = -y = e^{-i\frac{\pi}{4}}\frac{\sqrt{3}-1}{\sqrt{2}}\). Then \(\frac{1-x}{1+x} = ix = e^{i\frac{\pi}{4}}\frac{\sqrt{3}-1}{\sqrt{2}}\) and \(ix^2 = \rho\), so

\[ \begin{aligned} \operatorname{Li}_2(-x^2) &= \operatorname{Li}_2(x) + \operatorname{Li}_2(-x) + \operatorname{Li}_2(i)+\operatorname{Li}_2(ix^2)+\frac{1}{2}\log^2(ix) \\ \implies \operatorname{Li}_2(i\rho) &= \frac{1}{2}\operatorname{Li}_2(-i\rho)+ \operatorname{Li}_2(i) + \operatorname{Li}_2(\rho) +\frac{1}{2}\log^2\left(e^{i\frac{\pi}{4}}\frac{\sqrt{3}-1}{\sqrt{2}} \right). \end{aligned} \]

Using \(\operatorname{Re}\operatorname{Li}_2(\pm i\rho) = \frac{1}{4}\operatorname{Li}_2(-\rho^2) = \frac{1}{8}\operatorname{Li}_2(\rho^4) - \frac{1}{4}\operatorname{Li}_2(\rho^2)\), \(\operatorname{Re}\operatorname{Li}_2(i) = -\frac{\pi^2}{48}\), and \(\frac{\sqrt{3}-1}{\sqrt{2}} = \sqrt{\rho}\), extracting the real part of the above equation yields

\[ \frac{\operatorname{Li}_2(\rho^4)}{16}-\frac{\operatorname{Li}_2(\rho^2)}{8} -\operatorname{Li}_2(\rho) = -\frac{5\pi^2}{96}+\frac{1}{8}\log^2(\rho). \]

The inverse tangent integral at two minus root three

2026-10-24T10:10:00+05:30

The inverse tangent integral \(\operatorname{Ti}_2(z) = \int_0^z \frac{\arctan x}{x}\,dx\) equals Catalan's constant \(G = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2}\) at \(z = 1 = \tan\left(\frac{\pi}{4}\right)\). At the half-angle argument \(z = 2-\sqrt{3} = \tan\left(\frac{\pi}{12}\right)\), it satisfies a remarkable closed-form relation discovered by Ramanujan:

\[ \int_0^{2-\sqrt{3}}\frac{\arctan(x)}{x}\,dx = \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + \frac{2}{3}G. \]

Integration by parts and the Fourier series of \(\ln(\tan\theta)\)

Let \(I\) denote the integral. Integrating by parts with \(u = \arctan x\) and \(dv = \frac{dx}{x}\), and then substituting \(x = \tan\theta\) (using \(\tan\frac{\pi}{12} = 2-\sqrt{3}\)), we obtain

\[ \begin{aligned} I &= \Bigl[ \ln(x)\arctan(x) \Bigr]_0^{2-\sqrt{3}} - \int_0^{2-\sqrt{3}}\frac{\ln (x)}{1+x^2}\,dx \\ &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) - \int_0^{\frac{\pi}{12}} \ln(\tan\theta)\,d\theta. \end{aligned} \tag{1} \]

For \(\theta \in \left(0, \frac{\pi}{2}\right)\), recall the standard Fourier cosine expansions

\[ \begin{aligned} \ln(2\sin \theta) &= -\sum_{k=1}^\infty \frac{\cos(2k \theta)}{k}, \\ \ln(2\cos \theta) &= \sum_{k=1}^\infty (-1)^{k+1}\frac{\cos(2k\theta)}{k}. \end{aligned} \]

Subtracting the second series from the first cancels all even-indexed terms and doubles the odd-indexed terms (\(k = 2m+1\)):

\[ \ln(\tan\theta) = \ln(2\sin\theta) - \ln(2\cos\theta) = -2\sum_{k=0}^\infty \frac{\cos\bigl(2(2k+1)\theta\bigr)}{2k+1}. \]

Term-by-term integration and multisection of the series

Substituting the Fourier series of \(\ln(\tan\theta)\) into equation (1) and integrating term by term gives

\[ \begin{aligned} I &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + 2\int_0^{\frac{\pi}{12}}\left(\sum_{k=0}^\infty \frac{\cos\bigl(2(2k+1)\theta\bigr)}{2k+1} \right) d\theta \\ &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + 2\sum_{k=0}^\infty\frac{1}{2k+1}\int_0^{\frac{\pi}{12}} \cos\bigl(2(2k+1)\theta\bigr)\,d\theta \\ &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + \sum_{k=0}^\infty\frac{\sin\left( (2k+1)\frac{\pi}{6}\right)}{(2k+1)^2}. \end{aligned} \]

To evaluate the remaining Dirichlet-type series, examine the periodic numerator \(s_k = \sin\left((2k+1)\frac{\pi}{6}\right)\) as \(k\) varies modulo \(3\):

When \(k \equiv 0\) or \(2 \pmod 3\) (so \(2k+1\) is not divisible by \(3\)), \(s_k = \frac{1}{2}(-1)^k\).
When \(k = 3m+1\) (so \(2k+1 = 6m+3 = 3(2m+1)\) is an odd multiple of \(3\)), \(s_{3m+1} = \sin\left((2m+1)\frac{\pi}{2}\right) = (-1)^m = -(-1)^k\).

On the full alternating sum \(\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} = G\), separating out the subsequence \(k = 3m+1\) where \((-1)^{3m+1} = -(-1)^m\) gives

\[ \sum_{k \not\equiv 1 \pmod 3} \frac{(-1)^k}{(2k+1)^2} = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} + \sum_{m=0}^\infty \frac{(-1)^m}{(6m+3)^2} = G + \frac{G}{9}. \]

Multiplying those terms by \(\frac{1}{2}\) and adding the \(k = 3m+1\) terms (which have coefficient \((-1)^m\)) yields

\[ \begin{aligned} I &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + \frac{1}{2}\left( \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} + \sum_{m=0}^\infty \frac{(-1)^m}{(6m+3)^2}\right) + \sum_{m=0}^\infty \frac{(-1)^m}{(6m+3)^2} \\ &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + \frac{1}{2}\left(G + \frac{G}{9}\right) + \frac{G}{9} \\ &= \frac{\pi}{12}\ln\left(2-\sqrt{3}\right) + \frac{2}{3}G. \end{aligned} \]

An integral involving the squared arctangent of sine squared

2026-10-24T10:09:00+05:30

In this post we combine integration by parts with differentiation under the integral sign to evaluate

\[ \int_0^{\frac{\pi}{2}} \frac{\arctan^2 (\sin^2 \theta)}{\sin^2 \theta}\,d\theta = \pi \log\left( \frac{2+\sqrt{2}}{4}\right) \sqrt{\frac{\sqrt{2}+1}{2}}+\frac{\pi^2}{4}\sqrt{\frac{\sqrt{2}-1}{2}}. \]

Integration by parts and introducing a parameter

Let \(I\) denote the integral. Integrating by parts with \(u = \arctan^2(\sin^2 x)\) and \(dv = \frac{dx}{\sin^2 x}\) (so \(v = -\cot x\) and \(du = \frac{4\sin x \cos x \arctan(\sin^2 x)}{1+\sin^4 x}\,dx\)), the boundary term \(\bigl[-\cot x \arctan^2(\sin^2 x)\bigr]_0^{\pi/2}\) vanishes at both endpoints because \(\cot(\pi/2) = 0\) and \(\cot x \arctan^2(\sin^2 x) = O(x^3)\) as \(x \to 0^+\). Thus,

\[ I = \int_{0}^{\frac{\pi}{2}}\frac{\arctan^2(\sin^2 x)}{\sin^2 x}\,dx = 4\int_0^{\frac{\pi}{2}}\frac{\cos^2 x\arctan(\sin^2 x)}{1+\sin^4 x}\,dx. \]

To eliminate the remaining arctangent, introduce a parameter \(\alpha \in [0,1]\):

\[ f(\alpha) = \int_0^{\frac{\pi}{2}}\frac{\cos^2 x\arctan(\alpha \sin^2 x)}{1+\sin^4 x}\,dx, \]

so that \(f(0) = 0\) and \(I = 4f(1)\). Differentiating inside the integral with respect to \(\alpha\) and decomposing into partial fractions in \(\sin^4 x\) gives

\[ \begin{aligned} f'(\alpha) &= \int_0^{\frac{\pi}{2}}\frac{\cos^2 x}{1+\sin^4 x}\cdot\frac{\sin^2 x}{1+\alpha^2 \sin^4 x}\,dx \\ &= \frac{1}{1-\alpha^2}\int_0^{\frac{\pi}{2}}\frac{\cos^2 x\sin^2 x}{1+\sin^4 x}\,dx - \frac{\alpha^2}{1-\alpha^2}\int_0^{\frac{\pi}{2}}\frac{\cos^2 x\sin^2 x}{1+\alpha^2 \sin^4 x}\,dx. \end{aligned} \]

Define the auxiliary integral

\[ g(\alpha) = \int_0^{\frac{\pi}{2}}\frac{\cos^2 x\sin^2 x}{1+\alpha^2 \sin^4 x}\,dx. \]

Then \(f'(\alpha) = \frac{g(1) - \alpha^2 g(\alpha)}{1 - \alpha^2}\), and integrating from \(\alpha = 0\) to \(\alpha = 1\) yields

\[ I = 4f(1) = 4\int_0^1\frac{g(1)-\alpha^2 g(\alpha)}{1-\alpha^2}\,d\alpha. \tag{1} \]

Evaluating the auxiliary integral \(g(\alpha)\)

Substituting \(t = \tan x\) (so \(\sin^2 x = \frac{t^2}{1+t^2}\), \(\cos^2 x = \frac{1}{1+t^2}\), and \(dx = \frac{dt}{1+t^2}\)) and applying partial fractions, we obtain

\[ \begin{aligned} g(\alpha) &= \int_0^{\frac{\pi}{2}}\frac{\cos^2 x\sin^2 x}{1+\alpha^2 \sin^4 x}\,dx \\ &= \int_0^\infty \frac{t^2}{\bigl((1+\alpha^2)t^4+2t^2+1\bigr)(t^2+1)}\,dt \\ &= -\frac{1}{\alpha^2}\int_0^\infty\frac{1}{1+t^2}\,dt + \frac{1}{\alpha^2}\int_0^\infty\frac{1+(1+\alpha^2)t^2}{(1+\alpha^2)t^4+2t^2+1}\,dt. \end{aligned} \]

The first integral is \(\frac{\pi}{2}\). For the second integral, put \(r = \sqrt[4]{1+\alpha^2}\) and \(s = rt\). Then

\[ \begin{aligned} \int_0^\infty\frac{1+r^4 t^2}{r^4 t^4+2t^2+1}\,dt &= \frac{1}{r}\int_0^\infty \frac{1+r^2 s^2}{s^4 + \frac{2}{r^2}s^2 + 1}\,ds \\ &= \frac{1+r^2}{2r}\int_0^\infty \frac{1 + \frac{1}{s^2}}{\left(s - \frac{1}{s}\right)^2 + 2 + \frac{2}{r^2}}\,ds, \end{aligned} \]

where the last step averages the integral with its image under \(s \mapsto 1/s\). Substituting \(w = s - \frac{1}{s} \in (-\infty, \infty)\) gives a standard Cauchy integral:

\[ \frac{1+r^2}{2r}\int_{-\infty}^\infty \frac{dw}{w^2 + \frac{2(r^2+1)}{r^2}} = \frac{1+r^2}{2r}\cdot\frac{\pi}{\sqrt{\frac{2(r^2+1)}{r^2}}} = \frac{\pi\sqrt{1+r^2}}{2\sqrt{2}} = \frac{\pi\sqrt{1+\sqrt{1+\alpha^2}}}{2\sqrt{2}}. \]

Therefore,

\[ g(\alpha) = -\frac{\pi}{2\alpha^2} + \frac{\pi \sqrt{1+\sqrt{1+\alpha^2}}}{2\sqrt{2}\alpha^2}. \tag{2} \]

Substituting (2) into equation (1), the \(-\frac{\pi}{2}\) terms cancel between \(\alpha^2 g(\alpha)\) and \(g(1)\), leaving

\[ I = \pi\sqrt{2}\int_0^1\frac{\sqrt{1+\sqrt{2}}-\sqrt{1+\sqrt{1+\alpha^2}}}{1-\alpha^2}\,d\alpha. \tag{3} \]

Integrating with respect to \(\alpha\)

Integral (3) admits an elementary antiderivative via the two successive substitutions \(t = \sqrt{1+\alpha^2}\) (so \(\alpha\,d\alpha = t\,dt\) and \(\frac{d\alpha}{1-\alpha^2} = \frac{t\,dt}{(2-t^2)\sqrt{t^2-1}}\)) and \(u = \sqrt{t-1}\) (so \(t = u^2+1\) and \(dt = 2u\,du\)):

\[ \begin{aligned} &\int \frac{\sqrt{1+\sqrt{2}}-\sqrt{1+\sqrt{1+\alpha^2}}}{1-\alpha^2}\,d\alpha \\ &= \sqrt{1+\sqrt{2}}\operatorname{arctanh}(\alpha) - \int \frac{\sqrt{1+\sqrt{1+\alpha^2}}}{1-\alpha^2}\,d\alpha \\ &= \sqrt{1+\sqrt{2}}\operatorname{arctanh}(\alpha) - \int \frac{t\sqrt{1+t}}{(2-t^2)\sqrt{t^2-1}}\,dt \quad \left(t=\sqrt{1+\alpha^2}\right) \\ &= \sqrt{1+\sqrt{2}}\operatorname{arctanh}(\alpha) - \int \frac{t}{(2-t^2)\sqrt{t-1}}\,dt \\ &= \sqrt{1+\sqrt{2}}\operatorname{arctanh}(\alpha) - 2\int \frac{u^2+1}{2-(u^2+1)^2}\,du \quad \left(u=\sqrt{t-1}\right) \\ &= \sqrt{1+\sqrt{2}}\operatorname{arctanh}(\alpha) - 2\int \frac{u^2+1}{(\sqrt{2}-1-u^2)(\sqrt{2}+1+u^2)}\,du \\ &= \sqrt{1+\sqrt{2}}\operatorname{arctanh}(\alpha) - \int\frac{du}{\sqrt{2}-1-u^2} + \int\frac{du}{\sqrt{2}+1+u^2} \\ &= \sqrt{1+\sqrt{2}}\operatorname{arctanh}(\alpha) - \sqrt{\sqrt{2}+1}\operatorname{arctanh}\left( u \sqrt{\sqrt{2}+1}\right) + \sqrt{\sqrt{2}-1}\arctan\left(u\sqrt{\sqrt{2}-1} \right) + C. \end{aligned} \]

At the lower endpoint \(\alpha = 0\), we have \(u = 0\), so the antiderivative evaluates to \(0\). As \(\alpha \to 1^-\), we have \(u = \sqrt{\sqrt{1+\alpha^2}-1} \to \sqrt{\sqrt{2}-1}\). For the \(\arctan\) term, \(u\sqrt{\sqrt{2}-1} \to \sqrt{2}-1 = \tan\left(\frac{\pi}{8}\right)\), so \(\arctan(\sqrt{2}-1) = \frac{\pi}{8}\). For the difference of the two \(\operatorname{arctanh}\) terms, set \(c = \sqrt{\sqrt{2}+1}\) and use the factorization \(1-\alpha^2 = (\sqrt{2}-1-u^2)(\sqrt{2}+1+u^2) = c^{-2}(1-c^2 u^2)(c^2+u^2)\):

\[ \begin{aligned} \lim_{\alpha\to 1^-}\bigl(\operatorname{arctanh}(\alpha) - \operatorname{arctanh}(cu)\bigr) &= \frac{1}{2}\lim_{\alpha\to 1^-}\log\left(\frac{1+\alpha}{1+cu}\cdot\frac{1+\alpha}{1+cu}\cdot\frac{1-c^2 u^2}{1-\alpha^2}\right) \\ &= \frac{1}{2}\lim_{u^2\to \sqrt{2}-1}\log\left(\frac{c^2}{c^2+u^2}\right) \\ &= \frac{1}{2}\log\left(\frac{\sqrt{2}+1}{2\sqrt{2}}\right) = \frac{1}{2}\log\left(\frac{2+\sqrt{2}}{4}\right). \end{aligned} \]

Putting everything together, we conclude that

\[ \begin{aligned} I &= \pi\sqrt{2}\left\{\frac{\sqrt{\sqrt{2}+1}}{2}\log\left(\frac{\sqrt{2}+2}{4} \right) + \sqrt{\sqrt{2}-1}\arctan(\sqrt{2}-1)\right\} \\ &= \pi\sqrt{2}\left\{\frac{\sqrt{\sqrt{2}+1}}{2}\log\left(\frac{\sqrt{2}+2}{4} \right) + \frac{\pi}{8}\sqrt{\sqrt{2}-1}\right\} \\ &= \pi \log\left( \frac{2+\sqrt{2}}{4}\right) \sqrt{\frac{\sqrt{2}+1}{2}}+\frac{\pi^2}{4}\sqrt{\frac{\sqrt{2}-1}{2}}. \end{aligned} \]

In a later post on Lobachevsky-type integrals, we show that the improper integral \(\int_0^\infty \frac{\arctan^2(\sin^2 x)}{x^2}\,dx\) reduces to this exact same value.

Evaluating a logarithmic Fresnel integral by differentiating under the integral sign

2026-10-24T10:08:00+05:30

A factor of \(\ln x\) in an integrand can often be generated by differentiating a family of power integrals \(x^{s-1}\) with respect to the exponent \(s\). Combining this principle with the Mellin transform of \(\sin t\) proved in an earlier post, we will show that

\[ \int_0^\infty \frac{\sin\left(x^2\right)\ln(x)}{x}\,dx = -\frac{\gamma \pi}{8}, \]

where \(\gamma\) is the Euler-Mascheroni constant.

Reducing to a Mellin derivative

First, the substitution \(t = x^2\) (so that \(\frac{dx}{x} = \frac{dt}{2t}\) and \(\ln x = \frac{1}{2}\ln t\)) removes the quadratic argument inside the sine:

\[ \int_0^\infty \frac{\sin\left(x^2\right)\ln(x)}{x}\,dx = \frac{1}{4}\int_0^\infty \frac{\sin(t) \ln(t)}{t}\,dt. \tag{1} \]

Recall from our post on the Mellin transforms of sine and cosine that for \(s \in (-1, 1) \setminus \{0\}\),

\[ \int_0^\infty t^{s-1}\sin(t)\,dt = \sin\left(\frac{\pi s}{2}\right)\Gamma(s), \]

with removable singularity at \(s = 0\) matching the Dirichlet integral \(\frac{\pi}{2}\). Differentiating under the integral sign with respect to \(s\) at \(s = 0\) produces the logarithmic factor \(\ln t\):

\[ \begin{aligned} \int_0^\infty \frac{\sin(t)\ln(t)}{t}\,dt &= \lim_{s\to 0}\frac{d}{ds}\int_0^\infty t^{s-1} \sin(t)\,dt \\ &= \lim_{s\to 0} \frac{d}{ds}\left[\sin\left(\frac{\pi s}{2}\right) \Gamma(s)\right] \\ &= \lim_{s\to 0} \left[ \frac{\pi}{2}\cos\left(\frac{\pi s}{2}\right)\Gamma(s) + \sin\left(\frac{\pi s}{2}\right)\Gamma'(s)\right]. \end{aligned} \]

Taking the limit via Laurent expansions

Near \(s = 0\), the gamma function and its derivative have the Laurent expansions

\[ \Gamma(s) = \frac{1}{s} - \gamma + O(s), \qquad \Gamma'(s) = -\frac{1}{s^2} + \frac{6\gamma^2+\pi^2}{12} + O(s). \]

Substituting these into the limit gives

\[ \begin{aligned} \int_0^\infty \frac{\sin(t)\ln(t)}{t}\,dt &= \lim_{s\to 0} \left[ \frac{\pi}{2}\cos\left(\frac{\pi s}{2}\right)\left(\frac{1}{s}-\gamma+O(s) \right) + \sin\left(\frac{\pi s}{2}\right)\left(-\frac{1}{s^2}+\frac{6\gamma^2+\pi^2}{12}+O(s) \right) \right] \\ &= \lim_{s\to 0} \left[-\frac{\pi \gamma}{2} + \left( \frac{\pi}{2}\cos\left(\frac{\pi s}{2}\right) \frac{1}{s} - \sin\left(\frac{\pi s}{2}\right)\frac{1}{s^2}\right) + O(s)\right]. \end{aligned} \]

Using the Taylor series \(\cos\left(\frac{\pi s}{2}\right) = 1 - O(s^2)\) and \(\sin\left(\frac{\pi s}{2}\right) = \frac{\pi s}{2} - O(s^3)\), the singular \(1/s\) terms cancel identically:

\[ \frac{\pi}{2}\cos\left(\frac{\pi s}{2}\right) \frac{1}{s} - \sin\left(\frac{\pi s}{2}\right)\frac{1}{s^2} = \frac{\pi}{2s} - \frac{\pi}{2s} + O(s) = O(s). \]

Therefore,

\[ \int_0^\infty \frac{\sin(t)\ln(t)}{t}\,dt = -\frac{\pi \gamma}{2}. \tag{2} \]

A shortcut via \(\Gamma(s+1) = s\Gamma(s)\)

We can also see why the singular terms cancel without writing out the second-order Laurent coefficients by factoring \(\sin\left(\frac{\pi s}{2}\right)\Gamma(s)\) as

\[ \sin\left(\frac{\pi s}{2}\right)\Gamma(s) = \frac{\sin\left(\frac{\pi s}{2}\right)}{s}\Gamma(1+s). \]

Here \(g(s) = \frac{\sin(\pi s/2)}{s}\) is a smooth, even function with \(g(0) = \frac{\pi}{2}\) and \(g'(0) = 0\), while \(\Gamma(1) = 1\) and \(\Gamma'(1) = -\gamma\). Applying the product rule at \(s=0\) immediately yields \(g'(0)\Gamma(1) + g(0)\Gamma'(1) = -\frac{\pi\gamma}{2}\).

Substituting equation (2) into equation (1) completes the proof:

\[ \int_0^\infty \frac{\sin\left(x^2\right)\ln(x)}{x}\,dx = -\frac{\gamma \pi}{8}. \]

A logarithmic sine integral via a keyhole contour

2026-10-24T10:07:00+05:30

Integrals on \((0, \infty)\) containing \(\sin(\ln x)\) or \(\cos(\ln x)\) are the imaginary and real parts of \(\int_0^\infty x^i R(x)\,dx\), since \(x^i = e^{i\ln x} = \cos(\ln x) + i\sin(\ln x)\). By placing a keyhole branch cut along the negative real axis \((-\infty, 0]\) for an even rational function \(R(z)\), the integrals along the upper and lower banks of the cut both fold onto \((0, \infty)\) with different constant factors \(e^{-\pi}\) and \(e^{\pi}\). We will use this method to prove

\[ \int_0^\infty \frac{\sin(\ln x)}{x^2+4}\,dx = \frac{\pi \sin(\ln 2)}{4\cosh\left(\frac{\pi}{2}\right)}. \]

The keyhole contour

Consider the function

\[ f(z) = \frac{e^{i\log z}}{4+z^2}, \]

where the branch of the logarithm is chosen with \(-\pi < \arg z \leq \pi\), so the branch cut lies along the negative real axis \((-\infty, 0]\). We integrate \(f(z)\) around the positively oriented keyhole contour shown below:

Here \(C_R\) is a large circle of radius \(R > 2\) and \(C_\epsilon\) is a small semicircle of radius \(\epsilon < 2\), both centered at the origin. On the circle \(|z| = r\), we have \(\left|e^{i\log z}\right| = e^{-\arg z} \leq e^\pi\), so

\[ \left|\int_{C_R} f(z)\,dz\right| \leq \frac{2\pi R e^\pi}{R^2 - 4} \to 0 \quad (R \to \infty) \qquad\text{and}\qquad \left|\int_{C_\epsilon} f(z)\,dz\right| \leq \frac{\pi\epsilon e^\pi}{4 - \epsilon^2} \to 0 \quad (\epsilon \to 0^+). \]

Thus, in the limits \(R \to \infty\) and \(\epsilon \to 0^+\), only the two horizontal segments just above and just below the negative real axis contribute to the contour integral.

Computing the residues at \(\pm 2i\)

Inside the contour, \(f(z)\) has simple poles at \(z = 2i\) and \(z = -2i\). Using the branch \(-\pi < \arg z \leq \pi\), we have \(\log(2i) = \ln 2 + i\frac{\pi}{2}\) and \(\log(-2i) = \ln 2 - i\frac{\pi}{2}\). Hence,

\[ \begin{aligned} \mathop{\operatorname{Res}}_{z=2i} f(z) &= \lim_{z\to 2i} (z-2i) \frac{e^{i\log z}}{z^2+4} \\ &= \frac{e^{i\log(2i)}}{4i} \\ &= \frac{e^{i\ln 2 - \frac{\pi}{2}}}{4i}, \end{aligned} \]

and similarly,

\[ \mathop{\operatorname{Res}}_{z=-2i} f(z) = -\frac{e^{i\log(-2i)}}{4i} = -\frac{e^{i\ln 2 + \frac{\pi}{2}}}{4i}. \]

Evaluating the integral along the branch cut

On the upper bank of the negative real axis (traversed from \(-\infty\) to \(0\)), \(\arg z = +\pi\), so \(\log z = \ln|x| + i\pi\) and \(e^{i\log z} = e^{-\pi}e^{i\ln|x|}\). On the lower bank (traversed from \(0\) to \(-\infty\)), \(\arg z = -\pi\), so \(\log z = \ln|x| - i\pi\) and \(e^{i\log z} = e^{\pi}e^{i\ln|x|}\). Substituting \(t = -x > 0\) (and relabeling \(t\) as \(x\)) on both banks, the residue theorem gives

\[ \begin{aligned} \int_{-\infty}^0 \frac{e^{i(\ln|x|+i\pi)}}{4+x^2}\,dx + \int_0^{-\infty} \frac{e^{i(\ln|x|-i\pi)}}{4+x^2}\,dx &= 2\pi i \left(\mathop{\operatorname{Res}}_{z=2i} f(z) + \mathop{\operatorname{Res}}_{z=-2i} f(z)\right) \\ \implies e^{-\pi}\int_0^\infty \frac{e^{i\ln x}}{4+x^2}\,dx - e^{\pi}\int_0^\infty \frac{e^{i\ln x}}{4+x^2}\,dx &= 2\pi i \left( \frac{e^{i\ln 2}}{4i}e^{-\frac{\pi}{2}}- \frac{e^{i\ln 2}}{4i}e^{\frac{\pi}{2}}\right) \\ \implies -2\sinh(\pi) \int_0^\infty \frac{e^{i\ln x}}{4+x^2}\,dx &= -\sinh\left(\frac{\pi}{2}\right) \pi e^{i\ln 2}. \end{aligned} \]

Using the double-angle identity \(\sinh(\pi) = 2\sinh\left(\frac{\pi}{2}\right)\cosh\left(\frac{\pi}{2}\right)\) to simplify the ratio of hyperbolic sines, we obtain

\[ \int_0^\infty \frac{e^{i\ln x}}{4+x^2}\,dx = \frac{\pi e^{i\ln 2}}{4\cosh\left(\frac{\pi}{2}\right)}. \]

Taking the imaginary part of both sides yields the desired evaluation:

\[ \int_0^\infty \frac{\sin(\ln x)}{x^2+4}\,dx = \frac{\pi \sin(\ln 2)}{4\cosh\left(\frac{\pi}{2}\right)}. \]

Equating the real parts simultaneously gives the companion cosine integral:

\[ \int_0^\infty \frac{\cos(\ln x)}{x^2+4}\,dx = \frac{\pi \cos(\ln 2)}{4\cosh\left(\frac{\pi}{2}\right)}. \]

A trigonometric integral via a semicircular contour

2026-10-24T10:06:00+05:30

Integrals containing nested inverse trigonometric functions often simplify dramatically after a change of variables that unwraps the inner function, followed by a passage to the unit circle in the complex plane. We will prove that for every \(n \in \mathbb{N}\),

\[ \int_0^{\frac{\pi}{2}}\frac{\cos\left((2n-1)\arcsin\left(\frac{\sin x}{\sqrt{2}}\right) \right)}{\sqrt{1-\frac{\sin^2 x}{2}}}\,dx = \frac{\sqrt{\pi}}{2}\sin\left(\frac{n\pi}{2}\right)\frac{\Gamma\left(\frac{n}{2} \right)}{\Gamma\left(\frac{n+1}{2}\right)}. \]

Unwrapping the inverse sine

Let \(I\) denote the integral. We begin with the substitution \(\sin t = \frac{\sin x}{\sqrt{2}}\), so that \(\cos t\,dt = \frac{\cos x}{\sqrt{2}}\,dx\). Since \(\cos x = \sqrt{1 - 2\sin^2 t} = \sqrt{\cos(2t)}\) and \(t\) runs from \(0\) to \(\frac{\pi}{4}\) as \(x\) runs from \(0\) to \(\frac{\pi}{2}\), the factor \(\sqrt{1 - \frac{\sin^2 x}{2}} = \cos t\) cancels against \(\cos t\,dt\):

\[ I = \sqrt{2}\int_0^{\frac{\pi}{4}}\frac{\cos\bigl((2n-1) t\bigr)}{\sqrt{1-2\sin^2 t}}\,dt = \sqrt{2}\int_0^{\frac{\pi}{4}}\frac{\cos\bigl((2n-1) t\bigr)}{\sqrt{\cos(2t)}}\,dt. \]

Next, substitute \(\theta = 2t\) and use the evenness of the integrand on \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\):

\[ \begin{aligned} I &= \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}}\frac{\cos\left(\frac{2n-1}{2}\theta\right)}{\sqrt{\cos \theta}}\,d\theta \\ &= \frac{1}{2\sqrt{2}} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos\left(\frac{2n-1}{2}\theta\right)}{\sqrt{\cos \theta}}\,d\theta \\ &= \frac{1}{2\sqrt{2}} \operatorname{Re}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\exp\left(\frac{2n-1}{2}i\theta\right)}{\sqrt{\cos \theta}}\,d\theta. \end{aligned} \tag{1} \]

Integrating around the right half of the unit disk

Now integrate the principal branch of

\[ f(z) = \frac{z^{n-1}}{\sqrt{1+z^2}} \]

around the positively oriented contour in the right half-plane shown below:

The contour consists of the right half of the unit circle \(|z|=1\) (\(-\frac{\pi}{2} < \arg z < \frac{\pi}{2}\)), the segment of the imaginary axis between \(+i\) and \(-i\), and two small circular indents of radius \(\epsilon\) avoiding the branch points at \(z = \pm i\). Near \(z = \pm i\), we have \(|1+z^2| = |z-i||z+i| \sim 2\epsilon\), so \(|f(z)| = O(\epsilon^{-1/2})\) while the arc length of each indent is \(O(\epsilon)\). Thus the integrals along both indents are \(O(\epsilon^{1/2})\) and vanish as \(\epsilon \to 0^+\).

Since \(f(z)\) is analytic inside the indented region, Cauchy's theorem equates the integral over the right semicircle to the integral upwards along the imaginary axis from \(-i\) to \(+i\). Parameterizing the imaginary axis by \(z = iy\) with \(y \in (-1, 1)\) and \(dz = i\,dy\) gives

\[ \begin{aligned} \int_{\left\{z\in \mathbb{C} :\; |z|=1, \; -\frac{\pi}{2}<\arg z < \frac{\pi}{2} \right\}}\frac{z^{n-1}}{\sqrt{1+z^2}}\,dz &= \int_{-i}^i \frac{z^{n-1}}{\sqrt{1+z^2}}\,dz \\ &= i^{n}\int_{-1}^1 \frac{y^{n-1}}{\sqrt{1-y^2}}\,dy \\ &= \begin{cases} 2 i^n \int_0^1 \frac{y^{n-1}}{\sqrt{1-y^2}}\,dy, & n\text{ odd},\\ 0, & n\text{ even}. \end{cases} \end{aligned} \tag{2} \]

Notice that when \(n = 2k+1\) is odd, \(i^n = i(-1)^k = i\sin\left(\frac{n\pi}{2}\right)\), whereas when \(n\) is even, \(\sin\left(\frac{n\pi}{2}\right) = 0\). Hence, for every \(n \in \mathbb{N}\), equation (2) can be written in the unified form

\[ \int_{\left\{z\in \mathbb{C} :\; |z|=1, \; -\frac{\pi}{2}<\arg z < \frac{\pi}{2} \right\}}\frac{z^{n-1}}{\sqrt{1+z^2}}\,dz = 2i\sin\left(\frac{n\pi}{2}\right)\int_0^1 \frac{y^{n-1}}{\sqrt{1-y^2}}\,dy. \tag{2'} \]

Parameterizing the semicircular arc

On the right unit semicircle, set \(z = e^{i\theta}\) for \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\), with \(dz = i e^{i\theta}\,d\theta\). Since \(1 + e^{2i\theta} = 2e^{i\theta}\cos\theta\) and \(\cos\theta > 0\) on \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), the principal square root is \(\sqrt{1+e^{2i\theta}} = \sqrt{2\cos\theta}\,e^{\frac{i\theta}{2}}\). Therefore,

\[ \begin{aligned} \int_{\left\{z\in \mathbb{C} :\; |z|=1, \; -\frac{\pi}{2}<\arg z < \frac{\pi}{2} \right\}}\frac{z^{n-1}}{\sqrt{1+z^2}}\,dz &= i\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{e^{in \theta}}{\sqrt{1+e^{2i\theta}}}\,d\theta \\ &= i\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{e^{\frac{2n-1}{2}i \theta}}{\sqrt{e^{i\theta}+e^{-i\theta}}}\,d\theta \\ &= \frac{i}{\sqrt{2}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{e^{\frac{2n-1}{2}i \theta}}{\sqrt{\cos \theta}}\,d\theta. \end{aligned} \tag{3} \]

Combining equations (1), (2'), and (3), and finally setting \(\xi = y^2\) (\(d\xi = 2y\,dy\)) to reduce the remaining integral to Euler's beta function \(B\left(\frac{n}{2}, \frac{1}{2}\right)\), we conclude that

\[ \begin{aligned} I &= \sin\left(\frac{n\pi}{2}\right)\int_0^1 \frac{y^{n-1}}{\sqrt{1-y^2}}\,dy \\ &= \frac{1}{2}\sin\left(\frac{n\pi}{2}\right)\int_0^1 \xi^{\frac{n}{2}-1}(1-\xi)^{-\frac{1}{2}}\,d\xi \\ &= \frac{\sqrt{\pi}}{2}\sin\left(\frac{n\pi}{2}\right)\frac{\Gamma\left(\frac{n}{2} \right)}{\Gamma\left(\frac{n+1}{2}\right)}. \end{aligned} \]

Mellin transforms of sine and cosine via a quarter-circle contour

2026-10-24T10:05:00+05:30

Rotating the positive real axis onto the positive imaginary axis turns the oscillatory factor \(e^{ix}\) into the rapidly decaying factor \(e^{-y}\), converting oscillatory improper integrals into Euler's gamma integral. Using this idea, we will establish the formulas

\[ \begin{aligned} \int_0^\infty x^{s-1} \sin(x)\,dx &= \begin{cases} \sin\left(\frac{\pi s}{2}\right)\Gamma(s), & s\in (-1,0)\cup (0,1), \\ \frac{\pi}{2}, & s=0, \end{cases} \\ \int_0^\infty x^{s-1} \cos(x)\,dx &= \cos\left(\frac{\pi s}{2}\right)\Gamma(s), \qquad 0 < s < 1. \end{aligned} \]

Contour integration for \(0 < s < 1\)

Let \(f(z) = z^{s-1} e^{iz}\), where \(0 < s < 1\) and the principal branch of the logarithm (\(-\pi < \arg z \leq \pi\)) defines \(z^{s-1} = e^{(s-1)\log z}\). We integrate \(f(z)\) around the positively oriented indented quarter-circle contour shown below, and then let \(R \to \infty\) and \(\epsilon \to 0^+\):

Here \(C_R\) and \(C_\epsilon\) are quarter-circles centered at \(z=0\) of radii \(R\) and \(\epsilon\), respectively, and \(C_\epsilon\) avoids the branch point at \(z=0\). Since \(f(z)\) is analytic inside and on the contour, Cauchy's integral theorem gives

\[ \int_\epsilon^R f(z)\,dz + \int_{C_R} f(z)\,dz + \int_{iR}^{i\epsilon} f(z)\,dz + \int_{C_\epsilon}f(z)\,dz = 0. \tag{1} \]

On the vertical segment from \(iR\) down to \(i\epsilon\), parameterizing by \(z = iy\) (\(y \in [\epsilon, R]\), \(dz = i\,dy\)) and using \((iy)^{s-1} i = i^s y^{s-1} = e^{\frac{i\pi s}{2}} y^{s-1}\) gives

\[ \int_{iR}^{i\epsilon}f(z)\,dz = -i \int_\epsilon^R f(iy)\,dy = -e^{\frac{i\pi s}{2}}\int_\epsilon^R y^{s-1} e^{-y}\,dy. \tag{2} \]

Bounding the circular arcs

On \(C_R\), parameterize \(z = R e^{i\theta}\) for \(0 \leq \theta \leq \frac{\pi}{2}\). Then \(\left|e^{i R e^{i\theta}}\right| = e^{-R\sin\theta}\), and by Jordan's inequality \(\sin\theta \geq \frac{2\theta}{\pi}\) on \(\left[0, \frac{\pi}{2}\right]\), we obtain

\[ \begin{aligned} \left|\int_{C_R} f(z)\,dz\right| &= R^s \left| \int_0^{\frac{\pi}{2}} e^{i R e^{i\theta}} e^{i\theta s}\,d\theta\right| \\ &\leq R^s \int_0^{\frac{\pi}{2}} \left| e^{i R e^{i\theta}} \right|\,d\theta \\ &\leq R^s \int_0^{\frac{\pi}{2}} e^{-R \frac{2\theta}{\pi}}\,d\theta \\ &= \frac{\pi}{2}R^{s-1} \left(1-e^{-R} \right). \end{aligned} \]

Since \(s < 1\), we have \(R^{s-1} \to 0\) as \(R \to \infty\), so

\[ \lim_{R\to \infty}\left|\int_{C_R} f(z)\,dz\right| = 0. \tag{3} \]

By the exact same estimate on \(C_\epsilon\) (or simply noting \(\left|e^{i\epsilon e^{i\theta}}\right| \leq 1\) so the integral is bounded by \(\frac{\pi}{2}\epsilon^s\)), the condition \(s > 0\) ensures that \(1 - e^{-\epsilon} = O(\epsilon)\) and hence

\[ \left|\int_{C_\epsilon} f(z)\,dz\right| \leq \frac{\pi}{2}\epsilon^{s-1} \left(1-e^{-\epsilon} \right) = O(\epsilon^s) \to 0 \quad \text{as} \quad \epsilon \to 0^+. \tag{4} \]

Combining equations (1), (2), (3), and (4) in the limits \(R \to \infty\) and \(\epsilon \to 0^+\) yields

\[ \begin{aligned} \int_0^\infty x^{s-1} e^{ix}\,dx &= e^{\frac{i\pi s}{2}} \int_0^\infty y^{s-1} e^{-y}\,dy \\ &= \left(\cos\left(\frac{\pi s}{2}\right) + i \sin\left(\frac{\pi s}{2}\right) \right) \Gamma(s). \end{aligned} \]

Separating the imaginary and real parts gives, for \(0 < s < 1\),

\[ \int_0^\infty x^{s-1} \sin(x)\,dx = \sin\left(\frac{\pi s}{2}\right)\Gamma(s), \tag{5} \]

\[ \int_0^\infty x^{s-1} \cos(x)\,dx = \cos\left(\frac{\pi s}{2}\right)\Gamma(s). \tag{6} \]

Extending the sine integral to \(-1 < s \leq 0\)

Because \(\sin x = O(x)\) near \(x=0\), the sine integral \(I(s) = \int_0^\infty x^{s-1} \sin(x)\,dx\) converges on the wider strip \(-1 < s < 1\). We have already proved that \(I(s) = \sin\left(\frac{\pi s}{2}\right)\Gamma(s)\) for \(0 < s < 1\).

Now suppose \(-1 < s < 0\). Integrating by parts with \(u = \sin x\) and \(dv = x^{s-1}\,dx\) gives

\[ I(s) = \left[ \frac{x^s \sin(x)}{s} \right]_0^\infty - \frac{1}{s}\int_0^\infty x^s \cos(x)\,dx. \]

At the upper limit \(x \to \infty\), \(|x^s \sin x| \leq x^s \to 0\) because \(s < 0\); at the lower limit \(x \to 0^+\), \(x^s \sin x \sim x^{s+1} \to 0\) because \(s > -1\). Moreover, since \(0 < s+1 < 1\), we can apply equation (6) with \(s+1\) in place of \(s\), together with the functional equation \(\Gamma(s+1) = s\Gamma(s)\):

\[ \begin{aligned} I(s) &= -\frac{1}{s} \int_0^\infty x^{(s+1)-1} \cos(x)\,dx \\ &= -\frac{1}{s} \cos\left(\frac{\pi}{2}(s+1) \right)\Gamma(s+1) \\ &= \sin\left(\frac{\pi s}{2}\right)\Gamma(s). \end{aligned} \]

Finally, the case \(s=0\) is the classic Dirichlet integral

\[ I(0) = \int_0^\infty \frac{\sin(x)}{x}\,dx = \frac{\pi}{2}, \]

which matches the removable singularity of \(\sin\left(\frac{\pi s}{2}\right)\Gamma(s)\) at \(s=0\):

\[ \lim_{s\to 0} \left[ \sin\left(\frac{\pi s}{2}\right)\Gamma(s) \right] = \lim_{s\to 0} \left[ \frac{\sin\left(\frac{\pi s}{2}\right)}{s}\Gamma(s+1) \right] = \frac{\pi}{2}. \]

In a companion post, we differentiate this identity at \(s=0\) to evaluate a logarithmic Fresnel integral.

The Fourier transform of the hyperbolic secant

2026-10-24T10:04:00+05:30

A classic property of the hyperbolic secant \(\operatorname{sech}(\pi x) = \frac{1}{\cosh(\pi x)}\) is that, like the Gaussian, it is an eigenfunction of the Fourier transform. We will prove that for \(a > 0\),

\[ \int_{-\infty}^\infty \frac{e^{ia x}}{\cosh(\pi x)}\,dx = \frac{1}{\cosh\left(\frac{a}{2}\right)}. \]

Since \(\cosh(\pi x)\) is an even function, the imaginary part \(\int_{-\infty}^\infty \frac{\sin(ax)}{\cosh(\pi x)}\,dx\) vanishes by symmetry, so the formula is symmetric in \(a\) and holds for all real \(a \in \mathbb{R}\).

Method 1: An expanding rectangular contour

Let \(f(z) = \frac{e^{iaz}}{\cosh(\pi z)}\). The poles of \(f(z)\) occur where \(\cosh(\pi z) = 0\), namely at

\[ z_n = \frac{(2n+1)i}{2}, \qquad n \in \mathbb{Z}, \]

all of which lie on the imaginary axis with spacing \(i\). For each positive integer \(N \in \{1, 2, \dots\}\), define \(B_N\) to be the positively oriented rectangle with vertices \(-N\), \(N\), \(N+iN\), and \(-N+iN\):

Choosing the top edge at height \(\operatorname{Im} z = N\) ensures that the contour passes halfway between the poles \(z_{N-1} = \left(N-\frac{1}{2}\right)i\) and \(z_N = \left(N+\frac{1}{2}\right)i\), enclosing exactly the \(N\) simple poles \(z_0, z_1, \dots, z_{N-1}\).

At \(z_n = \frac{(2n+1)i}{2}\), we have \(\frac{d}{dz}\cosh(\pi z)\big|_{z=z_n} = \pi\sinh\left(i\pi\left(n+\frac{1}{2}\right)\right) = \pi i (-1)^n\), so the residue is

\[ \mathop{\operatorname{Res}}_{z=\frac{(2n+1)i}{2}} \frac{e^{iaz}}{\cosh(\pi z)} = \frac{e^{-a\left(n+\frac{1}{2}\right)}}{\pi i (-1)^n} = \frac{(-1)^n e^{-\frac{a}{2}} e^{-an}}{\pi i}. \]

Applying the residue theorem to \(B_N\) and taking \(N \to \infty\) gives a convergent geometric series (since \(a > 0\) implies \(e^{-a} < 1\)):

\[ \begin{aligned} \lim_{N\to \infty}\int_{B_N}\frac{e^{ia z}}{\cosh(\pi z)}\,dz &= 2\pi i \sum_{n=0}^\infty \mathop{\operatorname{Res}}_{z=\frac{(2n+1)i}{2}} \frac{e^{ia z}}{\cosh(\pi z)} \\ &= 2\pi i \left(\frac{e^{-\frac{a}{2}}}{\pi i}\sum_{n=0}^\infty (-1)^n e^{-an}\right) \\ &= \frac{2e^{-\frac{a}{2}}}{1 + e^{-a}} \\ &= \frac{1}{\cosh\left(\frac{a}{2}\right)}. \end{aligned} \]

To verify that only the integral along the bottom edge \([-N, N]\) survives as \(N \to \infty\), we estimate the other three sides:

Top edge (\(z = x + iN\), \(-N \leq x \leq N\)): Since \(\cosh(\pi(x+iN)) = (-1)^N\cosh(\pi x)\) and \(|e^{ia(x+iN)}| = e^{-aN}\),

\[ \left|\int_{-N}^{N} \frac{e^{ia(x+iN)}}{\cosh(\pi(x+iN))}\,dx\right| \leq e^{-aN}\int_{-\infty}^\infty \frac{dx}{\cosh(\pi x)} = e^{-aN} \to 0 \quad (N \to \infty). \]

Vertical edges (\(z = \pm N + iy\), \(0 \leq y \leq N\)): By the reverse triangle inequality, \(|\cosh(\pi(\pm N + iy))| \geq \sinh(\pi N)\), while \(|e^{ia(\pm N + iy)}| = e^{-ay} \leq 1\), so

\[ \left|\int_0^N \frac{e^{ia(\pm N + iy)}}{\cosh(\pi(\pm N + iy))}\,i\,dy\right| \leq \frac{N}{\sinh(\pi N)} \to 0 \quad (N \to \infty). \]

Consequently,

\[ \int_{-\infty}^\infty \frac{e^{ia x}}{\cosh(\pi x)}\,dx = \lim_{N\to \infty}\int_{B_N}\frac{e^{ia z}}{\cosh(\pi z)}\,dz = \frac{1}{\cosh\left(\frac{a}{2}\right)}. \]

Method 2: A fixed-height strip enclosing a single pole

We can also avoid summing an infinite series of residues by using a rectangle \(\Gamma_R\) of fixed height \(1\) with vertices \(-R\), \(R\), \(R+i\), and \(-R+i\). Since \(\cosh(\pi(x+i)) = -\cosh(\pi x)\), the top edge is a constant multiple of the bottom edge:

\[ \int_{R+i}^{-R+i} \frac{e^{iaz}}{\cosh(\pi z)}\,dz = e^{-a}\int_{-R}^R \frac{e^{iax}}{\cosh(\pi x)}\,dx. \]

The vertical sides are bounded by \(\frac{1}{\sinh(\pi R)} \to 0\) as \(R \to \infty\), and \(\Gamma_R\) encloses only the single pole at \(z_0 = \frac{i}{2}\). The residue theorem then gives

\[ (1 + e^{-a})\int_{-\infty}^\infty \frac{e^{iax}}{\cosh(\pi x)}\,dx = 2\pi i \mathop{\operatorname{Res}}_{z=\frac{i}{2}} \frac{e^{iaz}}{\cosh(\pi z)} = 2\pi i \left(\frac{e^{-\frac{a}{2}}}{\pi i}\right) = 2e^{-\frac{a}{2}}, \]

which immediately divides through to \(\frac{1}{\cosh(a/2)}\) for every \(a \in \mathbb{R}\) (and, by analytic continuation, for every complex \(a\) in the strip \(|\operatorname{Im} a| < \pi\)).

A real integral evaluated over a rectangular contour

2026-10-24T10:03:00+05:30

When an integrand combines exponential functions with a rational factor such as \(\frac{1}{x^2+\pi^2}\), a rectangular contour of height \(2\pi\) can exploit the periodicity of \(e^z\) in the imaginary direction while building the rational factor from the difference of two simple fractions. Using this technique, we will prove that for every \(a \in \mathbb{R}\),

\[ \int_{-\infty}^\infty \frac{e^x}{e^{2x}+e^{2a}}\frac{1}{x^2+\pi^2}\,dx = \frac{2\pi e^{-a}}{4a^2+\pi^2}-\frac{1}{1+e^{2a}}. \]

Setting up the rectangular contour

Consider the complex function

\[ f(z) = \frac{e^z}{(e^{2z}+e^{2a})z}. \]

Notice why \(f(z)\) is well suited to the horizontal lines \(\operatorname{Im} z = \pm \pi\): on those lines \(e^{x \pm i\pi} = -e^x\) and \(e^{2(x \pm i\pi)} = e^{2x}\), so the exponential factor \(\frac{e^z}{e^{2z}+e^{2a}}\) takes the exact same value \(-\frac{e^x}{e^{2x}+e^{2a}}\) on both horizontal edges, while the remaining factor \(\frac{1}{z}\) produces

\[ \frac{1}{x-i\pi} - \frac{1}{x+i\pi} = \frac{2\pi i}{x^2+\pi^2}. \]

Let \(\Gamma\) be the positively oriented rectangle in the complex plane with vertices \(R-i\pi\), \(R+i\pi\), \(-R+i\pi\), and \(-R-i\pi\), where \(R > |a|\):

Computing the residues inside the contour

Inside \(\Gamma\), the denominator \((e^{2z}+e^{2a})z\) vanishes at three points, each of which is a simple pole of \(f(z)\):

At \(z = 0\), the residue is

\[ \mathop{\operatorname{Res}}_{z=0} f(z) = \lim_{z \to 0} z f(z) = \frac{1}{1+e^{2a}}. \]

The solutions of \(e^{2z} = -e^{2a} = e^{2a \pm i\pi}\) with \(\operatorname{Im} z \in (-\pi, \pi)\) are \(z_\pm = a \pm \frac{i\pi}{2}\). At a simple zero \(z_0\) of \(e^{2z}+e^{2a}\), differentiating the denominator gives \(\frac{d}{dz}(e^{2z}+e^{2a})\big|_{z=z_0} = 2e^{2z_0} = -2e^{2a}\). Since \(e^{a \pm i\pi/2} = \pm i e^a\), the residues at \(z_\pm\) are

\[ \begin{aligned} \mathop{\operatorname{Res}}_{z=a+\frac{i\pi}{2}} f(z) &= \frac{e^{a+\frac{i\pi}{2}}}{-2e^{2a}\left(a+\frac{i\pi}{2}\right)} = -\frac{i e^{-a}}{2a+i\pi}, \\ \mathop{\operatorname{Res}}_{z=a-\frac{i\pi}{2}} f(z) &= \frac{e^{a-\frac{i\pi}{2}}}{-2e^{2a}\left(a-\frac{i\pi}{2}\right)} = \frac{i e^{-a}}{2a-i\pi}. \end{aligned} \]

By the residue theorem,

\[ \begin{aligned} \int_\Gamma f(z)\,dz &= 2\pi i \left( \mathop{\operatorname{Res}}_{z=0} f(z) + \mathop{\operatorname{Res}}_{z=a+\frac{i\pi}{2}} f(z) + \mathop{\operatorname{Res}}_{z=a-\frac{i\pi}{2}} f(z)\right) \\ &= 2\pi i \left( \frac{1}{1+e^{2a}} - \frac{i e^{-a}}{2a+i\pi} + \frac{i e^{-a}}{2a-i\pi}\right) \\ &= 2\pi i \left( \frac{1}{1+e^{2a}} - \frac{2\pi e^{-a}}{4a^2 +\pi^2}\right). \end{aligned} \tag{1} \]

Relating the contour integral to the real integral

Breaking \(\Gamma\) into its four sides and parameterizing the bottom and top edges by \(z = x \mp i\pi\) for \(x \in [-R, R]\), we have

\[ \begin{aligned} \int_\Gamma f(z)\,dz &= \int_{-R-i\pi}^{R-i\pi} f(z)\,dz + \int_{R-i\pi}^{R+i\pi} f(z)\,dz + \int_{R+i\pi}^{-R+i\pi} f(z)\,dz + \int_{-R+i\pi}^{-R-i\pi} f(z)\,dz \\ &= \int_{-R}^{R} \bigl(f(x-i\pi)-f(x+i\pi)\bigr)\,dx + \int_{R-i\pi}^{R+i\pi} f(z)\,dz + \int_{-R+i\pi}^{-R-i\pi} f(z)\,dz \\ &= -2\pi i \int_{-R}^{R} \frac{e^x}{e^{2x}+e^{2a}}\frac{1}{x^2+\pi^2}\,dx + \int_{R-i\pi}^{R+i\pi} f(z)\,dz + \int_{-R+i\pi}^{-R-i\pi} f(z)\,dz. \end{aligned} \tag{2} \]

Bounding the vertical segments

On the right vertical edge \(z = R + iy\) with \(y \in [-\pi, \pi]\), we have \(|R+iy| \geq R\) and, by the reverse triangle inequality for \(R > |a|\),

\[ |e^{2R+2iy} + e^{2a}| = e^{2R}|e^{2iy} + e^{2a-2R}| \geq e^{2R}(1 - e^{2a-2R}). \]

Hence

\[ \begin{aligned} \left| \int_{R-i\pi}^{R+i\pi} f(z)\,dz\right| &= \left|\int_{-\pi}^\pi f(R+iy)\,i\,dy \right| \\ &\leq e^R \int_{-\pi}^\pi \frac{dy}{|R+iy|\cdot |e^{2R+2iy}+e^{2a}|} \\ &\leq \frac{e^{-R}}{R} \int_{-\pi}^\pi \frac{dy}{|e^{2iy}+e^{2a-2R}|} \\ &\leq \frac{e^{-R}}{R} \frac{2\pi}{1-e^{2a-2R}}. \end{aligned} \tag{3} \]

Similarly, on the left vertical edge \(z = -R + iy\), we have \(|e^z| = e^{-R}\) and \(|e^{-2R+2iy} + e^{2a}| \geq e^{2a} - e^{-2R}\) for \(R > |a|\), so

\[ \left|\int_{-R+i\pi}^{-R-i\pi} f(z)\,dz \right| \leq \frac{e^{-R}}{R}\frac{2\pi}{e^{2a}-e^{-2R}}. \tag{4} \]

From bounds (3) and (4), both vertical integrals tend to \(0\) as \(R \to \infty\). Equating (1) and (2) in the limit \(R \to \infty\) and dividing by \(-2\pi i\) yields

\[ \int_{-\infty}^\infty \frac{e^x}{e^{2x}+e^{2a}}\frac{1}{x^2+\pi^2}\,dx = \frac{2\pi e^{-a}}{4a^2+\pi^2}-\frac{1}{1+e^{2a}}. \]

An algebraic integral with a fifth root via the beta function

2026-10-24T10:02:00+05:30

In the previous post, we showed that for \(a, b, p > 0\),

\[ \int_0^1 x^{a-1}(1-x)^{b-1} \frac{dx}{(x+p)^{a+b}} = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\frac{1}{(1+p)^{a} p^b}. \]

We can use this identity, together with Euler's reflection formula for the gamma function, to evaluate the algebraic integral

\[ \int_{0}^{1}\frac{1}{(2-x)\sqrt[5]{x^{2}(1-x)^{3}}}\,dx = \frac{2\pi \sqrt[10]{2}}{\sqrt{5+\sqrt{5}}}. \]

Matching the parametric form

Perform the reflection substitution \(t = 1-x\), so that \(dx = -dt\) and \(2-x = 1+t\). The integral becomes

\[ \begin{aligned} \int_0^1 \frac{1}{(2-x)\sqrt[5]{x^2(1-x)^3}}\,dx &= \int_0^1 \frac{1}{(1+t)\sqrt[5]{(1-t)^2 t^3}}\,dt \\ &= \int_0^1 t^{-\frac{3}{5}}(1-t)^{-\frac{2}{5}}\frac{dt}{1+t} \\ &= \int_0^1 t^{\frac{2}{5}-1}(1-t)^{\frac{3}{5}-1}\frac{dt}{(t+1)^{\frac{2}{5}+\frac{3}{5}}}. \end{aligned} \]

This is now the special case \(a = \frac{2}{5}\), \(b = \frac{3}{5}\), and \(p = 1\) of our parametric beta integral. Since \(a+b = 1\) and \(\Gamma(1) = 1\), we get

\[ \int_0^1 \frac{1}{(2-x)\sqrt[5]{x^2(1-x)^3}}\,dx = \frac{\Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right)}{2^{\frac{2}{5}}}. \]

Simplifying with Euler's reflection formula

By Euler's reflection formula, \(\Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin(\pi s)}\) for \(s \notin \mathbb{Z}\). Taking \(s = \frac{2}{5}\) gives

\[ \Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right) = \frac{\pi}{\sin\left(\frac{2\pi}{5}\right)}. \]

Recall the exact trigonometric value

\[ \sin\left(\frac{2\pi}{5}\right) = \frac{\sqrt{10+2\sqrt{5}}}{4} = \frac{\sqrt{5+\sqrt{5}}}{2\sqrt{2}}. \]

Substituting this into our expression and combining the powers of \(2\) via \(\frac{2\sqrt{2}}{2^{2/5}} = 2^{1 + \frac{1}{2} - \frac{2}{5}} = 2 \cdot 2^{\frac{1}{10}} = 2\sqrt[10]{2}\), we arrive at

\[ \begin{aligned} \int_0^1 \frac{1}{(2-x)\sqrt[5]{x^2(1-x)^3}}\,dx &= \frac{\pi}{2^{\frac{2}{5}} \sin\left(\frac{2\pi}{5}\right)} \\ &= \frac{2\pi \sqrt[10]{2}}{\sqrt{5+\sqrt{5}}}. \end{aligned} \]

A parametric beta integral with a shifted linear denominator

2026-10-24T10:01:00+05:30

Recall that Euler's beta function is defined for \(\operatorname{Re} x > 0\) and \(\operatorname{Re} y > 0\) by

\[ B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1}\,dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}. \]

When an extra linear factor \((x+p)^{a+b}\) appears in the denominator with its exponent matching the sum of the two beta parameters, a homographic (Möbius) substitution removes the denominator completely while preserving the unit interval \([0,1]\). We will prove that for \(a, b, p > 0\),

\[ \int_0^1 x^{a-1}(1-x)^{b-1} \frac{dx}{(x+p)^{a+b}} = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\frac{1}{(1+p)^{a} p^b}. \]

The homographic substitution

Before changing variables, rewrite the integrand by grouping factors of \(x+p\) with \(x^{a-1}\), \((1-x)^{b-1}\), and \(dx\):

\[ x^{a-1}(1-x)^{b-1}\frac{dx}{(x+p)^{a+b}} = \left(\frac{x}{x+p}\right)^{a-1}\left(\frac{1-x}{x+p}\right)^{b-1}\frac{dx}{(x+p)^2}. \]

Now perform the change of variables

\[ \frac{x}{x+p} = \frac{t}{p+1}. \]

Notice that \(x=0\) corresponds to \(t=0\), and \(x=1\) corresponds to \(t=1\), so the limits of integration remain \(0\) and \(1\). Differentiating both sides with respect to \(x\) gives

\[ \begin{aligned} \frac{p}{(x+p)^2} &= \frac{1}{p+1}\frac{dt}{dx} \\ \implies \frac{dx}{(x+p)^2} &= \frac{dt}{p(p+1)}. \end{aligned} \]

To express \(\frac{1-x}{x+p}\) in terms of \(t\), observe that

\[ \frac{1+p}{x+p} = 1 + \frac{1-x}{x+p} \quad\text{and}\quad 1 - \frac{x}{x+p} = \frac{p}{x+p} = 1 - \frac{t}{p+1}. \]

Subtracting \(\frac{x}{x+p} = \frac{t}{p+1}\) from \(\frac{1+p}{x+p} = \frac{1+p}{p}\left(1 - \frac{t}{p+1}\right) = \frac{p+1-t}{p}\) yields

\[ \frac{1-x}{x+p} = \frac{1-t}{p}. \]

Reduction to the standard beta integral

Substituting these three relations into the integral, we obtain

\[ \begin{aligned} \int_0^1 x^{a-1}(1-x)^{b-1} \frac{dx}{(x+p)^{a+b}} &= \int_0^1 \left(\frac{t}{p+1}\right)^{a-1}\left(\frac{1-t}{p}\right)^{b-1}\frac{dt}{p(p+1)} \\ &= \frac{1}{p^b(1+p)^a}\int_0^1 t^{a-1}(1-t)^{b-1}\,dt \\ &= \frac{1}{p^b(1+p)^a} B(a,b) \\ &= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\frac{1}{(1+p)^a p^b}. \end{aligned} \]

By analytic continuation, the same identity remains valid for all complex parameters \(\operatorname{Re} a > 0\) and \(\operatorname{Re} b > 0\), and for any \(p \in \mathbb{C} \setminus [-1,0]\) using the principal branches of the powers. For a concrete application of this formula, see an algebraic integral with a fifth root.

A proof of GR 3.255

2025-11-16T12:04:08+00:00

A quadratic denominator can make a beta-type integral look forbidding. For entry 3.255 in Gradshteyn and Ryzhik, a change of variables exposes an inversion symmetry and reduces the calculation to a standard beta integral. We will prove

\[ \int _0^1 \frac{x^{\mu+\frac{1}{2}} (1-x)^{\mu-\frac{1}{2}}}{(c+2bx-ax^2)^{\mu+1}}dx = \frac{\sqrt{\pi}}{\left\{a + \left(\sqrt{c+2b-a} + \sqrt{c}\right)^2\right\}^{\mu+\frac{1}{2}}\sqrt{c+2b-a}} \frac{\Gamma \left(\mu + \frac{1}{2}\right)}{\Gamma\left(\mu+1\right)} \]

Assume that \(a,b,c\) are real and

\[ c>0,\qquad d:=c+2b-a>0,\qquad D:=a+(\sqrt d+\sqrt c)^2>0, \qquad \operatorname{Re}\mu>-\frac12. \]

All powers of positive real numbers use the real logarithm. These hypotheses ensure that the denominator stays positive on \([0,1]\) and that the endpoint singularities are integrable.

Move the endpoints to zero and infinity

Write the integral as \(I\) and set \(t=(1-x)/x\), so that \(x=1/(1+t)\). The powers of \(1+t\) cancel, leaving

\[ I=\int_0^\infty\frac{t^{\mu-\frac12}}{\bigl(ct^2+2(c+b)t+d\bigr)^{\mu+1}}\,dt. \]

Now put \(r=\sqrt{d/c}\), \(t=rs\), and \(h=(c+b)/\sqrt{cd}\). Then

\[ I=\frac{r^{\mu+\frac12}}{d^{\mu+1}}J, \qquad J=\int_0^\infty\frac{s^{\mu-\frac12}}{(s^2+2hs+1)^{\mu+1}}\,ds. \]

Notice that \(D=2(c+b+\sqrt{cd})\), so \(D>0\) is equivalent to \(h>-1\). Consequently \(s^2+2hs+1=(s-1)^2+2(h+1)s>0\) for \(s>0\).

Use the inversion symmetry

Replacing \(s\) by \(1/s\) in \(J\) gives a second representation:

\[ J=\int_0^\infty\frac{s^{\mu+\frac12}}{(s^2+2hs+1)^{\mu+1}}\,ds. \]

Average the two representations and factor a power of \(s\) out of the denominator:

\[ J=\int_0^\infty \frac{s+1}{2s^{3/2}}\bigl(s+s^{-1}+2h\bigr)^{-\mu-1}\,ds. \]

The factor in front is exactly the derivative of \(u=\sqrt s-1/\sqrt s\). This substitution maps \((0,\infty)\) onto the real line and satisfies \(s+s^{-1}=u^2+2\). Hence

\[ J=\int_{-\infty}^{\infty}(u^2+2+2h)^{-\mu-1}\,du. \]

Finish with the beta integral

For \(A>0\) and \(\operatorname{Re}\mu>-1/2\), substituting \(u=\sqrt A\,v\) and then using \(v^2\) as a variable gives

\[ \int_{-\infty}^{\infty}(u^2+A)^{-\mu-1}\,du =A^{-\mu-\frac12}B\left(\frac12,\mu+\frac12\right) =\frac{\sqrt\pi\,\Gamma(\mu+\frac12)}{\Gamma(\mu+1)}A^{-\mu-\frac12}. \]

Take \(A=2+2h=D/\sqrt{cd}\) and restore the prefactor. The constants simplify as

\[ \frac{r^{\mu+\frac12}}{d^{\mu+1}} \left(\frac{D}{\sqrt{cd}}\right)^{-\mu-\frac12} =\frac{1}{\sqrt d\,D^{\mu+\frac12}}, \]

which is the claimed evaluation. The symmetry step is the key: averaging before substituting produces the exact differential needed for a single beta integral.

Integral representations of the reciprocal beta function

2023-11-27T18:28:58+00:00

Integrating a suitable branch of a complex power around a contour gives a useful representation of the reciprocal beta function. We will derive the identity below and then differentiate related formulas to obtain logarithmic integrals:

\[ \int_0^\pi \left(\sin(\theta) \right)^{\alpha-1} e^{i \beta \theta} \; d\theta = \frac{\pi e^{\frac{i \pi}{2} \beta}}{\alpha 2^{\alpha-1} B \left(\frac{\alpha+\beta+1}{2}, \frac{\alpha-\beta+1}{2} \right)} \]

We will first prove this for real parameters satisfying \(\beta + 1>\alpha>0\). An analytic-continuation argument at the end will show that the identity actually holds for every real \(\alpha>0\) and every complex \(\beta\). Here \(B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\) is the Beta function.

The idea is to integrate the principal branch of \(f(z) = (1-z^2)^{\alpha-1} (-iz)^{\beta - \alpha}\) around the following contour:

where

\(C_\epsilon\) is an arc parameterized by \(e^{it}\), where \(\varphi_\epsilon\leq t \leq \pi - \varphi_\epsilon\) and \(\varphi_\epsilon = \text{arctan}\left( \frac{\epsilon \sqrt{4-\epsilon^2}}{2-\epsilon^2}\right)\).
\(I_{1,\epsilon}, \; I_{2,\epsilon}, \; I_{3,\epsilon}\) are circular indents of radius \(\epsilon\) around the branch points -1, 1 and 0, respectively.
\(S_{1,\epsilon}\) is a line joining \(-1+\epsilon\) and \(-\epsilon\).
\(S_{2,\epsilon}\) is a line joining \(\epsilon\) and \(1-\epsilon\).

The condition \(\beta + 1 > \alpha > 0\) ensures that:

\[ \lim_{\epsilon\to 0^+}\int_{I_{1,\epsilon}}f(z)\; dz = \lim_{\epsilon\to 0^+}\int_{I_{2,\epsilon}}f(z)\; dz = \lim_{\epsilon\to 0^+}\int_{I_{3,\epsilon}}f(z)\; dz = 0 \]

We'll only prove that the integral around \(I_{3,\epsilon}\) tends to 0 as \(\epsilon \to 0^+\). The proof is quite similar for \(I_{1,\epsilon}\) and \(I_{2,\epsilon}\). We have

\[ \begin{aligned} \left| \int_{I_{3,\epsilon}} f(z) \; dz\right| &= \left|-i\epsilon \int_0^\pi e^{it} f(\epsilon e^{it})\; dt\right| \\ &\leq \epsilon\int_0^\pi |f(\epsilon e^{it})| \; dt \\ &\leq \epsilon^{\beta-\alpha+1} \int_0^\pi |1-\epsilon^2 e^{2it}|^{\alpha-1} \; dt. \end{aligned} \]

For \(0<\epsilon<1\) we have

\[ 1-\epsilon^2\leq |1-\epsilon^2e^{2it}|\leq 1+\epsilon^2, \]

we must distinguish the sign of the exponent \(\alpha-1\). More precisely,

\[ |1-\epsilon^2e^{2it}|^{\alpha-1} \leq M_\alpha(\epsilon) := \begin{cases} (1+\epsilon^2)^{\alpha-1}, & \alpha\geq 1,\\ (1-\epsilon^2)^{\alpha-1}, & 0<\alpha<1. \end{cases} \]

Therefore,

\[ \left| \int_{I_{3,\epsilon}} f(z) \; dz\right| \leq \pi\epsilon^{\beta-\alpha+1}M_\alpha(\epsilon). \]

As \(M_\alpha(\epsilon)\to1\), the indentation integral tends to zero whenever \(\beta-\alpha+1>0\). Next, by Cauchy's theorem we have:

\[ \begin{aligned} \int_{-1}^1 f(z) \; dz &= -i\int_0^\pi e^{it}f(e^{it})\; dt \end{aligned} \]

The integral on the left hand side is:

\[ \begin{aligned}\int_{-1}^1 f(z) \; dz &= 2\cos\left( (\beta-\alpha)\frac{\pi}{2}\right)\int_0^1 (1-z^2)^{\alpha-1} z^{\beta-\alpha}\; dz \\ &= \cos\left( (\beta-\alpha)\frac{\pi}{2}\right) \int_0^1 (1-w)^{\alpha-1} w^{\frac{\beta-\alpha-1}{2}}\; dw \\ &= \cos\left( (\beta-\alpha)\frac{\pi}{2}\right) B\left(\alpha, \frac{\beta-\alpha+1}{2} \right) \\ &= \cos\left( (\beta-\alpha)\frac{\pi}{2}\right) \frac{\Gamma(\alpha) \Gamma\left(\frac{\beta-\alpha+1}{2} \right)}{\Gamma\left(\frac{\beta+\alpha+1}{2} \right)} \\ &= \frac{\pi \Gamma(\alpha)}{\Gamma\left(\frac{\beta+\alpha+1}{2} \right) \Gamma\left(\frac{-\beta+\alpha+1}{2} \right)} \\ &= \frac{\pi}{\alpha B \left(\frac{\alpha+\beta+1}{2}, \frac{\alpha-\beta+1}{2} \right)}\end{aligned} \]

The integral on the right hand side is:

\[ \begin{aligned}-i\int_0^\pi e^{it}f(e^{it})\; dt &= \int_0^\pi (1-e^{2it})^{\alpha-1}(-i e^{it})^{\beta-\alpha+1} \; dt \\ &= \int_0^\pi (2\sin(t))^{\alpha-1}e^{i\beta t -\frac{i\pi}{2}\beta}\; dt \\ &= e^{-\frac{i\pi}{2}\beta} 2^{\alpha-1} \int_0^\pi (\sin (t))^{\alpha-1}e^{i\beta t}\; dt \end{aligned} \]

Therefore, we have:

\[ \begin{aligned}e^{-\frac{i\pi}{2}\beta} 2^{\alpha-1} \int_0^\pi (\sin (t))^{\alpha-1}e^{i\beta t}\; dt &= \frac{\pi}{\alpha B \left(\frac{\alpha+\beta+1}{2}, \frac{\alpha-\beta+1}{2} \right)} \\ \implies \int_0^\pi \left(\sin(t) \right)^{\alpha-1} e^{i \beta t} \; dt &= \frac{\pi e^{\frac{i \pi}{2} \beta}}{\alpha 2^{\alpha-1} B \left(\frac{\alpha+\beta+1}{2}, \frac{\alpha-\beta+1}{2} \right)}\end{aligned} \]

We can use a similar technique to show that:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\cos(t)\right)^{\alpha-1}e^{i\beta t}\; dt = \frac{\pi}{\alpha 2^{\alpha-1} B \left(\frac{\alpha+\beta+1}{2}, \frac{\alpha-\beta+1}{2} \right)}, \quad \beta+1 > \alpha > 0 \]

Removing the restriction on \(\beta\)

We now make the analytic-continuation argument precise. Fix a real \(\alpha>0\) and define

\[ F_\alpha(\beta) :=e^{-\frac{i\pi\beta}{2}} \int_0^\pi (\sin t)^{\alpha-1}e^{i\beta t}\;dt =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos u)^{\alpha-1}e^{i\beta u}\;du. \]

The function \(F_\alpha\) is entire in \(\beta\). Indeed, on every compact set of \(\beta\)-values, the integrand and each of its derivatives with respect to \(\beta\) are bounded by a constant times the integrable function \((\cos u)^{\alpha-1}\). Thus differentiation under the integral sign is valid to every order.

On the other hand, put

\[ G_\alpha(\beta) :=\frac{\pi\Gamma(\alpha)} {2^{\alpha-1} \Gamma\left(\frac{\alpha+\beta+1}{2}\right) \Gamma\left(\frac{\alpha-\beta+1}{2}\right)}. \]

This is also entire in \(\beta\), because \(1/\Gamma(z)\) is an entire function. The contour argument above proves \(F_\alpha(\beta)=G_\alpha(\beta)\) for every real \(\beta>\alpha-1\). This set has accumulation points in the complex plane, so the identity theorem gives

\[ F_\alpha(\beta)=G_\alpha(\beta),\qquad \beta\in\mathbb C. \]

Multiplying by \(e^{i\pi\beta/2}\) proves the original formula without the condition \(\beta+1>\alpha\). Notice that the integral on the left still converges for every complex \(\beta\) when \(\alpha>0\); it is only the auxiliary beta integral along the real diameter that required the original restriction. The apparent poles on the right are harmless: when one of the gamma factors has a pole, its reciprocal is zero.

The same identity is holomorphic in \(\alpha\) on the half-plane \(\operatorname{Re}(\alpha)>0\). Applying the identity theorem once more, now with \(\beta\) fixed and \(\alpha\) as the variable, extends the result to every complex \(\alpha\) with \(\operatorname{Re}(\alpha)>0\).

A lower-half-plane contour

There is a useful companion proof obtained by closing the contour in the lower half-plane. We cannot use exactly the same powers without crossing their branch cuts, so instead take principal branches in

\[ g(z)=(1-z^2)^{\alpha-1}(iz)^{-\beta-\alpha}. \]

Use the lower unit semicircle, with the same small indentations at \(-1\), \(0\), and \(1\). For real \(\alpha>0\) and \(\beta<1-\alpha\), the endpoint indentations are again \(O(\left(\epsilon^\alpha\right))\), while the indentation around the origin is

\[ O\left(\epsilon^{1-\alpha-\beta}\right), \]

and hence all three vanish. On the real diameter,

\[ \begin{aligned} \int_{-1}^{1}g(x)\;dx &=2\cos\left(\frac{\pi(\alpha+\beta)}{2}\right) \int_0^1(1-x^2)^{\alpha-1}x^{-\alpha-\beta}\;dx\\ &=\cos\left(\frac{\pi(\alpha+\beta)}{2}\right) B\left(\alpha,\frac{1-\alpha-\beta}{2}\right)\\ &=\frac{\pi\Gamma(\alpha)} {\Gamma\left(\frac{\alpha+\beta+1}{2}\right) \Gamma\left(\frac{\alpha-\beta+1}{2}\right)}. \end{aligned} \]

Take the diameter from \(-1\) to \(1\) and then traverse the lower arc from \(1\) back to \(-1\). Parameterizing that arc by \(z=e^{-it}\) with \(0<t<\pi\) gives

\[ g(z)\,dz =-2^{\alpha-1}e^{-\frac{i\pi\beta}{2}} (\sin t)^{\alpha-1}e^{i\beta t}\;dt. \]

Cauchy's theorem therefore yields the same integral representation as before. This lower contour proves it directly in the second range \(\beta<1-\alpha\), whereas the upper contour proves it for \(\beta>\alpha-1\). If \(0<\alpha<1\), these two ranges together cover the entire real \(\beta\)-axis. When \(\alpha=1\), they cover every real \(\beta\) except \(\beta=0\), which follows immediately by continuity (or direct evaluation). For \(\alpha>1\) there remains a gap \(1-\alpha\leq\beta\leq\alpha-1\), so the identity-theorem argument is still the clean way to obtain the unrestricted result.

Many interesting logarithmic integrals can be derived using these identities by differentiating both sides with respect to the parameters \(\alpha\) and \(\beta\). One such example is:

\[ \int_0^{\frac{\pi}{2}} \theta^2 \log^2\left(\cos \theta\right)d\theta = \frac{11\pi^5}{1440}+\frac{\pi^3}{24}\log^2(2)+\frac{\pi}{2} \log(2)\zeta(3) \]

Introduction to Theta Functions II

2021-05-27T18:36:01+00:00

In part 1, we developed the basic identities for Jacobi's theta functions. We now derive their infinite product representations and use them to connect theta functions with elliptic integrals. We retain the convention \(q=e^{i\pi\tau}\), with \(\operatorname{Im}\tau>0\).

Infinite product representations of theta functions

Let \(f(z) = \prod_{n=1}^{\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz})\). Each of these two products converges absolutely and uniformly in any bounded domain of values of \(z\). Hence \(f(z)\) is analytic throughout the finite part of the \(z\) plane. The zeros of \(f(z)\) are simple zeros at the points \(z=\frac{2n+1}{2}\pi\tau+m\pi\) where \(m,n\in \mathbb{Z}\). So, the zeros of \(f(z)\) coincide with the zeros of \(\vartheta_4(z)\). Therefore, the function \(\frac{\vartheta_4(z)}{f(z)}\) has no poles or zeros in the finite part of the plane. Also, it is easy to see that \(f(z+\pi) = f(z)\) and

\[ \begin{aligned} f(z+\pi \tau) &= \prod_{n=1}^\infty (1-q^{2n+1}e^{2iz})(1-q^{2n-3}e^{-2iz}) \\ &= f(z) \frac{1-q^{-1}e^{-2iz}}{1-qe^{2iz}} \\ &= -q^{-1}e^{-2iz} f(z) \end{aligned} \]

That is, \(\frac{\vartheta_4(z)}{f(z)}\) is a doubly periodic function with periods \(\pi,\pi\tau\) and has no poles and zeros. By section 20.12 of [2], it is simply a constant, say \(G\). We have

\[ \begin{equation}\label{introduction-to-theta-functions-ii-eq-1} \vartheta_4(z) = G \prod_{n=1}^{\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz}) \end{equation} \]

Incrementing \(z\) by the half periods \(\frac{\pi}{2}, \frac{\pi \tau}{2}\) and \(\frac{\pi +\pi\tau}{2}\), we get the product representations for the other theta functions:

\[ \begin{align} \vartheta_3(z) &= G \prod_{n=1}^{\infty} (1+q^{2n-1}e^{2iz})(1+q^{2n-1}e^{-2iz})\label{introduction-to-theta-functions-ii-eq-2}\\ \vartheta_1(z) &= 2G q^{\frac{1}{4}}\sin(z) \prod_{n=1}^{\infty} (1-q^{2n}e^{2iz})(1-q^{2n}e^{-2iz}) \label{introduction-to-theta-functions-ii-eq-3}\\ \vartheta_2(z) &= 2G q^{\frac{1}{4}}\cos(z) \prod_{n=1}^{\infty} (1+q^{2n}e^{2iz})(1+q^{2n}e^{-2iz}) \label{introduction-to-theta-functions-ii-eq-4} \end{align} \]

Now, all that is left is to find the constant \(G\). From identity \(\eqref{introduction-to-theta-functions-ii-eq-3}\), it is easy to see that

\[ \vartheta_1'(0) = 2G q^{\frac{1}{4}}\prod_{n=1}^{\infty} (1-q^{2n})^2 \]

Now, using the identity \(\vartheta_1'(0)=\vartheta_2(0)\vartheta_3(0)\vartheta_4(0)\), we get:

\[ \begin{aligned} 2G q^{\frac{1}{4}}\prod_{n=1}^{\infty} (1-q^{2n})^2 &= \left\{2G q^{\frac{1}{4}}\prod_{n=1}^{\infty}(1+q^{2n})^2 \right\} \times \left\{G\prod_{n=1}^{\infty}(1+q^{2n-1})^2 \right\} \\ &\quad \times \left\{G \prod_{n=1}^{\infty}(1-q^{2n-1})^2 \right\} \\ \implies G &= \prod_{n=1}^\infty (1-q^{2n}) \end{aligned} \]

Note that the rearrangements are justified since all products converge absolutely. Finally, we have

\[ \begin{align} \vartheta_1(z) &= 2 q^{\frac{1}{4}}\sin(z) \prod_{n=1}^{\infty} (1-q^{2n})(1-2q^{2n} \cos(2z)+ q^{4n}) \label{introduction-to-theta-functions-ii-eq-5}\\ \vartheta_2(z) &= 2 q^{\frac{1}{4}}\cos(z) \prod_{n=1}^{\infty} (1-q^{2n})(1+2q^{2n} \cos(2z) + q^{4n}) \label{introduction-to-theta-functions-ii-eq-6}\\ \vartheta_3(z) &= \prod_{n=1}^{\infty} (1-q^{2n})(1+2q^{2n-1} \cos(2z) + q^{4n-2}) \label{introduction-to-theta-functions-ii-eq-7}\\ \vartheta_4(z) &= \prod_{n=1}^{\infty} (1-q^{2n})(1-2q^{2n-1} \cos(2z) + q^{4n-2}) \label{introduction-to-theta-functions-ii-eq-8} \end{align} \]

Derivatives of Ratios of Theta Functions

Consider the function \(\phi(z) = \frac{\vartheta_1'(z)\vartheta_4(z) - \vartheta_1(z)\vartheta_4'(z)}{\vartheta_2(z)\vartheta_3(z)}\). Now, \(\phi(z)\) is doubly periodic with periods \(\pi\) and \(\frac{\pi\tau}{2}\). Relative to these periods, the only possible poles of \(\phi(z)\) are at points congruent to \(\pi\over 2\). It is easy to verify that \(\frac{\pi}{2}\) is a removable singularity. This means that \(\phi(z)\) is a constant. Letting \(z\to 0\), we see that \(\phi(z)=\vartheta_4^2(0)\). It is therefore established that

\[ \begin{equation}\label{introduction-to-theta-functions-ii-eq-9} \frac{d}{dz}\left(\frac{\vartheta_1(z)}{\vartheta_4(z)} \right)=\vartheta_4^2(0) \frac{\vartheta_2(z)\vartheta_3(z)}{\vartheta_4^2(z)} \end{equation} \]

Two other identities of this kind can be derived using the same technique:

\[ \begin{align} \frac{d}{dz}\left(\frac{\vartheta_2(z)}{\vartheta_4(z)} \right) &=-\vartheta_3^2(0) \frac{\vartheta_1(z)\vartheta_3(z)}{\vartheta_4^2(z)} \label{introduction-to-theta-functions-ii-eq-10}\\ \frac{d}{dz}\left(\frac{\vartheta_3(z)}{\vartheta_4(z)} \right) &=-\vartheta_2^2(0) \frac{\vartheta_1(z)\vartheta_2(z)}{\vartheta_4^2(z)} \label{introduction-to-theta-functions-ii-eq-11} \end{align} \]

By writing \(\xi = \frac{\vartheta_1(z)}{\vartheta_4(z)}\) and using identities (13) and (14) from part 1, we obtain:

\[ \left(\frac{d\xi}{dz}\right)^2 = (\vartheta_2^2(0) - \xi^2 \vartheta_3^2(0))(\vartheta_3^2(0)-\xi^2 \vartheta_2^2(0)) \]

Using the change of variables \(y = \frac{\vartheta_3(0)}{\vartheta_2(0)}\xi\) and \(u = \vartheta_3^2(0) z\), the above differential equation can be written as:

\[ \left(\frac{dy}{du}\right)^2 = (1-y^2)(1-k^2 y^2) \]

where \(k = \frac{\vartheta_2^2(0)}{\vartheta_3^2(0)}\). This differential equation has the particular solution:

\[ y = \frac{\vartheta_3(0)}{\vartheta_2(0)} \frac{\vartheta_1(u \vartheta_3^{-2}(0))}{\vartheta_4(u \vartheta_3^{-2}(0))} \]

We can write \(y\) as a function of \(u\) and \(k\) as \(y = \text{sn}(u,k)\) or simply \(\text{sn}(u)\). Clearly, \(\text{sn}(u,k)\) has periods \(2\pi \vartheta^2_3(0)\) and \(\pi \tau \vartheta_3^2(0)\). It has two simple poles congruent to \(\frac{1}{2}\pi\tau \vartheta_3^2(0)\) and \(\pi \vartheta_3^2(0) + \frac{1}{2}\pi\tau \vartheta_3^2(0)\). Also, it is easy to see that

\[ \int_0^1 \frac{dy}{\sqrt{(1-y^2)(1-k^2 y^2)}} = \text{sn}^{-1}(1) = \frac{\pi}{2}\vartheta_3^2(0) \]

The integral on the left is denoted by complete elliptic integral of the first kind and is denoted by \(K(k)\).

References

Derek F. Lawden (1989). Elliptic Functions and Applications. Springer-Verlag New York
E. T. Whittaker, G. N. Watson (1927). A Course of Modern Analysis. Cambridge University Press

Introduction to Theta Functions I

2021-05-12T18:01:49+00:00

Jacobi's theta functions connect rapidly convergent series with doubly periodic functions. In this first part, we will work from their series definitions to their periodicity, zeros, and fundamental identities.

The four theta functions

For complex \(z\) and a nonzero nome \(q\) with \(|q|<1\), the four functions are defined by the following series. Fractional powers of \(q\) are understood consistently by writing \(q=e^{i\pi\tau}\) with \(\operatorname{Im}\tau>0\).

\[ \begin{aligned} \vartheta_1 (z,q) &= -i \sum_{n=-\infty}^{\infty} (-1)^n q^{(n+1/2)^2} e^{i(2n+1)z} \\ \vartheta_2 (z,q) &= \sum_{n=-\infty}^{\infty} q^{(n+1/2)^2} e^{i(2n+1)z} \\ \vartheta_3 (z,q) &= \sum_{n=-\infty}^{\infty} q^{n^2} e^{2inz} \\ \vartheta_4 (z,q) &= \sum_{n=-\infty}^{\infty} (-1)^n q^{n^2} e^{2inz} \\ \end{aligned} \]

The parameter \(q\) is called the nome. Let \(\tau\) be a complex number whose imaginary part is positive and write \(q=e^{i\pi \tau}\) so that \(|q| < 1\). Sometimes, \(q\) will not be specified, so that \(\vartheta_r (z)\) is written for \(\vartheta_r(z,q)\) where \(r=1,2,3,4\). It is easy to see that

\[ \begin{aligned} \vartheta_1(z+\pi) &= -i e^{i \pi}\sum_{n=-\infty}^{\infty}(-1)^n q^{\left(n+\frac{1}{2}\right)^2}e^{i(2n+1)z} = -\vartheta_1(z) \\ \vartheta_1(z+\pi \tau) &= i e^{-i\pi \tau -2iz} \sum_{n=-\infty}^{\infty} (-1)^{n+1} e^{i\pi \tau \left( n+\frac{3}{2}\right)^2 + (2n+3) iz} = - (q e^{2iz})^{-1} \vartheta_1(z) \end{aligned} \]

Similarly, the periodicity laws for the other theta functions can be derived as well. The multipliers of the theta functions associated with the periods \(\pi\) and \(\pi \tau\) are summarized in the following table:

\[ \begin{array}{|c|c|c|c|c|} \hline \omega & \vartheta_1(z+\omega) & \vartheta_2(z+\omega) & \vartheta_3(z+\omega) & \vartheta_4(z+\omega) \\ \hline \hline \pi & -1 & -1 & 1 & 1 \\ \hline \pi \tau & -\lambda & \lambda & \lambda & -\lambda \\ \hline \end{array} \]

where \(\lambda = (q e^{2iz})^{-1}\). Furthermore, incrementation of \(z\) by the half periods \(\frac{\pi}{2}, \frac{\pi \tau}{2}, \frac{\pi +\pi \tau}{2}\) yields the following identities:

\[ \begin{array}{|c|c|c|c|c|} \hline \omega & \vartheta_1(z+\omega) & \vartheta_2(z+\omega) & \vartheta_3(z+\omega) & \vartheta_4(z+\omega) \\ \hline \hline \frac{\pi}{2} & \vartheta_2(z) & -\vartheta_1(z) & \vartheta_4(z) & \vartheta_3(z) \\ \hline \frac{\pi \tau}{2} & i \mu \vartheta_4(z) & \mu \vartheta_3(z) & \mu \vartheta_2(z) & i\mu \vartheta_1(z) \\ \hline \frac{\pi + \pi \tau}{2} & \mu \vartheta_3(z) & -i\mu \vartheta_4(z) & i\mu \vartheta_1(z) & \mu \vartheta_2(z) \\ \hline \end{array} \]

where \(\mu = (q^{\frac{1}{4}}e^{iz})^{-1}\).

Identities involving products of theta functions

Many theta function identities can be derived by multiplying their series and rearranging the terms. This is ok to do since both the series are absolutely convergent. For e.g., we have

\[ \begin{aligned} \vartheta_3(x,q)\vartheta_3(y,q) &= \left(\sum_{n=-\infty}^{\infty} q^{n^2} e^{2inx} \right)\left(\sum_{m=-\infty}^{\infty} q^{m^2} e^{2imy} \right) \\ &= \sum_{n=-\infty}^\infty \sum_{m=-\infty}^\infty q^{n^2+m^2}e^{2i(nx+my)} \end{aligned} \]

Now, we change the summation indices from \((m,n)\) to \((r,s)\) using the following transformation:

\[ \begin{aligned} r &= m+n \\ s &= m-n \end{aligned} \]

If \((m,n)\) are both even then \((r,s)\) will both be even and if \((m,n)\) have opposite parity then \((r,s)\) will both be odd. Therefore, we can rearrange the series as follows:

\[ \begin{equation}\label{introduction-to-theta-functions-i-eq-1} \begin{aligned} \vartheta_3(x,q)\vartheta_3(y,q) &= \sum_{r=-\infty}^{\infty} \sum_{s=-\infty}^{\infty} q^{2(r^2 +s^2)}e^{2i(r(x+y)+s(x-y))} \\ &\quad + \sum_{r=-\infty}^{\infty} \sum_{s=-\infty}^{\infty}q^{2\left[\left(r+\frac{1}{2} \right)^2 + \left(s+\frac{1}{2} \right)^2\right]} e^{i(2r+1)(x+y) + i(2s+1)(x-y)} \\ &= \vartheta_3(x+y,q^2)\vartheta_3(x-y,q^2) + \vartheta_2(x+y,q^2)\vartheta_2(x-y,q^2) \end{aligned} \end{equation} \]

Some similar identities that can be derived using this method are:

\[ \begin{align} \vartheta_1(x,q) \vartheta_1(y,q) &= \vartheta_3(x+y,q^2)\vartheta_2(x-y,q^2) - \vartheta_2(x+y,q^2)\vartheta_3(x-y,q^2) \label{introduction-to-theta-functions-i-eq-2}\\ \vartheta_2(x,q) \vartheta_2(y,q) &= \vartheta_2(x+y,q^2)\vartheta_3(x-y,q^2) + \vartheta_3(x+y,q^2)\vartheta_2(x-y,q^2) \label{introduction-to-theta-functions-i-eq-3}\\ \vartheta_4(x,q) \vartheta_4(y,q) &= \vartheta_3(x+y,q^2)\vartheta_3(x-y,q^2) - \vartheta_2(x+y,q^2)\vartheta_2(x-y,q^2) \label{introduction-to-theta-functions-i-eq-4}\\ \vartheta_1(x,q) \vartheta_2(y,q) &= \vartheta_1(x+y,q^2)\vartheta_4(x-y,q^2) + \vartheta_4(x+y,q^2)\vartheta_1(x-y,q^2) \label{introduction-to-theta-functions-i-eq-5}\\ \vartheta_3(x,q) \vartheta_4(y,q) &= \vartheta_4(x+y,q^2)\vartheta_4(x-y,q^2) - \vartheta_1(x+y,q^2)\vartheta_1(x-y,q^2) \label{introduction-to-theta-functions-i-eq-6} \end{align} \]

Squaring and subtracting identities \(\eqref{introduction-to-theta-functions-i-eq-4}\) and \(\eqref{introduction-to-theta-functions-i-eq-2}\) gives us:

\[ \begin{equation}\label{introduction-to-theta-functions-i-eq-7} \begin{aligned} &\; \vartheta_4^2(x,q) \vartheta_4^2(y,q) - \vartheta_1^2(x,q) \vartheta_1^2(y,q)\\ &= \left[\vartheta_3^2(x+y,q^2)-\vartheta_2^2(x+y,q^2) \right]\times \left[\vartheta_3^2(x-y,q^2)-\vartheta_2^2(x-y,q^2) \right] \end{aligned} \end{equation} \]

Putting \(y=0\) in the above equation gives the result:

\[ \vartheta_3^2(x,q^2)-\vartheta_2^2(x,q^2) = \vartheta_4(x,q)\vartheta_4(0,q) \]

Now, using the above result in equation \(\eqref{introduction-to-theta-functions-i-eq-7}\) gives us:

\[ \begin{equation}\label{introduction-to-theta-functions-i-eq-8} \vartheta_4(x+y,q)\vartheta_4(x-y,q)\vartheta_4^2(0,q) = \vartheta_4^2(x,q) \vartheta_4^2(y,q) - \vartheta_1^2(x,q) \vartheta_1^2(y,q) \end{equation} \]

Note that all theta functions in this equation have the same nome, \(q\). More identities of this type can be derived by incrementing \(x\) and/or \(y\) by the half periods \(\frac{\pi}{2},\frac{\pi \tau}{2},\frac{\pi+\pi\tau}{2}\):

\[ \begin{align} \vartheta_1(x+y)\vartheta_1(x-y)\vartheta_4^2(0) &= \vartheta_3^2(x) \vartheta_2^2(y) - \vartheta_2^2(x) \vartheta_3^2(y) \notag\\ &= \vartheta_1^2(x) \vartheta_4^2(y) - \vartheta_4^2(x) \vartheta_1^2(y)\label{introduction-to-theta-functions-i-eq-9}\\ \vartheta_2(x+y)\vartheta_2(x-y)\vartheta_4^2(0) &= \vartheta_4^2(x) \vartheta_2^2(y) - \vartheta_1^2(x) \vartheta_3^2(y) \notag\\ &= \vartheta_2^2(x) \vartheta_4^2(y) - \vartheta_3^2(x) \vartheta_1^2(y)\label{introduction-to-theta-functions-i-eq-10}\\ \vartheta_3(x+y)\vartheta_3(x-y)\vartheta_4^2(0) &= \vartheta_4^2(x) \vartheta_3^2(y) - \vartheta_1^2(x) \vartheta_2^2(y) \notag\\ &= \vartheta_3^2(x) \vartheta_4^2(y) - \vartheta_2^2(x) \vartheta_1^2(y)\label{introduction-to-theta-functions-i-eq-11}\\ \vartheta_4(x+y)\vartheta_4(x-y)\vartheta_4^2(0) &= \vartheta_3^2(x) \vartheta_3^2(y) - \vartheta_2^2(x) \vartheta_2^2(y) \notag\\ &= \vartheta_4^2(x) \vartheta_4^2(y) - \vartheta_1^2(x) \vartheta_1^2(y)\label{introduction-to-theta-functions-i-eq-12} \end{align} \]

One can keep on deriving similar identities by squaring and adding equations \(\eqref{introduction-to-theta-functions-i-eq-1}\) and \(\eqref{introduction-to-theta-functions-i-eq-2}\) and squaring and subtracting equations \(\eqref{introduction-to-theta-functions-i-eq-3}\) and \(\eqref{introduction-to-theta-functions-i-eq-2}\). For a complete list of such identities, refer to [1 sec 1.4].

Another way to derive theta function identities is to utilize properties of doubly periodic functions. More details on this method are presented in [2]. Consider the function:

\[ \frac{a \vartheta_1^2(z) + b \vartheta_4^2(z)}{\vartheta_2^2(z)} \]

This is a doubly periodic function with periods \(\pi\) and \(\pi \tau\). Furthermore, we can choose the constants \(a\) and \(b\) such that there are no poles in each cell. By Liouville's Theorem, such a function is a constant. By appropriately scaling the constants \(a\) and \(b\), we can make the function equal to 1. Therefore, there exists a relationship of the form:

\[ a \vartheta_1^2(z) + b \vartheta_4^2(z) = \vartheta_2^2(z) \]

To find \(a\) and \(b\), we can put \(z = 0, \frac{\pi\tau}{2}\):

\[ \begin{aligned} b \vartheta_4^2(0) &= \vartheta_2^2(0) \; \implies b = \frac{\vartheta_2^2(0)}{\vartheta_4^2(0)} \\ a (i \mu\vartheta_4(0))^2 &= (\mu \vartheta_3(0))^2 \; \implies a = -\frac{\vartheta_3^2(0)}{\vartheta_4^2(0)} \end{aligned} \]

Thus, we have obtained the identity:

\[ \begin{equation}\label{introduction-to-theta-functions-i-eq-13} \vartheta_2^2(0) \vartheta_4^2(z) -\vartheta_3^2(0) \vartheta_1^2(z) = \vartheta_4^2(0) \vartheta_2^2(z) \end{equation} \]

A similar technique yields the identity:

\[ \begin{equation}\label{introduction-to-theta-functions-i-eq-14} \vartheta_3^2(0) \vartheta_4^2(z) - \vartheta_2^2(0) \vartheta_1^2(z) = \vartheta_4^2(0) \vartheta_3^2(z) \end{equation} \]

Incrementing \(z\) by \(\frac{\pi}{2}\) in \(\eqref{introduction-to-theta-functions-i-eq-11}\) and \(\eqref{introduction-to-theta-functions-i-eq-12}\) gives two additional identities:

\[ \begin{align} \vartheta_4^2(0) \vartheta_1^2(z) &= \vartheta_2^2(0) \vartheta_3^2(z) - \vartheta_3^2(0) \vartheta_2^2(z) \label{introduction-to-theta-functions-i-eq-15}\\ \vartheta_4^2(0) \vartheta_4^2(z) &= \vartheta_3^2(0) \vartheta_3^2(z) - \vartheta_2^2(0) \vartheta_2^2(z) \label{introduction-to-theta-functions-i-eq-16} \end{align} \]

With these relations one can express any theta function in terms of any other pair of theta functions. Putting \(z=0\) in \(\eqref{introduction-to-theta-functions-i-eq-16}\) gives the famous identity:

\[ \vartheta_4^4(0)+\vartheta_2^4(0) = \vartheta_3^4(0) \]

The identity \(\vartheta_1'(0)=\vartheta_2(0)\vartheta_3(0)\vartheta_4(0)\)

This particular proof is from [1]. Differentiating equation \(\eqref{introduction-to-theta-functions-i-eq-5}\) with respect to \(x\) and substituting \(x=y=0\), we get:

\[ \begin{equation}\label{introduction-to-theta-functions-i-eq-17} \vartheta_1'(0,q) \vartheta_2(0,q) = 2 \vartheta_1'(0,q^2)\vartheta_4(0,q^2) \end{equation} \]

Substituting \(x=y=0\) in equations \(\eqref{introduction-to-theta-functions-i-eq-3}\) and \(\eqref{introduction-to-theta-functions-i-eq-6}\) gives:

\[ \begin{align} \vartheta_2^2(0,q) &= 2\vartheta_2(0,q^2) \vartheta_3(0,q^2) \label{introduction-to-theta-functions-i-eq-18}\\ \vartheta_3(0,q)\vartheta_4(0,q) &= \vartheta_4^2(0,q^2) \label{introduction-to-theta-functions-i-eq-19} \end{align} \]

Now, dividing \(\eqref{introduction-to-theta-functions-i-eq-17}\) by both \(\eqref{introduction-to-theta-functions-i-eq-18}\) and \(\eqref{introduction-to-theta-functions-i-eq-19}\) gives

\[ \frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)} = \frac{\vartheta_1'(0,q^2)}{\vartheta_2(0,q^2)\vartheta_3(0,q^2)\vartheta_4(0,q^2)} \]

The repeated application of this result gives:

\[ \frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)} = \frac{\vartheta_1'(0,q^{2^n})}{\vartheta_2(0,q^{2^n})\vartheta_3(0,q^{2^n})\vartheta_4(0,q^{2^n})} \]

for all positive integers \(n\). Letting \(n\to\infty\) in the above equation, we find that:

\[ \begin{equation}\label{introduction-to-theta-functions-i-eq-20} \frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)} = \lim_{q\to 0}\frac{\vartheta_1'(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)} \end{equation} \]

From the definitions of the theta functions, it is evident that

\[ \begin{aligned} \vartheta_1'(0,q) &= 2q^{\frac{1}{4}} + \mathcal{O}(q^{\frac{9}{4}}) \\ \vartheta_2(0,q) &= 2q^{\frac{1}{4}} + \mathcal{O}(q^{\frac{9}{4}}) \\ \vartheta_3(0,q) &= 1 + \mathcal{O}(q) \\ \vartheta_4(0,q) &= 1 + \mathcal{O}(q) \\ \end{aligned} \]

Therefore, the limit in equation \(\eqref{introduction-to-theta-functions-i-eq-20}\) equals 1 and

\[ \vartheta_1'(0)=\vartheta_2(0)\vartheta_3(0)\vartheta_4(0) \]

is established as desired.

References

Derek F. Lawden (1989). Elliptic Functions and Applications. Springer-Verlag New York
E. T. Whittaker, G. N. Watson (1927). A Course of Modern Analysis. Cambridge University Press

When is the group of units in the integers modulo n cyclic?

2020-12-12T13:22:20+00:00

We begin with the multiplicative group of units modulo an integer. By Bezout's identity that the integers co-prime to \(n\) from the set \(\{0,1,\cdots, n-1\}\) form a group under multiplication modulo \(n\). This group is denoted by \((\mathbb{Z}/n\mathbb{Z})^*\) and its order is given by the Euler's Totient function: \(\varphi(n) = |(\mathbb{Z}/n\mathbb{Z})^*|\).

The classification

Gauss showed that \((\mathbb{Z}/n\mathbb{Z})^*\) is a cyclic group if and only if \(n=1,2,4,p,p^k\) or \(2p^k\), where \(p\) is an odd prime and \(k > 0\). It is very easy to verify this for \(n=1,2\) and \(4\), as one can simply list out all positive integers less than and co-prime to \(n\). So, we will only focus on proving that \((\mathbb{Z}/n\mathbb{Z})^*\) is cyclic for the remaining three cases.

Theorem 1: \((\mathbb{Z}/p\mathbb{Z})^*\) is cyclic if \(p\) is a prime.

Proof: Let \(d_1, d_2, \cdots , d_r\) be all the possible orders of the elements of \((\mathbb{Z}/p\mathbb{Z})^*\). Let \(e=\text{lcm}(d_1, d_2, \cdots, d_r)\) and factor \(e=p_1^{a_1} p_2^{a_2}\cdots p_k^{a_k}\) as a product of distinct prime powers. By the definition of \(\text{lcm}\), for each \(p_i^{a_i}\) there is some \(d_j\) divisible by it. Since the \(d_j\)'s are orders of elements of \((\mathbb{Z}/p\mathbb{Z})^*\), there is an element \(x_i\) whose order is \(p_i^{a_i} t\). Then, the element \(y_i = x_i^t\) has order \(p_i^{a_i}\). Hence, \(y_1 y_2\cdots y_k\) has order \(e\). Thus, we have found an element of order \(e\). Therefore, \(e|p-1\). But the polynomial \(x^e-1\) has \(p-1\) roots \((\text{mod }p)\). Since \(\mathbb{Z}/p\mathbb{Z}\) is a field, any polynomial of degree \(e\) cannot have more than \(e\) roots. Therefore, \(p-1\leq e\) and we deduce that \(e=p-1\). Thus, we have found an element of order \(p-1\).

Theorem 2: \((\mathbb{Z}/p^a \mathbb{Z})^*\) is cyclic for any \(a\geq 1\) if \(p\) is an odd prime.

Proof: For \(a=1\), we are done by Theorem 1. Let \(g\) be a primitive root \((\text{mod }p)\). We first find a \(t\) such that \((g+pt)^{p-1}\not\equiv 1 \; (\text{mod }p^2)\). If \(g^{p-1}\not\equiv 1 \; (\text{mod }p^2)\), then we can take \(t=0\). Otherwise, we can choose \(t=1\). Indeed, we have \((g+p)^{p-1}\equiv 1+p(p-1) g^{p-2} \not \equiv 1 \; (\text{mod }p^2)\). Let \(g+pt\) have order \(d\; (\text{mod }p^a)\). Then, \(d|p^{a-1} (p-1)\). Since \(g\) is a primitive root modulo \(p\), \((p-1) | d\). So, \(d=p^{r-1} (p-1)\) for some \(r\leq a\). We also know that \((g+pt)^{p-1} = 1+p s\) where \(p\nmid s\). Thus,

\[ \begin{aligned} (g+pt)^{p(p-1)} &= (1+ps)^{p} \\ &= \sum_{i=0}^{p} \binom{p}{i} (ps)^i \\ &\equiv 1+p^2 s \; (\text{mod }p^3) \end{aligned} \]

By induction, it follows that

\[ (g+pt)^{p^{b-1}(p-1)} \equiv 1+p^b s \; (\text{mod }p^{b+1}) \]

Now, \(g+pt\) has order \(d=p^{r-1}(p-1)\; (\text{mod } p^a)\) which implies \((g+pt)^{p^{r-1}(p-1)} \equiv 1 \; (\text{mod }p^a)\). But then \(1+p^r s \equiv 1 \; (\text{mod } p^{r+1})\) if \(r\leq a-1\), which implies \(p | s\), a contradiction. Thus, \(r=a\).

Theorem 3: \((\mathbb{Z}/2p^a \mathbb{Z})^*\) is cyclic for any \(a\geq 1\) if \(p\) is an odd prime.

Proof: This follows immediately from Theorem 2 and the Chinese remainder theorem.

Why there are no other cases

Suppose first that \(n\) has two distinct odd prime factors, \(p\) and \(q\). Choosing a sign independently modulo each prime power and applying the Chinese remainder theorem gives at least four solutions of \(x^2\equiv1\pmod n\). A cyclic group has at most two elements whose square is the identity, so its unit group cannot be cyclic.

If \(4\mid n\) and an odd prime divides \(n\), the same argument applies by choosing signs independently modulo the power of \(2\) and the odd part. Finally, if \(8\mid n\), the four residues \(1,-1,1+2^{a-1},-1+2^{a-1}\) are distinct square roots of \(1\) modulo \(2^a\) for \(a\geq3\); the Chinese remainder theorem again transfers this obstruction to \(n\).

The only remaining possibilities are \(1,2,4,p^a\), and \(2p^a\) with \(p\) odd, exactly the cases already proved.

References

M. Ram Murty. Problems in analytic number theory. Springer New York, 01-Nov-2008

Evaluating very nasty logarithmic integrals: Part III

2020-09-07T18:58:20+00:00

This third part uses complex contours to evaluate logarithmic integrals involving inverse trigonometric functions. We continue the numbering and use results from part 1 and part 2. The contour diagrams below show the paths used in each argument.

Integral #5

The first integral that we will evaluate in this post is the following:

\[ I_1 = \int_0^1 \frac{\log^2(x) \arctan(x)}{1+x^2}dx \]

Of course, one can use brute force methods to find a closed-form antiderivative in terms of polylogarithms. Instead, a more elegant solution is possible by contour integration.

We'll integrate the principal branch of \(f(z) = \frac{\arctan(z)}{1+z^2}\left(\text{arctanh}^2(z) + \frac{\pi^2}{16} \right)\) around the following contour:

where

\(\gamma_{3,\epsilon}\) is an arc parameterized by \(e^{it}\), where \(\arctan\left(\frac{\epsilon \sqrt{4-\epsilon^2}}{2-\epsilon^2} \right)\leq t \leq \frac{\pi}{2} - \arctan\left(\frac{\epsilon \sqrt{4-\epsilon^2}}{2-\epsilon^2} \right)\).
\(\gamma_{2,\epsilon}\) and \(\gamma_{4,\epsilon}\) are circular indents of radius \(\epsilon\) around the branch points at \(z=1\) and \(z=i\), respectively.
\(\gamma_{1,\epsilon}\) is a straight line joining \(0\) and \(1-\epsilon\).
\(\gamma_{5,\epsilon}\) is a straight line joining \((1-\epsilon)i\) and \(0\).

Note that \(f\) is analytic on \(|z| < 1\). It is easy to see that

\[ \begin{aligned} \lim_{\epsilon\to 0^+} \int_{\gamma_{2,\epsilon}}f(z)\; dz &= 0 \\ \lim_{\epsilon\to 0^+} \int_{\gamma_{4,\epsilon}}f(z)\; dz &= 0 \end{aligned} \]

On \(\gamma_{1,\epsilon}\), we have

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-1} \begin{aligned} \lim_{\epsilon\to 0^+} \int_{\gamma_{1,\epsilon}} f(z)\; dz &= \int_0^1\frac{\arctan(x)}{1+x^2}\left(\text{arctanh}^2(x) + \frac{\pi^2}{16} \right)dx \\ &= \int_0^1\frac{\arctan(x) \text{arctanh}^2(x) }{1+x^2}dx + \frac{\pi^2}{16} \frac{\arctan^2(x)}{2}\Big|_0^1 \\ &= \frac{1}{4}\int_0^1 \frac{\arctan\left(\frac{1-x}{1+x} \right)\log^2(x)}{1+x^2}dx + \frac{\pi^4}{512} \quad \left(x\mapsto \frac{1-x}{1+x} \right) \\ &= \frac{1}{4}\int_0^1 \frac{\left(\frac{\pi}{4}-\arctan(x) \right)\log^2(x)}{1+x^2}dx + \frac{\pi^4}{512} \\ &= -\frac{I_1}{4} +\frac{\pi}{16}\int_0^1 \frac{\log^2(x)}{1+x^2}dx +\frac{\pi^4}{512} \\ &= -\frac{I_1}{4} + \frac{3\pi^4}{512} \end{aligned} \end{equation} \]

Here, we used the fact that \(\int_0^1 \frac{\log^2(x)}{1+x^2}dx = \frac{\pi^3}{16}\). Similarly, on \(\gamma_{5,\epsilon}\) we have

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-2} \begin{aligned} \lim_{\epsilon\to 0^+} \int_{\gamma_{5,\epsilon}} f(z) \; dz &= \int_i^0 \frac{\arctan(x)}{1+x^2}\left(\text{arctanh}^2(x) + \frac{\pi^2}{16} \right)dx \\ &= \int_0^1 \frac{\text{arctanh}(x)}{1-x^2}\left(\frac{\pi^2}{16}-\arctan^2(x) \right)dx \quad \left(x\mapsto i x \right) \\ &= \int_0^1\frac{\arctan(x) \text{arctanh}^2(x) }{1+x^2}dx \quad (\text{IBP}) \\ &= -\frac{I_1}{4}+\frac{\pi^4}{256} \end{aligned} \end{equation} \]

For the integral over \(\gamma_{3,\epsilon}\), we will take advantage of the following identities:

\[ \begin{aligned} \arctan(e^{i\theta}) &= \frac{\pi}{4}+\frac{i}{2} \log\left(\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}} \right), \quad 0\leq \theta \leq \frac{\pi}{2} \\ \text{arctanh}(e^{i\theta}) &= -\frac{1}{2}\log \left(\tan\frac{\theta}{2} \right)+\frac{i\pi}{4}, \quad 0\leq \theta \leq \frac{\pi}{2}\\ \end{aligned} \]

We have

\[ \begin{aligned} \lim_{\epsilon\to 0^+} \int_{\gamma_{3,\epsilon}}f(z) \; dz &= i \int_0^{\frac{\pi}{2}}f(e^{i\theta}) e^{i\theta} \; d\theta \\ &= \frac{i}{2} \int_0^{\frac{\pi}{2}}\frac{\left( \frac{\pi}{4}+\frac{i}{2} \log\left(\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}} \right)\right)\left( -\frac{i\pi}{4}\log \left(\tan\frac{\theta}{2} \right)+\frac{1}{4}\log^2\left(\tan \frac{\theta}{2} \right) \right)}{\cos \theta}d\theta \end{aligned} \]

The real part of the above integral is

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-3} \begin{aligned} \text{Re}\left[\lim_{\epsilon\to 0^+} \int_{\gamma_{3,\epsilon}}f(z) \; dz \right] &= \int_0^\frac{\pi}{2} \frac{\frac{1}{16}\log^2\left(\tan\frac{\theta}{2} \right)\log\left(\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}} \right)+\frac{\pi^2}{32}\log\left(\tan\frac{\theta}{2} \right)}{\cos \theta} d\theta \\ &= \int_0^1 \frac{\frac{1}{8}\log^2\left(u \right)\log\left(\frac{1-u}{1+u} \right)+\frac{\pi^2}{16}\log\left(u \right)}{1-u^2} du \quad \left(u= \tan\frac{\theta}{2} \right) \\ &= -\frac{1}{8} \int_0^1 \frac{\log^2(u)\log\left(\frac{1+u}{1-u}\right)}{1-u^2} + \frac{\pi^2}{16}\left(-\frac{\pi^2}{8} \right) \\ &= \frac{1}{16}\int_0^1 \frac{\log(u) \log^2\left(\frac{1-u}{1+u} \right)}{u}du - \frac{\pi^4}{128} \end{aligned} \end{equation} \]

Using the results from part 1, we get

\[ \begin{aligned} \int_0^1 \frac{\log(u)\log^2\left(\frac{1-u}{1+u} \right)}{u}du &= \frac{7}{4}\int_0^1 \frac{\log(u)\log^2(1-u)}{u}du + 2\int_0^1 \frac{\log(u)\log^2(1+u)}{u}du \\ &= \frac{7}{4}\left(2\zeta(4) - 2\sum_{n=1}^\infty \frac{H_n}{n^3} \right)+2\left(2\text{Li}_4(-1)+2\sum_{n=1}^\infty\frac{(-1)^{n+1} H_n}{n^3} \right) \\ &= -\frac{7\pi^4}{144}+4\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n^3} \end{aligned} \]

By Cauchy's integral theorem, the sum of \(\eqref{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-1}\), \(\eqref{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-2}\) and \(\eqref{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-3}\) is equal to zero. Therefore,

\[ \begin{aligned} &\; -\frac{I_1}{2} + \frac{5\pi^4}{512}+\frac{1}{16}\left(-\frac{7\pi^4}{144}+4 \sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n^3}\right)-\frac{\pi^4}{128} = 0 \\ &\implies I_1 = -\frac{5\pi^4}{2304}+\frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^3} \\ &\implies I_1 = -\frac{5\pi^4}{2304}+\frac{1}{2}\left(\frac{11\pi^4}{360}+\frac{\pi^2}{12}\log^2(2)-\frac{\log^4(2)}{12} -\frac{7}{4}\log(2)\zeta(3)-2\text{Li}_4\left(\frac{1}{2} \right)\right) \\ &\implies \boxed{I_1 = \frac{151 \pi ^4}{11520}+\frac{\pi ^2}{24} \log ^2(2)-\frac{\log ^4(2)}{24} -\frac{7}{8} \zeta (3) \log (2)-\text{Li}_4\left(\frac{1}{2}\right)} \end{aligned} \]

Refer to this post for the evaluation of \(\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^3}\).

Integral #6

The next integral that we'll evaluate is

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-4} I_2 = \int_0^1 \frac{\log^3(x) \arctan(x)}{1+x^2}dx \end{equation} \]

Using the transformation \(x\mapsto \frac{1}{x}\), we write the integral as:

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-5} \begin{aligned} I_2 &= -\int_1^\infty\frac{\log^3(x)\left(\frac{\pi}{2}-\arctan(x) \right)}{1+x^2}dx \end{aligned} \end{equation} \]

Adding up equations \(\eqref{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-4}\) and \(\eqref{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-5}\) and dividing both sides by 2 gives us:

\[ \begin{aligned} I_2 &= -\frac{\pi}{4}\int_1^\infty \frac{\log^3(x)}{1+x^2}dx + \frac{1}{2}\int_0^\infty \frac{\log^3(x) \arctan(x)}{1+x^2}dx\\ &= \frac{\pi}{4}\int_0^1 \frac{\log^3(x)}{1+x^2}dx + \frac{1}{2}\int_0^\infty \frac{\log^3(x) \arctan(x)}{1+x^2}dx \end{aligned} \]

Note that

\[ \begin{aligned} \int_0^1 \frac{\log^3(x)}{1+x^2}dx &= \sum_{n=0}^\infty (-1)^n \int_0^1 x^{2n}\log^3(x) \; dx \\ &= -6\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^4} \\ &= -6\beta(4) \end{aligned} \]

So, we have

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-6} I_2 = - \frac{3\pi}{2}\beta(4) + \frac{1}{2}\int_0^\infty \frac{\log^3(x) \arctan(x)}{1+x^2}dx \end{equation} \]

To evaluate the integral in the above equation, we will use the Feynman technique. Define the function \(F:[0,\infty) \to \mathbb{R}\) as \(F(s) = \int_0^\infty \frac{\log^3(x)\arctan(s x)}{1+x^2}dx\). Now, we have

\[ F'(s) = \int_0^\infty \frac{x \log^3(x)}{(1+x^2)(1+s^2 x^2)}dx \]

Before evaluating \(F'(s)\), we will evaluate the simpler integral \(\int_0^\infty \frac{x \log(x)}{(1+x^2)(1+s^2 x^2)}dx\). To do this, integrate the principal branch of \(g(z) = \frac{z\log^2(-z)}{(1+z^2)(1+s^2 z^2)}\) around the following "key hole" contour:

where:

\(\gamma_{1,\epsilon, R}\) is a line joining the points \(\sqrt{R^2-\epsilon^2}-i\epsilon\) and \(-i\epsilon\).
\(\gamma_{2,\epsilon, R}\) is a line joining the points \(i\epsilon\) and \(\sqrt{R^2-\epsilon^2}+i\epsilon\).
\(\gamma_{3,\epsilon}\) is parameterized by \(\epsilon e^{-i t}\) where \(\frac{\pi}{2} \leq t \leq \frac{3\pi}{2}\).
\(\gamma_{4,R}\) is parameterized by \(R e^{i t}\) where \(\arctan\left(\frac{\epsilon}{\sqrt{R^2-\epsilon^2}} \right) \leq t \leq 2\pi - \arctan\left(\frac{\epsilon}{\sqrt{R^2-\epsilon^2}} \right)\).

As \(\epsilon\to 0^+\) and \(R\to \infty\), the integrals along \(\gamma_{3,\epsilon}\) and \(\gamma_{4,R}\) tend to 0. Therefore, we are only left with the integrals above and below the branch cut. The residues of \(g(z)\) at it's poles is given by:

\[ \begin{aligned} \mathop{\text{Res}}\limits_{z=i} \; g(z) &= -\frac{\pi^2}{8(1-s^2)} \\ \mathop{\text{Res}}\limits_{z=-i} \; g(z) &= -\frac{\pi^2}{8(1-s^2)} \\ \mathop{\text{Res}}\limits_{z=i/s} \; g(z) &= -\frac{\left(\log(s)+\frac{i\pi}{2} \right)^2}{2(1-s^2)} \\ \mathop{\text{Res}}\limits_{z=-i/s} \; g(z) &= -\frac{\left(\log(s)-\frac{i\pi}{2} \right)^2}{2(1-s^2)} \end{aligned} \]

Now, the residue theorem gives us:

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-7} \begin{aligned} \int_0^\infty \frac{x\left((\log (x)-i\pi)^2 - (\log (x)+i\pi)^2 \right)}{(1+x^2)(1+s^2 x^2)}dx &= 2i\pi \Big(\mathop{\text{Res}}\limits_{z=i} \; g(z)+ \mathop{\text{Res}}\limits_{z=-i}\; g(z) + \mathop{\text{Res}}\limits_{z=i/s}\; g(z) \\ &\quad + \mathop{\text{Res}}\limits_{z=-i/s}\; g(z) \Big) \\ \implies -4i\pi \int_0^\infty \frac{x\log(x)}{(1+x^2)(1+s^2 x^2)}dx &= -2i\pi \frac{\log^2(s)}{1-s^2} \\ \implies \int_0^\infty \frac{x\log(x)}{(1+x^2)(1+s^2 x^2)}dx &= \frac{\log^2(s)}{2(1-s^2)} \end{aligned} \end{equation} \]

To evaluate \(F'(s)\), we integrate the principal branch of \(h(z) = \frac{z \log^4(-z)}{(1+z^2)(1+s^2 z^2)}\) around the same contour. This time, the residues are:

\[ \begin{aligned} \mathop{\text{Res}}\limits_{z=i} \; h(z) &= \frac{\pi^4}{32(1-s^2)} \\ \mathop{\text{Res}}\limits_{z=-i} \; h(z) &= \frac{\pi^4}{32(1-s^2)} \\ \mathop{\text{Res}}\limits_{z=i/s} \; h(z) &= -\frac{\left(\log(s)+\frac{i\pi}{2} \right)^4}{2(1-s^2)} \\ \mathop{\text{Res}}\limits_{z=-i/s} \; h(z) &= -\frac{\left(\log(s)-\frac{i\pi}{2} \right)^4}{2(1-s^2)} \end{aligned} \]

The residue theorem gives us:

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-8} \begin{aligned} \int_0^\infty \frac{x\left((\log (x)-i\pi)^4 - (\log (x)+i\pi)^4 \right)}{(1+x^2)(1+s^2 x^2)}dx &= 2i\pi \Big(\mathop{\text{Res}}\limits_{z=i} \; h(z)+ \mathop{\text{Res}}\limits_{z=-i}\; h(z) + \mathop{\text{Res}}\limits_{z=i/s}\; h(z) \\ &\quad + \mathop{\text{Res}}\limits_{z=-i/s}\; h(z) \Big) \\ \implies -8i\pi \int_0^\infty \frac{x\left(\log^3(x) -\pi^2 \log(x) \right)}{(1+x^2)(1+s^2 x^2)}dx &= 2i\pi \frac{\frac{3\pi^2}{2}\log^2(s)-\log^4(s)}{1-s^2} \\ \implies \int_0^\infty \frac{x\log^3(x)}{(1+x^2)(1+s^2 x^2)}dx &= \frac{\pi^2\log^2(s)+2\log^4(s)}{8(1-s^2)} \end{aligned} \end{equation} \]

Note that we used equation \(\eqref{evaluating-very-nasty-logarithmic-integrals-part-iii-eq-7}\) to the get the above result. We can now calculate our original integral as follows:

\[ \begin{aligned} I_2 &= -\frac{3\pi}{2}\beta(4) + \frac{1}{2}\int_0^1 F'(s) ds \\ &= -\frac{3\pi}{2}\beta(4) + \frac{1}{2}\int_0^1 \frac{\pi^2\log^2(s)+2\log^4(s)}{8(1-s^2)} ds \\ &= -\frac{3\pi}{2}\beta(4) + \frac{\pi^2}{16}\int_0^1\frac{\log^2(s)}{1-s^2}ds+\frac{1}{8}\int_0^1 \frac{\log^4(s)}{1-s^2} ds \\ &= -\frac{3\pi}{2}\beta(4) + \frac{\pi^2}{16}\sum_{k=0}^\infty\int_0^1 s^{2k}\log^2(s)ds+\frac{1}{8}\sum_{k=0}^\infty\int_0^1 s^{2k}\log^4(s) ds \\ &= -\frac{3\pi}{2}\beta(4) + \frac{\pi^2}{8}\sum_{k=0}^\infty\frac{1}{(2k+1)^3}+3\sum_{k=0}^\infty\frac{1}{(2k+1)^5} \\ &= \boxed{-\frac{3\pi}{2}\beta(4) + \frac{7\pi^2}{64}\zeta(3) +\frac{93}{32}\zeta(5)} \end{aligned} \]

Interestingly, \(I_2\) can be reduced into an Euler sum which can be evaluated using contour integration.

Evaluating very nasty logarithmic integrals: Part II

2020-08-06T10:27:24+00:00

We continue the evaluations from part 1, keeping the integral numbering from that post. Here the resulting Euler sums lead us toward polylogarithmic identities rather than a direct residue calculation.

Integral #3

We'll start by finding a closed form for the integral:

\[ I_1 = \int_0^1 \frac{\log^2(1+x^2)}{1+x^2}dx \]

This integral can be reduced to Euler sums like our previous problem. But this time, the resulting Euler sums cannot be evaluated using the method of residues. Therefore, we'll have to use a different approach.

Let us first consider the following integral:

\[ I_2 = \int_0^1 \frac{\log^2(1+ix)}{1+x^2}dx \]

Throughout this post, \(\log\) denotes the principal branch of the logarithmic function defined by \(\log z = \log|z| + i\text{arg}(z)\), with \(-\pi < |\text{arg}(z)| \leq \pi\). We have

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-1} \begin{aligned} I_2 &= \frac{1}{2}\int_0^1\log^2(1+ix)\left(\frac{1}{1+ix}+\frac{1}{1-ix} \right)dx \\ &= \frac{\log^3(1+ix)}{6i}\Big|_0^1 + \frac{1}{2}\int_0^1 \frac{\log^2(1+ix)}{1-ix}dx \\ &= \frac{\log^3(1+i)}{6i} + \frac{i}{2}\int_{\frac{1}{2}}^{\frac{1-i}{2}}\frac{\log^2(2(1-x))}{x}dx \\ &= \frac{\log^3(1+i)}{6i} + \frac{i}{2}\int_{\frac{1}{2}}^{\frac{1-i}{2}}\frac{\log^2(1-x)+\log^2(2)+2\log(2)\log(1-x)}{x}dx \\ &= \frac{\log^3(1+i)}{6i} + \frac{i\log^2(2)\log\left(1-i\right)}{2} + i\log(2) \left[\text{Li}_2\left(\frac{1}{2}\right)-\text{Li}_2\left(\frac{1-i}{2}\right) \right] \\ &\quad + \frac{i}{2}\int_{\frac{1}{2}}^{\frac{1-i}{2}}\frac{\log^2(1-x)}{x}dx \end{aligned} \end{equation} \]

We can use equation (2) to evaluate \(\int_{\frac{1}{2}}^{\frac{1+i}{2}}\frac{\log^2(1-x)}{x}dx\).

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-2} \begin{aligned} \int_{\frac{1}{2}}^{\frac{1-i}{2}}\frac{\log^2(1-x)}{x}dx &= \log^2\left( \frac{1+i}{2}\right)\log\left(\frac{1-i}{2}\right) + 2\log\left(\frac{1+i}{2} \right)\text{Li}_2\left(\frac{1+i}{2}\right) \\ &\quad -2\text{Li}_3\left(\frac{1+i}{2}\right) + \log^3(2) + 2\log(2)\text{Li}_2\left(\frac{1}{2}\right) + 2\text{Li}_3\left(\frac{1}{2} \right) \end{aligned} \end{equation} \]

To simplify \(\text{Li}_2\left(\frac{1+i}{2} \right)\) and \(\text{Li}_2\left(\frac{1-i}{2} \right)\), we can use the following Dilogarithm identity:

\[ \text{Li}_2(1-z) + \text{Li}_2\left(1-z^{-1} \right) = -\frac{1}{2}\log^2(z) \]

This is easy to verify by differentiating both sides of the above equation with respect to \(z\). Plugging in \(z=\frac{1+i}{2}\) gives

\[ \begin{aligned} \text{Li}_2\left(\frac{1-i}{2}\right) &= -\text{Li}_2(i) - \frac{1}{2}\log^2\left(\frac{1+i}{2}\right) = \frac{5\pi^2}{96}-\frac{\log^2(2)}{8}+i\left(-G + \frac{\pi}{8}\log(2)\right) \\ \text{Li}_2\left(\frac{1+i}{2}\right) &= \overline{\text{Li}_2\left(\frac{1-i}{2}\right)} = \frac{5\pi^2}{96}-\frac{\log^2(2)}{8}-i\left(-G + \frac{\pi}{8}\log(2)\right) \end{aligned} \]

Now, we have everything needed to simplify equation \(\eqref{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-1}\). This is a cumbersome task so I will only present the final result:

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-3} \begin{aligned} I_2 &= -\frac{3\pi^3}{128}- \frac{G \log(2) }{2} + \frac{7\pi \log^2(2)}{32} + i\Bigg( -\frac{ G \pi }{4} + \frac{7\pi^2 \log(2)}{192} + \frac{\log^3(2)}{48}+\frac{7}{8}\zeta(3) \\ &\quad - \text{Li}_3\left(\frac{1+i}{2} \right)\Bigg) \end{aligned} \end{equation} \]

To the best of my knowledge, \(\text{Li}_3\left(\frac{1+i}{2} \right)\) can not be simplified further. So, we'll leave it as it is. We can now extract \(I_1\) from the real part of \(I_2\).

\[ \begin{aligned} \text{Re }I_2 &= \int_0^1 \frac{\frac{1}{4}\log^2(1+x^2) - \arctan^2(x)}{1+x^2}dx \\ &= \frac{1}{4}\int_0^1 \frac{\log^2(1+x^2)}{1+x^2}dx - \frac{\arctan^3(x)}{3}\Big|_0^1 \\ &= \frac{1}{4}I_1 - \frac{\pi^3}{192} \end{aligned} \]

Therefore,

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-4} \boxed{I_1 = -2G\log(2) - \frac{7\pi^3}{96} + \frac{7\pi \log^2(2)}{8} + 4\text{ Im }\text{Li}_3\left(\frac{1+i}{2} \right)} \end{equation} \]

Now, let's turn our attention to another integral:

\[ I_3 = \int_0^1 \frac{\log(x)\log(1+x^2)}{1+x^2}dx \]

Notice that with the help of some algebra, we can write:

\[ I_3 = -\frac{1}{2}\int_0^1 \frac{\log^2\left(\frac{x}{1+x^2} \right)}{1+x^2}dx + \frac{1}{2}\int_0^1 \frac{\log^2(1+x^2)}{1+x^2}dx + \frac{1}{2}\int_0^1 \frac{\log^2(x)}{1+x^2}dx \]

The leftmost integral can be dealt with the trigonometric substitution \(x=\tan \theta\):

\[ \begin{aligned} \int_0^1 \frac{\log^2\left(\frac{x}{1+x^2} \right)}{1+x^2}dx &= \int_0^{\frac{\pi}{4}} \log^2\left(\sin \theta \cos\theta \right) \; d\theta \\ &= \int_0^{\frac{\pi}{4}} \log^2\left(\frac{\sin(2\theta)}{2} \right)\; d\theta \\ &= \frac{1}{2}\int_0^{\frac{\pi}{2}}\log^2\left(\frac{\sin \theta}{2} \right)\; d\theta \\ &= \frac{1}{2}\lim_{s\to 1}\frac{d^2}{ds^2}\int_0^{\frac{\pi}{2}}\left(\frac{\sin\theta}{2} \right)^{s-1}d\theta \\ &= \frac{1}{2}\lim_{s\to 1}\frac{d^2}{ds^2}\left[\frac{2^{-s}\sqrt{\pi}\Gamma\left(\frac{s}{2}\right)}{\Gamma\left(\frac{1+s}{2} \right)} \right] \\ &= \frac{\pi^3}{48}+ \pi \log^2(2) \end{aligned} \]

The middle integral has already been evaluated. As for the rightmost integral, we have:

\[ \begin{aligned} \int_0^1 \frac{\log^2(x)}{1+x^2}dx &= \sum_{n=0}^\infty (-1)^{n}\int_0^1 x^{2n}\log^2(x)\; dx \\ &= 2\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3} \\ &= \frac{\pi^3}{16} \end{aligned} \]

This gives us

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-5} \boxed{I_3 =-\frac{\pi^3}{64} -G \log(2) - \frac{\pi \log^2(2)}{16} +2 \text{ Im }\text{Li}_3\left(\frac{1+i}{2}\right)} \end{equation} \]

One can also evaluate \(I_4 = \int_0^1\frac{\log(1+x^2)\arctan(x)}{x}dx\) by noting that

\[ I_4 = \text{Im}\int_0^1 \frac{\log^2(1+ix)}{x}dx = \text{Im}\int_0^{-i}\frac{\log^2(1-x)}{x}dx \]

and using equation (3). The end result is:

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-6} \boxed{I_4 =-\frac{3\pi^3}{64}+G\log(2)-\frac{\pi \log^2(2)}{16}+2\text{ Im }\text{Li}_3\left(\frac{1+i}{2}\right) } \end{equation} \]

Using \(I_3\) and \(I_4\), we can evaluate \(I_5 = \int_0^1 \frac{\log(x)\arctan(x)}{x(1+x^2)}dx\) as follows:

\[ \begin{aligned} I_5 &= \int_0^1 \log(x)\arctan(x)\left(\frac{1}{x}-\frac{x}{1+x^2} \right) dx \\ &= \int_0^1 \frac{\log(x)\arctan(x)}{x}dx - \int_0^1 \frac{x\log(x)\arctan(x)}{1+x^2}dx \\ &= -\frac{1}{2}\int_0^1 \frac{\log^2(x)}{1+x^2}dx + \frac{1}{2} \int_0^1 \frac{\log(x)\log(1+x^2)}{1+x^2} dx + \frac{1}{2}\int_0^1 \frac{\log(1+x^2)\arctan(x)}{x}dx \quad (\text{IBP}) \\ &= -\frac{\pi^3}{32} + \frac{I_3 + I_4}{2} \\ &= -\frac{\pi^3}{16} - \frac{\pi \log^2(2)}{16} + 2\text{ Im }\text{Li}_3\left(\frac{1+i}{2}\right) \end{aligned} \]

On the other hand, we have:

\[ \begin{aligned} I_5 &= \int_0^1 \log(x) \left(\sum_{n=0}^\infty (-1)^n \tilde{H}_n x^{2n} \right) dx\\ &= \sum_{n=0}^\infty (-1)^n \tilde{H}_n \int_0^1 x^{2n}\log(x) \; dx \\ &= -\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2} \end{aligned} \]

where \(\tilde{H}_n = \sum_{j=0}^n \frac{1}{2j+1}\). This gives us an interesting Euler sum:

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-7} \boxed{\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2} = \frac{\pi^3}{16} + \frac{\pi \log^2(2)}{16} - 2\text{ Im }\text{Li}_3\left(\frac{1+i}{2}\right)} \end{equation} \]

Of course, one can proceed in a similar manner to create more crazy integrals. The following problem is left as an exercise for the reader.

Exercise 1: Using the method of residues, show that

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-8} \sum_{n=0}^\infty \frac{(-1)^n\tilde{H}_n}{2n+1} = \frac{G}{2}+\frac{\pi \log(2)}{8} \end{equation} \]

Integral #4

Many years ago, I encountered the following integral:

\[ I_6 = \int_0^1 \frac{x \arctan(x)\log(1-x^2)}{1+x^2}dx \]

At that time, I couldn't find a solution to this problem. Hence, I ended up asking it on math.stackexchange.com. The answers that I received there involved evaluating complex logarithmic integrals by brute force. Recently, I discovered a much simpler way to solve it using the method of residues.

Let's start by breaking down \(I_6\) into Euler sums.

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-9} \begin{aligned} I_6 &= \int_0^1 x\log(1-x^2) \left(\sum_{n=0}^\infty (-1)^n \tilde{H}_n x^{2n+1} \right)dx \\ &= \sum_{n=0}^\infty (-1)^n \tilde{H}_n \int_0^1 x^{2n+2}\log(1-x^2) dx \\ &= \sum_{n=0}^\infty (-1)^{n+1} \tilde{H}_n \left(\frac{\psi_0\left(n+\frac{5}{2} \right)+\gamma}{2n+3} \right) \\ &= \sum_{n=0}^\infty (-1)^{n+1}\left(\tilde{H}_{n+1}-\frac{1}{2n+3} \right)\left(\frac{-2\log(2)+2\tilde{H}_{n+1}}{2n+3} \right) \\ &= \sum_{n=0}^\infty (-1)^n \left(\tilde{H}_{n}-\frac{1}{2n+1} \right)\left(\frac{-2\log(2)+2\tilde{H}_{n}}{2n+1} \right) \\ &= -2\log(2)\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{2n+1}+2G\log(2) + 2\sum_{n=0}^\infty \frac{(-1)^n (\tilde{H}_n)^2}{2n+1} -2\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2} \\ &= G\log(2) - \frac{\pi \log^2(2)}{4} + 2\sum_{n=0}^\infty \frac{(-1)^n (\tilde{H}_n)^2}{2n+1} -2\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2} \end{aligned} \end{equation} \]

The result of exercise 1 was used in the last step.

Now, integrate the function \(f(z) = \pi\csc(\pi z) \frac{\left( \gamma + \psi_0 \left(-z+\frac{3}{2} \right)\right)^2}{-2z+1}\) over the positively oriented square, \(C_N\), with vertices \(\pm \left(N+\frac{1}{4}\right)\pm i\left(N+\frac{1}{4} \right)\). It takes a bit of effort to see that

\[ \lim_{N\to \infty}\int_{C_N} f(z)\; dz = 0 \]

This implies that the sum of residues of \(f(z)\) at it's poles is equal to \(0\). We have

\[ \begin{aligned} \mathop{\text{Res}}\limits_{z=-n} f(z) &= (-1)^n \frac{\left(\gamma +\psi_0\left(n+\frac{3}{2}\right) \right)^2}{2n+1} = (-1)^n \frac{\left(-2\log(2)+2\tilde{H}_n \right)^2}{2n+1}, \quad n\in\{0,1,2,\cdots\} \\ \mathop{\text{Res}}\limits_{z=n} f(z) &= (-1)^{n-1} \frac{\left(\gamma+\psi_0\left(-n+\frac{3}{2} \right) \right)^2}{2n-1} = (-1)^{n-1} \frac{\left(-2\log(2)+2\tilde{H}_{n-1} -\frac{2}{2n-1}\right)^2}{2n-1}, \quad n\in\{1,2,3,\cdots\} \\ \mathop{\text{Res}}\limits_{z=\frac{2n+1}{2}} f(z) &= (-1)^{n-1} \frac{\pi H_n}{n} - (-1)^{n-1} \frac{3\pi}{2n^2}, \quad n\in\{1,2,3,\cdots\} \end{aligned} \]

Summing up the residues and performing some algebraic simplifications gives:

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-10} \begin{aligned} &\; 8\sum_{n=0}^\infty \frac{(-1)^n (\tilde{H}_n)^2}{2n+1}-8\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2}-16\log(2)\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{2n+1}+ 4\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3} \\ &\quad + 8\log(2)\sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)^2} + \pi\sum_{n=1}^\infty \frac{(-1)^{n-1}H_n}{n} - \frac{3\pi}{2}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} = 0 \\ &\implies 8\left(\sum_{n=0}^\infty \frac{(-1)^n (\tilde{H}_n)^2}{2n+1}-\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2} \right) + \frac{\pi^3}{8} + \pi \left(\frac{\pi^2}{12}-\frac{\log^2(2)}{2}\right)-\frac{3\pi}{2}\left(\frac{\pi^2}{12} \right) = 0 \\ &\implies \sum_{n=0}^\infty \frac{(-1)^n (\tilde{H}_n)^2}{2n+1}-\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+1)^2} = -\frac{\pi^3}{96}+\frac{\pi \log^2(2)}{16} \end{aligned} \end{equation} \]

In the above calculation, we used the result of exercise 1 and that

\[ \sum_{n=1}^\infty \frac{(-1)^{n-1}H_n}{n} = \frac{\pi^2}{12}-\frac{\log^2(2)}{2} \]

This follows from (A). Finally, plugging equation \(\eqref{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-10}\) into \(\eqref{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-9}\), gives

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-ii-eq-11} \boxed{I_6 = -\frac{\pi^3}{48}-\frac{\pi}{8}\log^2 (2) +G\log (2)} \end{equation} \]

Evaluating very nasty logarithmic integrals: Part I

2020-08-01T09:15:41+00:00

Logarithmic integrals often become more manageable when we express them as Euler sums. This first part develops that approach and then introduces contour integration as another way to evaluate the resulting sums. We will use identities from An alternating Euler sum (A) and Euler sums involving squared harmonic numbers (B).

Integral #1

Our first integral is the following:

\[ \begin{aligned} I &= \int_0^1 \frac{\log(t)\log(1-t)\log(1+t)}{t}dt \end{aligned} \]

We will use the integral \(J = \int_0^1 \frac{\log(t)\log^2(1-t)}{t}dt\) as a starting point. Make the substitution \(t\mapsto t^2\) in \(J\) to obtain:

\[ \begin{aligned} J &= 2\int_0^1 \frac{\log(t^2) \log^2(1-t^2)}{t}dt \\ &= 4\int_0^1 \frac{\log(t)\log^2(1-t)}{t}dt + 4\int_0^1 \frac{\log(t)\log^2(1+t)}{t}dt + 8\int_0^1 \frac{\log(t)\log(1-t) \log(1+t)}{t}dt \\ &= 4J + 4\int_0^1 \frac{\log(t)\log^2(1+t)}{t}dt + 8I \end{aligned} \]

Solving the above equation for \(I\) gives us:

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-i-eq-1} I = -\frac{3}{8}J - \frac{1}{2} \int_0^1 \frac{\log(t)\log^2(1+t)}{t}dt \end{equation} \]

Applying integration by parts and using the generating function of the Harmonic number yields:

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-i-eq-2} \begin{aligned} \int_0^1 \frac{\log(t)\log^2(1+t)}{t}dt &= -\int_0^1 \frac{\log^2(t)\log(1+t)}{1+t}dt \\ &= -\sum_{n=1}^\infty (-1)^{n+1} H_n \int_0^1 t^{n} \log^2(t) \; dt \\ &= -2\sum_{n=1}^\infty (-1)^{n+1} \frac{H_n}{(n+1)^3} \\ &= -2\sum_{n=0}^\infty (-1)^{n+1} \frac{H_{n+1}-\frac{1}{n+1}}{(n+1)^3} \\ &= 2\sum_{n=1}^\infty (-1)^{n+1} \frac{H_n}{n^3} - 2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^4} \\ &= 2\text{Li}_4(-1) + 2\sum_{n=1}^\infty (-1)^{n+1} \frac{H_n}{n^3} \end{aligned} \end{equation} \]

A similar calculation shows that:

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-i-eq-3} J = 2\zeta(4) - 2\sum_{n=1}^\infty \frac{H_n}{n^3} \end{equation} \]

From (A), we know how to evaluate \(\sum_{n=1}^\infty (-1)^{n+1} \frac{H_n}{n^3}\) and \(\sum_{n=1}^\infty \frac{H_n}{n^3}\).

\[ \begin{aligned} \sum_{n=1}^\infty \frac{H_n}{n^3} &= \frac{\pi^4}{72} \\ \sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^3} &= \frac{11\pi^4}{360}+\frac{\pi^2}{12}\log^2(2)-\frac{\log^4(2)}{12}-\frac{7}{4}\log(2)\zeta(3) -2\text{Li}_4\left(\frac{1}{2}\right) \end{aligned} \]

Finally, putting everything together gives us:

\[ \boxed{I = -\frac{3\pi^4}{160}-\frac{\pi^2}{12}\log^2(2)+\frac{\log^4(2)}{12}+\frac{7}{4}\log(2)\zeta(3) +2\text{Li}_4\left(\frac{1}{2}\right)} \]

Integral #2

The next integral on our list is

\[ K = \int_0^1 \frac{\log(x)\log(1+x^2)\arctan(x)}{x}dx \]

This integral was originally posted by the user FDP on math.stackexchange.com. My solution is posted there as well.

Using integration by parts gives us:

\[ K = - \frac{1}{2}\int_0^1 \frac{\log^2(x)\log(1+x^2)}{1+x^2}dx - \int_0^1 \frac{x \log^2(x) \arctan(x)}{1+x^2}dx \]

Now, we can use the following series expansions to reduce \(K\) into Euler sums:

\[ \begin{aligned} \frac{\arctan (x)}{1+x^2} &= \sum_{n=0}^\infty (-1)^n \tilde{H}_n x^{2n+1} , \quad |x| < 1\\ \frac{\log(1+x^2)}{1+x^2} &= \sum_{n=1}^\infty (-1)^{n+1} H_n x^{2n} , \quad |x| < 1 \end{aligned} \]

where \(\tilde{H}_n = \sum_{i=0}^n \frac{1}{2i+1}\). This gives us:

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-i-eq-4} \begin{aligned} K &= -\sum_{n=0}^\infty (-1)^n \tilde{H}_n\int_0^1 x^{2n+2} \log^2(x)\; dx-\frac{1}{2}\sum_{n=1}^\infty (-1)^{n+1} H_n\int_0^1 x^{2n}\log^2(x)\; dx \\ &= -2\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+3)^3} - \sum_{n=1}^\infty \frac{(-1)^{n+1} H_n}{(2n+1)^3} \\ &= -2\sum_{n=1}^\infty \frac{(-1)^{n+1} H_{2n}}{(2n+1)^3} \end{aligned} \end{equation} \]

We'll employ the contour integration method to evaluate the above Euler sum. It is a very powerful tool that can handle a large class of Euler sums (see, for e.g. Euler sums and contour integral representations by Philippe Flajolet and Bruno Salvy).

We'll integrate the function \(f(z) = \pi \csc(\pi z) \frac{\gamma+\psi_0(-2z+1)}{(-2z+1)^3}\) around the positively oriented square, \(C_N\), with vertices \(\pm \left(N+\frac{1}{4} \right)\pm \left(N+\frac{1}{4} \right)i\). It is easy to see that

\[ \lim_{N\to \infty}\int_{C_N}f(z)\; dz = 0 \]

Hence, the sum of all residues of \(f(z)\) at its poles is equal to \(0\). A straightforward computation shows that the residues are:

\[ \begin{aligned} \mathop{\text{Res}}\limits_{z=-n} f(z) &= (-1)^n \frac{H_{2n}}{(2n+1)^3} , \quad n\in \{0,1,2,\cdots\} \\ \mathop{\text{Res}}\limits_{z=\frac{2n+1}{2}} f(z) &= \frac{(-1)^{n+1} \pi}{16 n^3} , \quad n\in \{1,2,3,\cdots\} \\ \mathop{\text{Res}}\limits_{z=n} f(z) &= \frac{(-1)^{n+1}H_{2n-1}}{(2n-1)^3}- 3\frac{(-1)^{n+1}}{(2n-1)^4}, \quad n\in \{1,2,3,\cdots \}\\ \mathop{\text{Res}}\limits_{z=\frac{1}{2}} f(z) &= \frac{\pi \zeta(3)}{2} \end{aligned} \]

The list of local expansions of basic kernels given on page 6 of the above mentioned paper are quite useful for carrying out these computations. Now, adding up all the residues gives us:

\[ \begin{equation}\label{evaluating-very-nasty-logarithmic-integrals-part-i-eq-5} \begin{aligned} \frac{\pi \zeta(3)}{2}+\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{(2n+1)^3} + \frac{\pi}{16}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3} + \sum_{n=1} ^\infty \frac{(-1)^{n+1}H_{2n-1}}{(2n-1)^3} -3\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-1)^4}&= 0\\ \implies \frac{\pi \zeta(3)}{2}+\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{(2n+1)^3} + \frac{\pi}{16}\left(\frac{3\zeta(3)}{4} \right) + \sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{(2n+1)^3} -2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-1)^4} &= 0 \\ \implies -2\sum_{n=1}^\infty \frac{(-1)^{n+1}H_{2n}}{(2n+1)^3} + \frac{35\pi \zeta(3)}{64} -2\beta(4) = 0 \\ \implies \boxed{\sum_{n=1}^\infty \frac{(-1)^{n+1}H_{2n}}{(2n+1)^3} = -\beta(4) + \frac{35\pi \zeta(3)}{128}} \end{aligned} \end{equation} \]

where \(\beta(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}\) is the Dirichlet beta function. Finally, plugging equation \(\eqref{evaluating-very-nasty-logarithmic-integrals-part-i-eq-5}\) into \(\eqref{evaluating-very-nasty-logarithmic-integrals-part-i-eq-4}\) gives us:

\[ \boxed{K = 2\beta(4) - \frac{35\pi \zeta(3)}{64}} \]

Following are some more examples of Euler sums that can be evaluated using the contour integration method:

\[ \begin{aligned} \sum_{n=0}^\infty\frac{(-1)^n\psi_2(n+1)}{2n+1} &= 8\beta(4)+\frac{\pi^2}{3}G-\frac{7\pi}{2}\zeta(3) \\ \sum_{n=0}^\infty\frac{(-1)^n\psi_1(n+1)}{(2n+1)^2} &= 6\beta(4)+\frac{\pi^2}{4}G-\frac{7\pi}{4}\zeta(3) \\ \sum_{n=1}^\infty \frac{(-1)^{n+1} H_n}{(2n+1)^3} &= -3\beta(4)+\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\log(2) \end{aligned} \]

where \(G=\beta(2)\) denotes the Catalan's constant. Some integrals that can be evaluated with the above Euler sums are:

\[ \begin{aligned} \int_0^1 \frac{\log^2(x)\arctan(x)}{x\left(1-x^2 \right)}dx &=\beta(4)+\frac{7\pi \zeta(3)}{64}+\frac{\pi^3 \log(2)}{32} \\ \int_0^1\frac{\log(x)\arctan(x)\text{arctanh}(x)}{x}dx &= \frac{\pi^2}{16}G-\frac{7\pi\zeta(3)}{32} \\ \int_0^{\frac{\pi}{2}}\frac{x}{\sin x}\log^2\left(\frac{1+\cos x-\sin x}{1+\cos x+\sin x}\right)dx &= \frac{\pi^2}{6}G +4\beta(4) \end{aligned} \]

Euler sums involving squared harmonic numbers

2020-07-23T17:33:33+00:00

The previous post on Euler sums evaluated sums containing \(H_n\) and \(H_n^{(2)}\). We now turn to squares of harmonic numbers. Their generating function leads to the integral \(\int_0^x \frac{\log^3(1-t)}{t}dt\), from which several further evaluations follow.

A generating function for squared harmonic numbers

Our starting point is the identity

\[ \sum_{n=1}^\infty (H_n)^2 x^n = \frac{\log^2(1-x)+\text{Li}_2(x)}{1-x} ,\quad -1\leq x < 1 \]

This can be derived by plugging \(H_n = \int_0^1\frac{1-t^n}{1-t}dt\) and interchanging the sum and the integral. We can rewrite the above equation using the fact that \(\sum_{n=1}^\infty H_n^{(2)} x^n = \frac{\text{Li}_2(x)}{1-x}\).

\[ \begin{equation}\label{euler-sums-involving-square-of-harmonic-numbers-eq-1} \sum_{n=1}^\infty (H_n)^2 x^n = \frac{\log^2(1-x)}{1-x} + \sum_{n=1}^\infty H_n^{(2)} x^n , \quad -1\leq x < 1 \end{equation} \]

1. Evaluation of \(\int_0^x \frac{\log^n(1-t)}{t}dt\)

In this section, we will derive a formula for the integral \(\int_0^x \frac{\log^n(1-t)}{t}dt\) where \(n\) is a positive integer. First, we'll consider the case when \(0\leq x < 1\). We have:

\[ \begin{equation}\label{euler-sums-involving-square-of-harmonic-numbers-eq-2} \begin{aligned} \int_0^x \frac{\log^n (1-t)}{t}dt &= \int_0^{-\log(1-x)} \frac{t^n e^{-t}}{1-e^{-t}}dt \quad (t\mapsto 1-e^{-t}) \\ &= \sum_{j=1}^\infty \int_0^{-\log(1-x)}t^n e^{-jt} dt \\ &= -\sum_{j=1}^\infty \left[e^{-jt}\sum_{i=0}^n \frac{(-1)^{n-i}\log^{n-i}(1-x)}{j^{i+1}}n^{\underline{i}} \right]_0^{-\log(1-x)} \\ &= - \sum_{j=1}^\infty \left( (1-x)^j \sum_{i=0}^n \frac{(-1)^{n-i}\log^{n-i}(1-x)}{j^{i+1}}n^{\underline{i}}-\frac{n!}{j^{n+1}} \right) \\ &= n! \zeta(n+1) + \sum_{i=0}^n (-1)^{n-i+1} n^{\underline{i}}\log^{n-i}(1-x) \text{Li}_{i+1}(1-x) \end{aligned} \end{equation} \]

A similar calculation shows that for \(-1\leq x < 0\), we have:

\[ \begin{equation}\label{euler-sums-involving-square-of-harmonic-numbers-eq-3} \begin{aligned} \int_x^0 \frac{\log^n(1-t)}{t}dt &= -\frac{\log^{n+1}(1-x)}{n+1} - n! \zeta(n+1) + \sum_{i=0}^{n} n^{\underline{i}}\log^{n-i}(1-x)\text{Li}_{i+1}\left(\frac{1}{1-x}\right) \end{aligned} \end{equation} \]

2. Sums with \((H_n)^2\)

We can divide equation \(\eqref{euler-sums-involving-square-of-harmonic-numbers-eq-1}\) by \(x\) and integrate both sides to get some interesting results.

\[ \begin{align} \sum_{n=1}^\infty \frac{(H_n)^2}{n}x^n &= -\frac{\log^3(1-x)}{3}-\log(1-x)\text{Li}_2(x)+\text{Li}_3(x), \quad -1\leq x < 1\label{euler-sums-involving-square-of-harmonic-numbers-eq-4}\\ \sum_{n=1}^\infty \frac{(H_n)^2}{n^2}x^n &= \text{Li}_4(x) + \frac{\text{Li}_2^2(x)}{2}-\frac{1}{3}\int_0^x \frac{\log^3(1-t)}{t}dt, \quad -1\leq x \leq 1\label{euler-sums-involving-square-of-harmonic-numbers-eq-5} \end{align} \]

Equation \(\eqref{euler-sums-involving-square-of-harmonic-numbers-eq-4}\) was derived using results from section 4 of this post.

One can now plug in \(x=-1\) in equation \(\eqref{euler-sums-involving-square-of-harmonic-numbers-eq-5}\) to get:

\[ \boxed{\sum_{n=1}^\infty \frac{(H_n)^2}{n^2}(-1)^{n+1} = \frac{41\pi^4}{1440} + \frac{\pi^2 \log^2(2)}{12}-\frac{\log^4(2)}{12}-\frac{7}{4}\log(3)\zeta(3)-2\text{Li}_4\left(\frac{1}{2}\right)} \]

Of course, equation \(\eqref{euler-sums-involving-square-of-harmonic-numbers-eq-3}\) was used to evaluate \(\int_{-1}^0 \frac{\log^3(1-t)}{t}dt\).

A contour integration approach to infinite series

2020-07-14T20:12:16+00:00

We will evaluate the trigonometric series \(\sum_{n=0}^\infty \frac{\cot\left(\frac{2n+1}{2}\pi\sqrt{2} \right)}{(2n+1)^3}\) by integrating a meromorphic function around a sequence of expanding squares. The residues encode the desired sum, while equidistribution helps us choose a useful subsequence of contours. Define the meromorphic function

\[ f(z) = \frac{\pi \tan(\pi z)\tan(\pi z \theta)}{z^3} \]

where the parameter \(\theta\) is a positive irrational number. Let \(C_N\) denote the positively oriented square with vertices \((N+1)(1+i),\; (N+1)(-1+i),\; (N+1)(-1-i)\) and \((N+1)(1-i)\). The irrationality of \(\theta\) ensures that no pole on the real axis lies on the boundary of this square.

Estimating the contour integral

Put \(m=N+1\). We will estimate the horizontal and vertical sides separately. We use the elementary identities

\[ |\tan(u+iv)|^2 =\frac{\sin^2u+\sinh^2v}{\cos^2u+\sinh^2v} \]

and, whenever \(\tan u\) is finite,

\[ |\tan(u+iv)|^2 =\frac{\tan^2u+\tanh^2v} {1+\tan^2u\,\tanh^2v}. \]

The first identity, including its limiting cases when \(\cos u=0\), shows that for \(v\neq0\),

\[ \tanh|v|\leq|\tan(u+iv)|\leq\coth|v|, \]

and, if \(u\) is fixed while \(v\) varies,

\[ |\tan(u+iv)|\leq\max\{1,|\tan u|\}. \]

For the second bound, put \(a=\tan^2u\) and \(s=\tanh^2v\). Then the square of the modulus is \((a+s)/(1+as)\), which lies between \(a\) and \(1\) as \(s\) ranges from \(0\) to \(1\). This proves the claimed estimate.

On either horizontal side, write \(z=x\pm im\) with \(-m\leq x\leq m\). Since \(|z|\geq m\), we have

\[ |\tan(\pi z)|\leq\coth(\pi m), \qquad |\tan(\pi\theta z)|\leq\coth(\pi\theta m). \]

The combined length of the two horizontal sides is \(4m\). Therefore,

\[ \int_{\text{horizontal sides}}|f(z)|\,|dz| \leq \frac{4\pi}{m^2}\coth(\pi m)\coth(\pi\theta m). \]

On either vertical side, write \(z=\pm m+iy\) with \(-m\leq y\leq m\). Because \(m\) is an integer,

\[ |\tan(\pi z)|=|\tan(\pm m\pi+i\pi y)| =|\tanh(\pi y)|\leq1. \]

Set

\[ T_m:=|\tan(\pi m\theta)|. \]

The second consequence of the tangent identity gives

\[ |\tan(\pi\theta(\pm m+iy))| \leq\max\{1,T_m\}. \]

Again \(|z|\geq m\), and the two vertical sides have combined length \(4m\), so

\[ \int_{\text{vertical sides}}|f(z)|\,|dz| \leq\frac{4\pi}{m^2}\max\{1,T_m\}. \]

Combining the four sides yields the explicit estimate

\[ \begin{aligned} \left|\int_{C_N} f(z)\,dz\right| &\leq\int_{C_N}|f(z)|\,|dz|\\ &\leq\frac{4\pi}{m^2} \left( \max\{1,T_m\} +\coth(\pi m)\coth(\pi\theta m) \right). \end{aligned} \]

Choosing the contours by equidistribution

It remains to keep \(T_m\) away from the poles of the tangent. By Weyl's equidistribution theorem, the fractional parts \(\{m\theta\}\) are equidistributed in \([0,1)\). In particular,

\[ \lim_{M\to\infty} \frac{1}{M} \#\left\{1\leq m\leq M:\{m\theta\}\in \left[0,\frac14\right]\right\} =\frac14. \]

Thus there are infinitely many integers \(m\) for which \(\{m\theta\}\in[0,1/4]\). Enumerate them as an increasing sequence \(m_1<m_2<\cdots\). Since tangent has period \(\pi\),

\[ T_{m_k} =|\tan(\pi\{m_k\theta\})| \leq1. \]

Let \(N_k=m_k-1\). Substituting this bound into the contour estimate gives

\[ \left|\int_{C_{N_k}}f(z)\,dz\right| \leq\frac{4\pi}{m_k^2} \left(1+\coth(\pi m_k)\coth(\pi\theta m_k)\right). \]

Because \(m_k\to\infty\) and both hyperbolic cotangents tend to \(1\), the expression on the right is asymptotic to \(8\pi/m_k^2\). Hence

\[ \lim_{k\to \infty}\int_{C_{N_k}} f(z)\; dz = 0. \]

On the other hand, the residue theorem gives

\[ \frac{1}{2\pi i}\int_{C_{N_k}} f(z)\,dz = \substack{\displaystyle \operatorname{Res} \\ z=0}f(z) + \sum_{|i+\frac12|<m_k} \substack{\displaystyle \operatorname{Res} \\ z=\frac{2i+1}{2}}f(z) + \sum_{|j+\frac12|<\theta m_k} \substack{\displaystyle \operatorname{Res} \\ z=\frac{2j+1}{2\theta}}f(z). \]

A simple calculation gives the following residues:

\[ \begin{aligned} \substack{\displaystyle \text{Res} \\ z=0}f(z) &= \pi^3 \theta \\ \substack{\displaystyle \text{Res} \\ z=\frac{2n+1}{2}}f(z) &= -8\frac{\tan\left(\frac{\pi \theta}{2}(2n+1) \right)}{(2n+1)^3}, \quad n\in \{0,1,2,\cdots\}\\ \substack{\displaystyle \text{Res} \\ z=\frac{2n+1}{2\theta}}f(z) &= -8\theta^2 \frac{\tan\left(\frac{\pi}{2\theta}(2n+1) \right)}{(2n+1)^3} , \quad n\in \{0,1,2,\cdots\} \end{aligned} \]

The nonzero poles occur in pairs \(z\) and \(-z\), and the two residues in each pair are equal. Letting \(k\to\infty\) in the residue identity therefore gives the following relation, provided the two series converge:

\[ \sum_{n=0}^\infty \frac{\tan\left(\frac{\pi \theta}{2}(2n+1) \right)}{(2n+1)^3} + \theta^2 \sum_{n=0}^\infty \frac{\tan\left(\frac{\pi}{2\theta}(2n+1) \right)}{(2n+1)^3} = \frac{\pi^3 \theta}{16} \]

Equidistribution was used only to choose contours on which the boundary integral is small; by itself it does not guarantee convergence of the two series. For the value needed here, \(\theta=\sqrt2+1\), convergence presents no problem. A quadratic irrational is badly approximable, so there is a constant \(c>0\) such that

\[ |q\theta-p|\geq\frac{c}{q} \]

for all integers \(p\) and positive integers \(q\). The same is true of \(1/\theta=\sqrt2-1\). Consequently, the tangent factors in the two series are \(O(2n+1)\), even when their arguments approach a pole. After division by \((2n+1)^3\), the summands are \(O((2n+1)^{-2})\), and both series converge absolutely.

Finally, substitute \(\theta=\sqrt2+1\). Since \(2n+1\) is odd and \(1/\theta=\sqrt2-1\), periodicity gives

\[ \tan\left(\frac{\pi\theta}{2}(2n+1)\right) =\tan\left(\frac{\pi}{2\theta}(2n+1)\right) =-\cot\left(\frac{\pi\sqrt2}{2}(2n+1)\right). \]

Moreover,

\[ \frac{\theta}{1+\theta^2} =\frac{1}{\theta+\theta^{-1}} =\frac{1}{2\sqrt2}. \]

The relation above therefore reduces to

\[ \boxed{\sum_{n=0}^\infty \frac{\cot\left(\frac{2n+1}{2}\pi\sqrt{2} \right)}{(2n+1)^3} = -\frac{\pi^3}{32\sqrt{2}}} \]

An alternating Euler Sum

2020-07-05T19:55:22+00:00

Our goal is an alternating Euler sum involving harmonic numbers. The route to its evaluation passes through a generating function, logarithmic integrals, and polylogarithmic identities:

\[ \sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^3} \]

where \(H_n = \sum_{k=1}^n \frac{1}{k}=\int_0^1 \frac{1-t^n}{1-t}dt\) is the \(n\)-th harmonic number. This series resists contour integration techniques which makes its computation quite challenging. We will start with the following well known generating function identity:

\[ \sum_{n=1}^\infty H_n t^n = -\frac{\log(1-t)}{1-t} ,\quad -1 \leq t < 1 \]

Dividing the above equation by \(t\) and integrating both sides, gives us:

\[ \begin{aligned} \sum_{n=1}^\infty H_n \frac{x^n}{n} &= -\int_0^x\frac{\log(1-t)}{t(1-t)}dt \\ &= -\int_0^x \left(\frac{1}{t} + \frac{1}{1-t} \right)\log(1-t) \; dt \\ &= \text{Li}_2(x) + \frac{1}{2}\log^2(1-x) \end{aligned} \]

where \(-1\leq x < 1\).

1. Computation of \(\sum_{n=1}^\infty \frac{H_n}{n^2}x^n\)

Now, we will use the same process to evaluate \(\sum_{n=1}^\infty \frac{H_n}{n^2}x^n\) but we'll have to divide the evaluation into two parts.

(I) First consider the case when \(0 \leq x < 1\). We don't need this for evaluating the original series but I'll do it anyway for completeness. We have:

\[ \begin{equation}\label{an-alternating-euler-sum-eq-1} \begin{aligned} \sum_{n=1}^\infty \frac{H_n}{n^2} x^n &= \int_0^x\left(\frac{\text{Li}_2(t)}{t} + \frac{1}{2}\frac{\log^2(1-t)}{t}\right)dt \\ &= \text{Li}_3(x) + \frac{1}{2}\int_0^x \frac{\log^2(1-t)}{t} dt \\ &= \text{Li}_3(x) + \frac{1}{2}\int_{\log(1-x)}^0 \frac{t^2 e^t}{1-e^t}dt \quad (t\mapsto 1-e^t) \\ &= \text{Li}_3(x) + \frac{1}{2}\sum_{n=1}^\infty \int_{\log(1-x)}^0 t^2 e^{nt} \; dt \\ &= \text{Li}_3(x) + \frac{1}{2}\sum_{n=1}^\infty \left(\frac{2}{n^3} -\frac{\log^2(1-x)}{n}(1-x)^n+\frac{2\log(1-x)}{n^2}(1-x)^n-2\frac{(1-x)^n}{n^3}\right) \\ &= \text{Li}_3(x) + \frac{\log^2(1-x)\log(x)}{2} + \log(1-x)\text{Li}_2(1-x) - \text{Li}_3(1-x) + \zeta(3) \end{aligned} \end{equation} \]

(II) Similarly, for the case \(-1\leq x < 0\), we obtain:

\[ \begin{equation}\label{an-alternating-euler-sum-eq-2} \begin{aligned} \sum_{n=1}^\infty \frac{H_n}{n^2}x^n &= \text{Li}_3(x) -\frac{1}{2}\int_x^0 \frac{\log^2(1-t)}{t}dt \\ &= \text{Li}_3(x) +\frac{1}{2}\int_{0}^{-x} \frac{\log^2(1+t)}{t}dt \quad (t\mapsto -t) \\ &= \text{Li}_3(x) +\frac{1}{2}\int_{0}^{\log(1-x)} \frac{t^2 e^t}{e^t-1}dt \quad (t\mapsto e^t-1) \\ &= \text{Li}_3(x) +\frac{1}{2} \sum_{n=0}^\infty \int_0^{\log(1-x)} t^2 e^{-nt} dt \\ &= \text{Li}_3(x) +\frac{\log^3(1-x)}{6} + \frac{1}{2}\sum_{n=1}^\infty \int_0^{\log(1-x)} t^2 e^{-nt} dt \\ &= \text{Li}_3(x) +\frac{\log^3(1-x)}{6} + \frac{1}{2}\sum_{n=1}^\infty \left(\frac{2}{n^3} -\log^2(1-x)\frac{(1-x)^{-n}}{n}-2\log(1-x)\frac{(1-x)^{-n}}{n^2}-2\frac{(1-x)^{-n}}{n^3}\right) \\ &= \text{Li}_3(x) -\frac{\log^3(1-x)}{3} +\frac{\log^2(1-x)\log(-x)}{2} -\log(1-x)\text{Li}_2\left( \frac{1}{1-x}\right)-\text{Li}_3\left(\frac{1}{1-x}\right)+\zeta(3) \end{aligned} \end{equation} \]

2. Evaluation of \(\sum_{n=1}^\infty \frac{H_n}{n^3}x^n\)

This is the hardest step in the evaluation. Once again, we will divide the evaluation into two parts.

(I) Similar to section 1, we'll first consider the case when \(0\leq x < 1\).

Using integration by parts, we obtain:

\[ \begin{equation}\label{an-alternating-euler-sum-eq-3} \begin{aligned} \int_0^x \frac{\zeta(3)-\text{Li}_3(1-t)}{t}dt &= \log(t)\left(\zeta(3)-\text{Li}_3(1-t) \right) \Big|_{0}^x - \int_0^x \frac{\log(t)\text{Li}_2(1-t)}{1-t}dt \\ &= \log(x)\left(\zeta(3)-\text{Li}_3(1-x) \right) - \frac{1}{2}\text{Li}_2^2 (1-x) + \frac{\zeta^2 (2)}{2} \end{aligned} \end{equation} \]

\[ \begin{equation}\label{an-alternating-euler-sum-eq-4} \begin{aligned} \int_0^x \frac{\log(1-t)\text{Li}_2(1-t)}{t}dt &= \log(t)\log(1-t)\text{Li}_2(1-t) \Big|_0^x - \int_0^x \log(t) \left(\frac{-\text{Li}_2(1-t)}{1-t} + \frac{\log(1-t)\log(t)}{1-t}\right)dt \\ &= \log(x)\log(1-x)\text{Li}_2(1-x) + \frac{1}{2}\text{Li}_2^2(1-x)-\frac{\zeta^2(2)}{2} - \int_0^x \frac{\log^2(t) \log(1-t)}{1-t}dt \end{aligned} \end{equation} \]

\[ \begin{equation}\label{an-alternating-euler-sum-eq-5} \begin{aligned} \int_0^x \frac{\log(t)\log^2(1-t)}{2t}dt &= \frac{1}{4}\log^2(t)\log^2(1-t)\Big|_0^x +\frac{1}{2}\int_0^x \frac{\log^2(t)\log(1-t)}{1-t}dt \\ &= \frac{1}{4}\log^2(x)\log^2(1-x) +\frac{1}{2}\int_0^x \frac{\log^2(t)\log(1-t)}{1-t}dt \end{aligned} \end{equation} \]

Putting together the results of equations \(\eqref{an-alternating-euler-sum-eq-3}\), \(\eqref{an-alternating-euler-sum-eq-4}\) and \(\eqref{an-alternating-euler-sum-eq-5}\) gives us:

\[ \begin{equation}\label{an-alternating-euler-sum-eq-6} \begin{aligned} \sum_{n=1}^\infty \frac{H_n}{n^3}x^n &= \text{Li}_4(x) +\frac{1}{4}\log^2(x)\log^2(1-x) + \log(x)\log(1-x)\text{Li}_2(1-x)\\ &\quad + \log(x)\left(\zeta(3)-\text{Li}_3(1-x) \right) -\frac{1}{2}\int_0^x \frac{\log^2(t)\log(1-t)}{1-t}dt \end{aligned} \end{equation} \]

(II) Now, let \(-1\leq x < 0\). We have:

\[ \begin{equation}\label{an-alternating-euler-sum-eq-7} \begin{aligned} -\int_x^0 \frac{\zeta(3)-\text{Li}_3\left(\frac{1}{1-t}\right)}{t}dt &= \int_0^{\frac{-x}{1-x}}\frac{\zeta(3)-\text{Li}_3(1-t)}{t(1-t)}dt \quad \left(t\mapsto\frac{-t}{1-t} \right)\\ &= \int_0^{\frac{-x}{1-x}}\frac{\zeta(3)-\text{Li}_3(1-t)}{t}dt + \int_0^{\frac{-x}{1-x}}\frac{\zeta(3)-\text{Li}_3(1-t)}{1-t}dt \\ &= \log\left(\frac{-x}{1-x}\right)\left(\zeta(3) - \text{Li}_3\left(\frac{1}{1-x}\right) \right) - \frac{1}{2}\text{Li}_2^2\left(\frac{1}{1-x}\right) + \frac{\zeta^2(2)}{2} \\ &\quad + \zeta(3)\log(1-x) + \text{Li}_4\left(\frac{1}{1-x} \right) - \zeta(4) \end{aligned} \end{equation} \]

Here, we used equation \(\eqref{an-alternating-euler-sum-eq-3}\) to simplify the integral after the transformation. Similarly, we obtain:

\[ \begin{equation}\label{an-alternating-euler-sum-eq-8} \begin{aligned} \int_x^0 \frac{\log(1-t)\text{Li}_2\left(\frac{1}{1-t}\right)}{t}dt &= \int_0^{\frac{-x}{1-x}}\frac{\log(1-t)\text{Li}_2(1-t)}{t(1-t)}dt \\ &= \int_0^{\frac{-x}{1-x}}\frac{\log(1-t)\text{Li}_2(1-t)}{t}dt + \int_0^{\frac{-x}{1-x}}\frac{\log(1-t)\text{Li}_2(1-t)}{1-t}dt \\ &= -\log\left(\frac{-x}{1-x} \right)\log(1-x)\text{Li}_2\left(\frac{1}{1-x}\right)+\frac{1}{2}\text{Li}_2^2\left(\frac{1}{1-x}\right)-\frac{\zeta^2(2)}{2} \\ &\quad +\text{Li}_4\left(\frac{1}{1-x}\right)+\log(1-x)\text{Li}_3\left(\frac{1}{1-x}\right) -\zeta(4) \\ &\quad -\int_0^{\frac{-x}{1-x}}\frac{\log^2(t)\log(1-t)}{1-t}dt \end{aligned} \end{equation} \]

\[ \begin{equation}\label{an-alternating-euler-sum-eq-9} \begin{aligned} \int_x^0 \left( \frac{\log^3(1-t)}{3t} - \frac{\log^2(1-t)\log(-t)}{2t}\right) dt &= \frac{1}{6}\log^3(1-x)\log\left(\frac{-x}{1-x}\right) + \frac{1}{24}\log^4(1-x) \\ &\quad +\frac{1}{4}\log^2\left(\frac{-x}{1-x}\right)\log^2(1-x) + \frac{1}{2}\int_0^{\frac{-x}{1-x}}\frac{\log^2(t)\log(1-t)}{1-t}dt \end{aligned} \end{equation} \]

Finally, putting these results together gives:

\[ \begin{equation}\label{an-alternating-euler-sum-eq-10} \begin{aligned} \sum_{n=1}^\infty \frac{H_n}{n^3}x^n &= \text{Li}_4(x) + 2\text{Li}_4\left( \frac{1}{1-x}\right) + \log\left(\frac{-x}{1-x} \right)\left(-\text{Li}_3\left(\frac{1}{1-x}\right) +\frac{\log^3(1-x)}{6}-\log(1-x)\text{Li}_2\left(\frac{1}{1-x} \right)\right) \\ &\quad + \log(1-x)\text{Li}_3\left(\frac{1}{1-x}\right) + \frac{\log^4(1-x)}{24} +\frac{1}{4}\log^2 \left(\frac{-x}{1-x}\right)\log^2(1-x) +\zeta(3) \log(-x)- 2\zeta(4) \\ &\quad -\frac{1}{2}\int_0^{\frac{-x}{1-x}}\frac{\log^2(t)\log(1-t)}{1-t}dt \end{aligned} \end{equation} \]

The integral \(\int \frac{\log^2(t)\log(1-t)}{1-t}dt\) can be evaluated in terms of Polylogarithms (refer to section 7.6 of "Polylogarithms and Associated Functions" by Leonard Lewin). Luckily for us, there is a much simpler was to evaluate this integral between the limits \(0\) and \(\frac{1}{2}\).

3. Evaluation of \(\int_0^{\frac{1}{2}}\frac{\log^2(t)\log(1-t)}{1-t}dt\)

Let \(I=\int_0^{1}\frac{\log^2(t)\log(1-t)}{1-t}dt\) and \(J=\int_0^{\frac{1}{2}}\frac{\log^2(t)\log(1-t)}{1-t}dt\). The integral, \(I\), can be evaluated by the differentiation under the integral technique.

\[ \begin{aligned} I &= \lim_{y\to 0^+}\lim_{x\to 1}\frac{\partial^2 }{\partial x^2} \frac{\partial }{\partial y} \int_0^1 t^{x-1} (1-t)^{y-1} dt \\ &= \lim_{y\to 0^+}\lim_{x\to 1}\frac{\partial^2 }{\partial x^2} \frac{\partial }{\partial y} \left\{\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \right\} \end{aligned} \]

It takes a bit of effort to compute the derivatives in terms of Polygamma functions so I won't write the details here. One can easily verify that the end result is:

\[ I = -\frac{\pi^4}{180} \]

Next, apply integration by parts to the integral \(J\):

\[ \begin{aligned} J &= \int_0^{\frac{1}{2}}\frac{\log^2(t)\log(1-t)}{1-t}dt \\ &= -\frac{\log^2(t)\log^2(1-t)}{2}\Big|_0^{\frac{1}{2}} + \int_0^{\frac{1}{2}} \frac{\log(t)\log^2(1-t)}{t}dt \\ &= -\frac{\log^4(2)}{2} + \int_{\frac{1}{2}}^1 \frac{\log^2(t)\log(1-t)}{1-t}dt \quad (t\mapsto 1-t)\\ &= -\frac{\log^4(4)}{2} + I-J \end{aligned} \]

Now, one can solve the above equation for \(J\) to get:

\[ \begin{aligned} J &= -\frac{\log^4(2)}{4}+\frac{I}{2} \\ &= -\frac{\log^4(2)}{4}-\frac{\pi^4}{360} \end{aligned} \]

Finally, we have everything needed to evaluate the original series. Plugging in \(x=-1\) in equation \(\eqref{an-alternating-euler-sum-eq-10}\) and performing some simplifications gives us the beautiful result:

\[ \boxed{\sum_{n=1}^\infty \frac{H_n}{n^3}(-1)^{n+1} = \frac{11\pi^4}{360}+\frac{\pi^2}{12}\log^2(2)-\frac{\log^4(2)}{12}-\frac{7}{4}\log(2)\zeta(3) -2\text{Li}_4\left(\frac{1}{2}\right)} \]

A similar result is obtained by plugging \(x=\frac{1}{2}\) into equation \(\eqref{an-alternating-euler-sum-eq-6}\):

\[ \boxed{\sum_{n=1}^\infty \frac{H_n}{n^3 2^n}= \frac{\pi^4}{720}+\frac{\log^4(2)}{24}-\frac{1}{8}\log(2)\zeta(3) +\text{Li}_4\left(\frac{1}{2}\right)} \]

4. Further Results

Let \(H_n^{(2)} = \zeta(2) - \psi_1(n+1) = \sum_{k=1}^n \frac{1}{k^2}\). Then, it is easy to verify that:

\[ \sum_{n=1}^\infty t^n H_n^{(2)} = \frac{\text{Li}_2(t)}{1-t}, \quad -1\leq t < 1 \]

Now, dividing the above equation by \(t\) and integrating both sides, gives us:

\[ \begin{aligned} \sum_{n=1}^\infty \frac{H_n^{(2)}}{n}x^n &= \int_0^x \text{Li}_2(t)\left( \frac{1}{t}+\frac{1}{1-t}\right) \\ &= \text{Li}_3(x) + \int_0^x \frac{\text{Li}_2(t)}{1-t} dt \\ &= \text{Li}_3(x) -\log(1-x)\text{Li}_2(x) - \int_0^x \frac{\log^2(1-t)}{t}dt \end{aligned} \]

From our previous calculation, we know that

\[ \int_0^x \frac{\log^2(1-t)}{t}dt = 2\sum_{n=1}^\infty \frac{H_n}{n^2}x^n -2\text{Li}_3(x) \]

Hence, we obtain the following relation:

\[ \begin{equation}\label{an-alternating-euler-sum-eq-11} \sum_{n=1}^\infty \frac{H_n^{(2)}}{n}x^n =3\text{Li}_3(x)-\log(1-x)\text{Li}_2(x) - 2\sum_{n=1}^\infty \frac{H_n}{n^2}x^n \end{equation} \]

Once again, divide the above equation by \(x\) and integrate both sides to get:

\[ \begin{equation}\label{an-alternating-euler-sum-eq-12} \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^2}x^n = 3\text{Li}_4(x) + \frac{1}{2}\text{Li}_2^2(x) - 2\sum_{n=1}^\infty \frac{H_n}{n^3}x^n \end{equation} \]

Equations \(\eqref{an-alternating-euler-sum-eq-11}\) and \(\eqref{an-alternating-euler-sum-eq-12}\) are valid for \(-1\leq x < 1\). Plugging in \(x=\frac{1}{2}\) in equation \(\eqref{an-alternating-euler-sum-eq-12}\) and using the known closed form for \(\sum_{n=1}^\infty \frac{H_n}{n^3 2^n}\) gives us:

\[ \boxed{ \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^2 2^n} = \frac{\pi^4}{1440}-\frac{\pi^2}{24}\log^2(2) + \frac{\log^4(2)}{24}+ \frac{1}{4}\log(2)\zeta(3) + \text{Li}_4\left(\frac{1}{2}\right) } \]

Ramanujan's Master Theorem: Part I

2020-07-02T12:38:30+00:00

Ramanujan's master theorem (RMT) connects the coefficients of a power series with a Mellin transform. We will use this connection to evaluate integrals involving one Bessel function and a product of two Bessel functions.

The master-theorem principle

Suppose that \(f(z)\) has the expansion

\[ f(z) = \sum_{n=0}^\infty \frac{\varphi(n)}{n!}(-z)^n \]

Under suitable analytic continuation and growth hypotheses on the coefficient function \(\varphi\), the master theorem gives, in a strip where the integral converges,

\[ \int_0^\infty x^{s-1} f(x)\; dx = \Gamma(s) \varphi(-s) \]

The hypotheses matter: the values \(\varphi(n)\) at nonnegative integers alone do not determine \(\varphi(-s)\), and a local power series does not justify integrating term by term over the entire positive axis. Here we use the principle to derive the Bessel identities; for a discussion of its hypotheses, see González, Moll, and Schmidt.

A single Bessel function

One application is the following Mellin transform:

\[ \begin{equation}\label{ramanujans-master-theorem-part-i-eq-1} \int_0^\infty x^{\sigma-1} J_\nu (x) \; dx = \frac{2^{\sigma-1}\Gamma\left(\frac{\sigma+\nu}{2}\right)}{\Gamma\left(\frac{2-\sigma+\nu}{2} \right)} \; ,\quad -\nu < \sigma < \frac{3}{2} \end{equation} \]

where \(J_\nu (x)\) is the Bessel function of the first kind.

Indeed, \(J_\nu(x)\) has the following series expansion around \(x=0\):

\[ J_\nu(x) =\sum_{n=0}^\infty \frac{(-1)^n }{n! \Gamma(n+\nu+1)}\left(\frac{x}{2} \right)^{\nu+2n} \]

Hence, we may apply the RMT to the function \(x^{-\nu/2}J_\nu(2\sqrt{x})\) to obtain:

\[ \int_0^\infty x^{s-\nu/2-1} J_\nu(2\sqrt{x}) dx=\frac{\Gamma(s)}{\Gamma(\nu+1-s)} \]

Now, one can perform the substitutions \(x=t^2/4\) and \(s=\frac{\sigma+\nu}{2}\) to transform the above integral into the integral of equation \(\eqref{ramanujans-master-theorem-part-i-eq-1}\).

It turns out that a similar approach also works for the product \(J_\nu (x) J_\mu(x)\). First, we need to compute its series expansion. This can be done by multiplying the individual series expansions and rearranging/grouping certain terms.

\[ \begin{equation}\label{ramanujans-master-theorem-part-i-eq-2} \begin{aligned} J_{\nu}(x)J_{\mu}(x) &= \left(\sum_{n=0}^\infty \frac{(-1)^n }{n! \Gamma(n+\nu+1)}\left(\frac{x}{2} \right)^{\nu+2n}\right) \left(\sum_{m=0}^\infty \frac{(-1)^m }{m! \Gamma(m+\nu+1)}\left(\frac{x}{2} \right)^{\nu+2m} \right) \\ &= \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{n+m} x^{\nu+\mu+2(m+n)}}{2^{\nu+\mu+2(m+n)}} \cdot \frac{1}{m! n!\Gamma(n+\nu+1)\Gamma(m+\mu+1)} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} x^{\nu+\mu+2r}}{2^{\nu+\mu+2r}}\sum\limits_{\substack{0\leq m,n \leq r \\ m+n=r}}\frac{1}{m! n!\Gamma(n+\nu+1)\Gamma(m+\mu+1)} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} x^{\nu+\mu+2r}}{2^{\nu+\mu+2r} r! \Gamma(r+\nu+1)\Gamma(r+\mu+1)}\sum\limits_{\substack{0\leq m,n \leq r \\ m+n=r}}\binom{r}{m,n} \left(\prod_{i=m+1}^r (\nu+i)\right) \left(\prod_{j=n+1}^r (\mu+j)\right) \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} x^{\nu+\mu+2r}}{2^{\nu+\mu+2r} r! \Gamma(r+\nu+1)\Gamma(r+\mu+1)}\sum _{n=0}^r\binom{r}{n}(\nu +r)^{\underline{n}}(\mu +r)^{\underline{r-n}} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} x^{\nu+\mu+2r}}{2^{\nu+\mu+2r} r! \Gamma(r+\nu+1)\Gamma(r+\mu+1)} (\nu +\mu+2r)^{\underline{r}} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r} \Gamma(\nu+\mu+2r+1)}{ r! \Gamma(r+\nu+1)\Gamma(r+\mu+1)\Gamma(\mu+\nu+r+1)} \left( \frac{x}{2}\right)^{\nu+\mu+2r} \quad \; \end{aligned} \end{equation} \]

To simplify the inner summation, we used the binomial theorem for falling factorials.

Therefore, applying RMT to the function \(x^{-\frac{\nu+\mu}{2}}J_{\nu}(2\sqrt{x})J_{\mu}(2\sqrt{x})\), gives us:

\[ \int_0^\infty x^{s-\frac{\nu+\mu}{2}-1} J_{\nu}(2\sqrt{x})J_{\mu}(2\sqrt{x}) \; dx = \frac{\Gamma(s)\Gamma(\nu+\mu+1-2s)}{\Gamma(\nu+1-s)\Gamma(\mu+1-s)\Gamma(\nu+\mu+1-s)} \]

Finally, make the substitutions \(x=t^2/4\) and \(s=\frac{\sigma+\nu+\mu}{2}\) in the above integral to get:

\[ \begin{equation}\label{ramanujans-master-theorem-part-i-eq-3} \boxed{ \int_0^\infty x^{\sigma-1}J_\nu(x)J_\mu(x)\; dx = \frac{2^{\sigma-1} \Gamma\left(\frac{\nu+\mu+\sigma}{2} \right)\Gamma\left(1-\sigma\right)}{\Gamma\left(\frac{\nu-\mu-\sigma}{2} +1\right)\Gamma\left(\frac{\mu-\nu-\sigma}{2} +1\right)\Gamma\left(\frac{\nu+\mu-\sigma}{2} + 1\right)} } \quad \; \end{equation} \]

The above equation holds provided that the condition \(-(\nu+\mu)< \sigma < 1\) is satisfied.

Fourier Series of the Log-Gamma Function and Vardi's Integral

2020-06-01T12:11:38+00:00

Kummer's Fourier series for \(\log\Gamma\) provides a useful bridge between special functions and logarithmic integrals. We will derive the series and then evaluate it at a suitable point to obtain Vardi's integral:

\[ \int_0^1 \frac{\log\log \left(\frac{1}{x}\right)}{1+x^2}dx = \frac{\pi}{2}\log\left(\sqrt{2\pi}\frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)} \right) \]

Fourier Series of the Log-Gamma Function:

For \(s\in (0,1)\), we have

\[ \begin{equation}\label{fourier-series-of-the-log-gamma-function-and-vardis-integral-eq-1} \log \Gamma(s) = \left(\frac{1}{2}-s \right)(\gamma + \log 2)-\frac{1}{2}\log(\sin (\pi s)) + (1-s) \log(\pi) + \frac{1}{\pi}\sum_{n=1}^\infty \frac{\sin(2\pi n s)\log n}{n} \end{equation} \]

Proof: It suffices to do the following integrals:

\[ \begin{aligned} \int_0^1 \log \Gamma (s) \; ds &= \frac{1}{2}\log(2\pi) \\ \int_0^1 \log \Gamma(s) \cos(2\pi n s)\; ds &= \frac{1}{4n} \quad \forall n\in \mathbb{Z}^+ \\ \int_0^1 \log \Gamma(s) \sin(2\pi n s)\; ds &= \frac{\gamma + \log(2n\pi )}{2n\pi}\quad \forall n\in \mathbb{Z}^+ \end{aligned} \]

The first integral can be evaluated with the help of Euler's reflection formula:

\[ \begin{aligned} \int_0^1 \log \Gamma(s) \; ds &= \frac{1}{2} \int_0^1 \log \Gamma(s) \; ds + \frac{1}{2}\int_0^1 \log \Gamma(1-s) \; ds \\ &= \frac{1}{2}\int_0^1 \log\left(\frac{\pi}{\sin(\pi s)} \right)\; ds \\ &= \frac{\log (\pi)}{2} - \frac{1}{2}\int_0^1 \log(\sin(\pi s))\; ds \\ &= \frac{\log(2\pi)}{2} \end{aligned} \]

The second integral can also be dealt in a similar way. We have:

\[ \begin{aligned} \int_0^1 \log \Gamma(s) \cos(2\pi n s)\; ds &= \frac{1}{2}\int_0^1 \log \Gamma(s) \cos(2\pi n s)\; ds + \frac{1}{2}\int_0^1 \log \Gamma(1-s) \cos(2\pi n (1-s))\; ds \\ &= \frac{1}{2}\int_0^1 \log\left(\frac{\pi}{\sin(\pi s)} \right)\cos(2\pi n s)\; ds \\ &= -\frac{1}{2}\int_0^1 \log(2\sin(\pi s))\cos(2\pi n s)\; ds \\ &= \frac{1}{2}\int_0^1 \left(\sum_{k=1}^\infty \frac{\cos(2\pi k s)}{k} \right)\cos(2\pi n s)\; ds \\ &= \frac{1}{2}\sum_{k=1}^\infty \frac{1}{k}\int_0^1 \cos(2\pi k s)\cos(2\pi n s)\; ds \\ &= \frac{1}{4n} \end{aligned} \]

In the above calculation, we used the well known Fourier series:

\[ \begin{equation}\label{fourier-series-of-the-log-gamma-function-and-vardis-integral-eq-2} \log(2\sin (\pi s)) = -\sum_{k=1}^\infty \frac{\cos(2\pi k s)}{k} \quad \forall s\in (0,1) \quad \end{equation} \]

The third integral is the most troublesome of all since the trick involving Euler's reflection formula does not work. We will instead use the following integral representation of the Log-Gamma function:

\[ \log \Gamma(s) = \int_0^\infty \left(\frac{s-1}{t e^t} -\frac{1-e^{t(1-s)}}{t(e^t-1)}\right) dt \quad \forall s>0 \]

One can easily verify the above equation via the differentiation under the integral technique. Therefore, upon changing the order of integration we get:

\[ \begin{aligned} \int_0^1 \log \Gamma(s) \sin(2\pi n s)\; ds &= \int_0^\infty \left(\frac{1}{t e^t}\int_0^1 s\sin(2\pi ns)\; ds +\frac{1}{t(e^t-1)}\int_0^1e^{t(1-s)}\sin(2\pi n s)\; ds \right)dt \\ &= \int_0^\infty \left[\frac{1}{t e^t}\left(-\frac{1}{2\pi n}\right) +\frac{1}{t(e^t-1)}\left(\frac{2\pi n (e^t-1)}{t^2 + (2\pi n)^2} \right) \right]dt \\ &= \int_0^\infty \left(-\frac{1}{2\pi n t e^t} + \frac{2\pi n}{t\left( t^2 + (2\pi n)^2 \right)} \right)dt \\ &= \frac{1}{2\pi n}\int_0^\infty\left(-\frac{1}{te^t} +\frac{1}{t}-\frac{t}{t^2 + (2\pi n)^2} \right) dt \\ &= \frac{1}{2\pi n}\left(\gamma + \left[ \log t - \log (t) e^{-t} - \frac{1}{2}\log\left( t^2 + (2\pi n)^2 \right)\right]_{t=0}^\infty \right) \\ &= \frac{\gamma + \log(2\pi n)}{2\pi n} \end{aligned} \]

To get equation \(\eqref{fourier-series-of-the-log-gamma-function-and-vardis-integral-eq-1}\), one needs to piece together all these calculations. Equation \(\eqref{fourier-series-of-the-log-gamma-function-and-vardis-integral-eq-2}\) along with the result:

\[ \pi \left(\frac{1}{2}- s\right) = \sum_{k=1}^\infty \frac{\sin(2\pi k s)}{k} \quad \forall s\in (0,1) \]

are needed to perform simplifications.

Now, we turn our attention to the Vardi's integral:

\[ \begin{equation}\label{fourier-series-of-the-log-gamma-function-and-vardis-integral-eq-3} \begin{aligned} I &= \int_0^1 \frac{\log\log \left(\frac{1}{x}\right)}{1+x^2}dx \\ &= \int_0^\infty \frac{\log(t) e^{-t}}{1 + e^{-2t}}dt \\ &= \int_0^\infty \log(t)\sum_{n=0}^\infty (-1)^n e^{-(2n+1)t} \; dt\\ &= \sum_{n=0}^\infty (-1)^n \int_0^\infty \log (t) e^{-(2n+1)t}\; dt \\ &= -\sum_{n=0}^\infty (-1)^n \left(\frac{\log(2n+1)}{2n+1}+\frac{\gamma}{2n+1} \right) \\ &= -\frac{\gamma \pi}{4}-\sum_{n=1}^\infty \frac{\sin\left(\frac{\pi n}{2}\right)\log n}{n} \end{aligned} \end{equation} \]

To get the value of the sum, plug \(s=\frac{1}{4}\) in equation \(\eqref{fourier-series-of-the-log-gamma-function-and-vardis-integral-eq-1}\).

\[ \sum_{n=1}^\infty \frac{\sin\left(\frac{\pi n}{2}\right)\log n}{n} = \pi \log \Gamma\left(\frac{1}{4}\right) - \frac{\gamma \pi}{4} - \frac{\pi \log 2}{2}- \frac{3\pi}{4}\log(\pi) \]

Substituting the above into equation \(\eqref{fourier-series-of-the-log-gamma-function-and-vardis-integral-eq-3}\) and performing some simplifications gives the desired result.

Bernoulli numbers and a related integral

2020-04-19T10:32:48+00:00

Bernoulli polynomials link a simple differential recursion to Fourier series and definite integrals. We begin with their defining conditions, derive the identities we need, and then apply them to an integral.

Bernoulli polynomials

Define the sequence \(\{B_r(x)\}_{r=0}^{\infty}\) by

\[ \begin{aligned} B_0(x) &= 1 \\ B_r^\prime(x) &= r B_{r-1}(x) \quad \forall r\geq 1 \\ \int_0^1 B_r(x) dx &= 0 \quad \forall r\geq 1 \end{aligned} \]

The first few Bernoulli Polynomials are:

\[ \begin{aligned} B_0(x) & =1 \\ B_1(x) & =x-\frac{1}{2} \\ B_2(x) & =x^2-x+\frac{1}{6} \\ B_3(x) & =x^3-\frac{3}{2}x^2+\frac{1}{2}x \\ B_4(x) & =x^4-2x^3+x^2-\frac{1}{30} \\ \end{aligned} \]

The numbers \(B_n = B_n(0)\) are called the Bernoulli numbers. Integrating the relation \(B_r^\prime(x)=rB_r(x)\) between 0 and 1 gives:

\[ B_r(1)-B_r(0) = \int_0^1 B_r^\prime (x) dx = r\int_0^1 B_{r-1}(x) dx = 0 \quad \forall r\geq 2 \]

This motivates us to define the periodic Bernoulli polynomials by

\[ \tilde{B}_r(x) = B_r(\langle x\rangle), \quad x\in\mathbb{R}, \; r\geq 2 \]

where \(\langle x\rangle\) denotes the fractional part of \(x\). We will now compute the Fourier series of \(\tilde{B}_r(x)\), where \(r\geq 2\). The \(n\)-th Fourier coefficient is given by:

\[ \begin{aligned} a_n &= \int_0^1 \tilde{B}_r(x) e^{-2\pi i n x} dx= \int_0^1 B_r(x) e^{-2\pi i n x} dx \end{aligned} \]

Let's first consider the case when \(n\neq 0\). Integration by parts, gives us:

\[ \begin{equation}\label{bernoulli-numbers-and-a-related-integral-eq-1} \begin{aligned} a_n &= -\frac{e^{-2\pi i n x}}{2\pi i n}B_r(x)\Big|_0^1 + \frac{1}{2\pi i n}\int_0^1 B_r^\prime (x) e^{-2\pi i n x} dx \\ &= \frac{1}{2\pi i n}\int_0^1 B_r^\prime (x) e^{-2\pi i n x} dx \\ &= \frac{r}{2\pi i n}\int_0^1 B_{r-1}(x) e^{-2\pi i n x} dx \end{aligned} \end{equation} \]

The repeated use of equation \(\eqref{bernoulli-numbers-and-a-related-integral-eq-1}\) gives:

\[ \begin{aligned} a_n &= \frac{r!}{(2\pi i n)^{r-1}} \int_0^1 B_1(x) e^{-2\pi i nx} dx \\ &= \frac{r!}{(2\pi i n)^{r-1}} \int_0^1 \left(x-\frac{1}{2} \right) e^{-2\pi i nx} dx \\ &= \frac{r!}{(2\pi i n)^{r-1}}\int_0^1 x e^{-2\pi i n x} dx \\ &= -\frac{r!}{(2\pi i n)^r} \end{aligned} \]

When \(n=0\), we have \(a_0 = \int_0^1 B_r(x)dx = 0\). Note that the Fourier series

\[ -r! \sum_{\substack{n=-\infty \\ n\neq 0}} \frac{e^{2\pi i n x}}{(2\pi i n)^r} \]

converges absolutely for all \(r\geq 2\). Therefore, it converges uniformly to \(\tilde{B}_r(x)\) for all \(r\geq 2\). This leads to the following bound:

\[ \begin{equation}\label{bernoulli-numbers-and-a-related-integral-eq-2} |\tilde{B}_r(x)| \leq \frac{2r!}{(2\pi)^r}\sum_{n=1}^\infty \frac{1}{n^r} < \frac{4r!}{(2\pi)^r} \;\; \forall r\geq 2 \end{equation} \]

Note that the above inequality also remains valid for \(r=0\) and \(r=1\). Now, let's consider the generating function:

\[ F(x,t) = \sum_{n=0}^\infty \frac{\tilde{B}_n(x) t^n}{n!} \]

The inequality \(\eqref{bernoulli-numbers-and-a-related-integral-eq-2}\) implies that:

\[ |F(x,t)| < 4 \sum_{n=0}^\infty \left(\frac{t}{2\pi}\right)^n \]

Therefore, the series converges uniformly for all \(t\in [0,2\pi]\) and all \(x\). We, may, therefore differentiate term by term to obtain:

\[ \frac{\partial F(x,t)}{\partial x} = \sum_{n=1}^\infty \frac{\tilde{B}_{n-1}(x)}{(n-1)!}t^n = t F(x,t) \]

Solving the above differential equation, we get \(F(x,t) = G(t) e^{xt}\) where \(G\) is some arbitrary function of \(t\). Next, we integrate \(F(x,t)\) between 0 and 1:

\[ \begin{aligned} \int_0^1 F(x,t) dx &= G(t) \int_0^1 e^{xt} dx \\ &= G(t) \frac{e^t-1}{t} \end{aligned} \]

On the other hand, note that:

\[ \begin{aligned} \int_0^1 F(x,t) dx &= \int_0^1 \sum_{n=0}^\infty \frac{\tilde{B}_n(x) t^n}{n!} dx \\ &= 1 + \sum_{n=1}^\infty \frac{t^n}{n!}\int_0^1 B_n(x) dx \\ &= 1 \end{aligned} \]

Therefore, we obtain \(G(t) = \frac{t}{e^t - 1}\) and

\[ \boxed{F(x,t) = \frac{t e^{xt}}{e^t-1}} \]

An interesting property of the Bernoulli numbers is that \(B_{2n+1}=0\) for all \(n\geq 1\). To see this, consider:

\[ \frac{t}{e^t -1} + \frac{t}{2}= 1+\sum_{n=2}^\infty \frac{B_n t^n}{n!} \]

Now, on the left hand side we have an even function of \(t\). Therefore, the coefficients of the odd powers of \(t\) on the right hand side are equal to 0. Using the Fourier series expansion, we can express the even-index Bernoulli numbers in terms of the Riemann zeta function:

\[ B_{2n} = \frac{2 (-1)^{n-1} (2n)!}{(2\pi)^{2n}} \zeta(2n) \]

The Bernoulli polynomials satisfy the following recursive equation:

\[ {B}_n(x) = \sum_{k=0}^n \binom{n}{k} B_{n-k} x^k \]

This can be proved by noting that:

\[ \begin{aligned} \sum_{n=0}^\infty \frac{{B}_n(x) t^n}{n!} &= \frac{te^{xt}}{e^t-1} \\ &= e^{xt} \sum_{n=0}^\infty \frac{B_n t^n}{n!} \\ &= \sum_{m=0}^\infty \frac{(xt)^m}{m!} \sum_{n=0}^\infty \frac{B_n t^n}{n!} \\ &= \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{B_n x^m t^{n+m}}{n! m!} \\ &= \sum_{n=0}^\infty \sum_{k=0}^n \frac{B_k t^n x^{n-k}}{k! (n-k)!} \\ &= \sum_{n=0}^\infty \frac{t^n}{n!}\sum_{k=0}^n \binom{n}{k} B_k x^{n-k} \end{aligned} \]

where \(x\in [0,1]\). Now, compare the coefficients of \(t^n\) to get the desired result. Plugging in \(x=1\), gives the identity:

\[ \sum_{k=0}^{n-1} \binom{n}{k} B_k = 0 \]

Now, let's turn our attention to the integral:

\[ I=\int_0^{\frac{\pi}{2}}\frac{\sin(2nx)}{\sin^{2n+2}(x)}\cdot \frac{1}{e^{2\pi \cot x}-1} dx \]

where \(n\in\mathbb{N}\). We will use the following trigonometric identity:

\[ \frac{\sin(2nx)}{\sin^{2n}(x)} =(-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r}\cot^{2r-1}(x) \]

Substituting the above into the integral, gives:

\[ \begin{aligned} I &= (-1)^n \int_0^{\pi\over 2}\left( \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r}\cot^{2r-1}(x)\right)\frac{\csc^2(x)}{e^{2\pi \cot x}-1}dx \\ &= (-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r} \int_0^{\pi \over 2}\cot^{2r-1}(x)\frac{\csc^2(x)}{e^{2\pi \cot x}-1}dx \\ &= (-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r} \int_0^\infty \frac{t^{2r-1}}{e^{2\pi t}-1}dt \\ &= (-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r}\frac{(2r-1)! \zeta(2r)}{(2\pi)^r }\\ &= (-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r} (-1)^{r-1} \frac{B_{2r}}{4r}\\ &= \frac{(-1)^{n-1}}{4}\sum_{r=1}^{n}\binom{2n}{2r-1}\frac{B_{2r}}{r} \\ &= \frac{(-1)^{n-1}}{2(2n+1)}\sum_{r=1}^n \binom{2n+1}{2r} B_{2r} \\ &= \frac{(-1)^{n-1}}{2(2n+1)} \left[\sum_{r=0}^{2n} \binom{2n+1}{r} B_r - \binom{2n+1}{0}B_0 - \binom{2n+1}{1} B_1\right] \\ &= \frac{(-1)^{n-1}}{2(2n+1)} \left[-\binom{2n+1}{0}B_0 - \binom{2n+1}{1} B_1\right] \\ &= \frac{(-1)^{n-1}}{4}\cdot \frac{2n-1}{2n+1} \end{aligned} \]

Weyl's Equidistribution Theorem

2020-02-05T16:39:57+00:00

Weyl's criterion lets us test uniform distribution by checking exponential sums. We will prove the equivalence between this test, interval counts, and averages of Riemann-integrable functions. A sequence of real numbers \(x_1, x_2, \cdots\) is said to be equidistributed (mod 1) if for every sub-interval \((a,b)\subset [0,1]\), we have

\[ \lim_{N\to \infty}\frac{|\{1\leq n\leq N:\; \langle x_n \rangle\in (a,b)\}|}{N} = b-a \]

where \(\langle x \rangle\) denotes the fractional part of \(x\). Weyl's equidistribution criterion states that the following statements are equivalent:

\(x_1, x_2, \cdots\) are equidistributed (mod 1).
For each non-zero integer \(k\), we have

\[ \lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N e^{2\pi i k x_n}=0 \]
For each Riemann integrable function \(f:[0,1]\to\mathbb{C}\), we have

\[ \lim_{N\to \infty}\frac{1}{N}\sum_{n=1}^N f(\langle x_n \rangle) = \int_0^1 f(x) dx \]

Proof: (1) ⇒ (3)

Let \(I=[a,b)\subseteq [0,1]\) and note that

\[ \frac{|\{1 \leq n \leq N: \langle x_n \rangle\in [a,b) \}|}{N}=\frac{1}{N}\sum_{n=1}^N \chi_{[a,b)}(\langle x_n \rangle) \]

where \(\chi_{[a,b)}(x)\) equals 1 if \(x\in [a,b)\) and 0 otherwise. This shows that (3) holds for the case when \(f\) is a characteristic function. Now, let \(\lambda_1, \lambda_2\in \mathbb{R}\) and \(f_1\), \(f_2\) be functions for which (3) holds. Then,

\[ \begin{aligned}\lim_{N\to \infty}\frac{1}{N}\sum_{n=1}^N (\lambda_1 f_1 + \lambda_2 f_2)(\langle x_n \rangle) &= \lim_{N\to \infty} \frac{\lambda_1}{N}\sum_{n=1}^N f_1(\langle x_n\rangle) + \lim_{N\to \infty}\frac{\lambda_2}{N}\sum_{n=1}^Nf_2(\langle x_n\rangle) \\ &= \lambda_1\int_0^1 f_1(x) dx + \lambda_2 \int_0^1 f_2(x) dx \\ &= \int_0^1 (\lambda_1 f_1 + \lambda_2 f_2)(x) dx\end{aligned} \]

Thus, (3) holds for all linear combinations of characteristic functions of subintervals of \([0,1]\).

Now, let \(f:[0,1]\to \mathbb{R}\) be an integrable function, and let \(\epsilon >0\). Choose step functions \(f_1\) and \(f_2\) such that:

\(f_1\leq f\leq f_2\) pointwise
\(\int_0^1 (f_2(x)-f_1(x))dx < \frac{\epsilon}{2}\)
There exists \(N_0\) such that \(\left|\int_0^1 f_1(x)dx - \frac{1}{N}\sum_{n=1}^N f_1(\langle x_n\rangle) \right| < \frac{\epsilon}{2}\) and \(\left|\int_0^1 f_2(x)dx - \frac{1}{N}\sum_{n=1}^N f_2(\langle x_n\rangle) \right| < \frac{\epsilon}{2}\) for all \(N\geq N_0\)

It follows that for \(N\geq N_0\),

\[ \begin{aligned} \int_0^1 f(x) dx - \frac{1}{N}\sum_{n=1}^N f(\langle x_n\rangle) &\leq \int_0^1 f(x) dx - \frac{1}{N}\sum_{n=1}^N f_1(\langle x_n\rangle) \\ &< \int_0^1 f(x) dx -\int_0^1 f_1(x) dx +\frac{\epsilon}{2} \\ &< \int_0^1 (f_2(x)-f_1(x)) dx + \frac{\epsilon}{2} \\ &< \epsilon \end{aligned} \]

In a similar way, we can prove that

\[ \int_0^1 f(x) dx - \frac{1}{N}\sum_{n=1}^N f(\langle x_n\rangle) > -\epsilon \quad \forall\; N\geq N_0 \]

Therefore, we have

\[ \left|\int_0^1 f(x) dx - \frac{1}{N}\sum_{n=1}^N f(\langle x_n\rangle) \right| < \epsilon \quad \forall\; N\geq N_0 \]

To see that (3) holds when \(f\) is complex valued, we need only consider the real and imaginary parts separately.

(2) ⇒ (3)

First let \(f\) be continuous and \(1\)-periodic. Trigonometric polynomials uniformly approximate such functions on the circle. For any \(\epsilon>0\), choose

\[ p(x)=\sum_{|k|\leq m}c_k e^{2\pi i kx},\qquad \|f-p\|_\infty<\epsilon. \]

By (2), the average of \(p(x_n)\) tends to its constant coefficient \(c_0=\int_0^1p(x)\,dx\). Thus

\[ \limsup_{N\to\infty}\left|\frac1N\sum_{n=1}^N f(\langle x_n\rangle)-\int_0^1 f(x)\,dx\right|\leq2\epsilon. \]

Letting \(\epsilon\to0\) proves the assertion for continuous periodic \(f\).

To pass to interval indicators, fix an arc \(I\) of the circle. For any \(\epsilon>0\), choose continuous periodic functions \(g_-\) and \(g_+\), with short linear ramps near the endpoints, such that

\[ g_-\leq\chi_I\leq g_+,\qquad \int_0^1(g_+-g_-)\,dx<\epsilon. \]

Their integrals bracket the length of \(I\). Applying the result just proved to \(g_-\) and \(g_+\) squeezes the limiting interval frequency to that length. The same construction with a small neighborhood of a single point shows that endpoint visits have density zero, so the choice of open, closed, or half-open endpoints does not affect the result. This proves (1). The argument in (1) ⇒ (3) now extends the conclusion to all Riemann-integrable functions.

Periodicity is essential in the first approximation step: a continuous function on \([0,1]\) whose endpoint values differ cannot be uniformly approximated there by trigonometric polynomials.

The implications (3) ⇒ (1) and (3) ⇒ (2) are obvious.

References

Hannigan-Daley, Brad. Equidistribution and Weyl’s criterion. Retrieved from http://individual.utoronto.ca/hannigandaley/equidistribution.pdf. Accessed 5 Feb. 2020.
Stein, Elias M. and Shakarchi, Rami. Fourier Analysis: An Introduction. Princeton University Press, 2003

Eisenstein's Proof of Quadratic Reciprocity

2020-01-25T16:47:43+00:00

Eisenstein's proof of quadratic reciprocity turns a question about quadratic residues into a symmetry between two products of sines. We will first establish Euler's criterion and Gauss's lemma, then use them to identify the sign of that product.

Definition: For an odd prime \(p\), the Legendre symbol \(\left(\frac{a}{p}\right)\) records whether \(a\) is a nonzero square modulo \(p\); it is zero when \(p\) divides \(a\).

\[ \left(\frac{a}{p}\right) = \begin{cases}0 \quad \text{if }p|a \\ 1 \quad \text{if }a\text{ is a quadratic residue modulo }p \\ -1 \quad \text{if }a\text{ is a quadratic non-residue modulo }p\end{cases} \]

Theorem (Quadratic Reciprocity Law): If \(p\) and \(q\) are distinct odd primes, then the quadratic reciprocity theorem states that the congruences

\[ \begin{aligned} x^2 \equiv q \quad (\text{mod }p) \\ x^2 \equiv p \quad (\text{mod }q) \end{aligned} \]

are both solvable or both unsolvable unless both \(p\) and \(q\) leave the remainder \(3\) when divided by \(4\). Written symbolically,

\[ \left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = (-1)^{(p-1)(q-1)/4} \]

We will start by proving two important results about quadratic residues that will be useful later on.

Lemma 1: Let \(n\not\equiv 0 \;(\text{mod }p)\). Then \(n\) is a quadratic residue modulo \(p\) iff \(n^{\frac{p-1}{2}}\equiv 1 \; (\text{mod }p)\).

Proof: Fermat's little theorem tells us that \(n^{p-1}\equiv 1 \;(\text{mod }p)\) whenever \(p \not| n\). Since,

\[ n^{p-1}-1\equiv (n^{\frac{p-1}{2}}-1)(n^{\frac{p-1}{2}}+1)\equiv 0 \; (\text{mod }p) \]

we have \(n^{\frac{p-1}{2}}\equiv \pm 1\; (\text{mod }p)\). Therefore, it suffices to prove that \(n\) is a quadratic residue modulo \(p\) if and only if \(n^{\frac{p-1}{2}}\equiv 1 \; (\text{mod }p)\).

Suppose that \(\left(\frac{n}{p} \right)=1\). Then, there is an integer \(x\) such that \(n\equiv x^2 \; (\text{mod p})\). Now, we have

\[ n^{\frac{p-1}{2}}\equiv x^{p-1} \equiv 1 \equiv \left(\frac{n}{p} \right) \; (\text{mod }p) \]

Conversely, assume that \(n^{\frac{p-1}{2}}\equiv 1 \; (\text{mod }p)\). Let \(g\) be a primitive root of \(p\). Then, we have \(n\equiv g^k \; (\text{mod }p)\) for some integer \(k\). Since \(n^{\frac{p-1}{2}}\equiv g^{\frac{k(p-1)}{2}} \equiv 1 \; (\text{mod }p)\), the order of \(g\) must divide the exponent \(\frac{k(p-1)}{2}\). Therefore, \(p-1 | \frac{k(p-1)}{2}\) and thus \(k\) is an even integer. Let's say \(k=2j\) for some integer \(j\). Then, we have

\[ n \equiv g^k \equiv g^{2j} \equiv (g^j)^2 \; (\text{mod }p) \]

This proves that \(n\) is a quadratic residue modulo \(p\).

Lemma 2 (Gauss's Lemma): For any odd prime \(p\), let \(a\) be an integer that is co-prime to \(p\). Consider the integers

\[ S = \left\{ a,\ 2a,\ 3a,\cdots, \frac{p-1}{2}a \right\} \]

and their least positive residues modulo \(p\). Let \(n\) be the number of these residues that are greater than \(p/2\). Then

\[ \left(\frac{a}{p}\right)=(-1)^n \]

Proof: Since \(p\not | a\), none of the integers in \(S\) are congruent to 0 and no two of them are congruent to each other modulo \(p\). Let \(r_1, \cdots, r_m\) be the residues modulo \(p\) smaller than \(\frac{p}{2}\), and let \(s_1, \cdots, s_n\) be the residues modulo \(p\) greater than \(\frac{p}{2}\). Then \(m+n=\frac{p-1}{2}\) and the integers

\[ r_1, \cdots, r_m, p-s_1, \cdots , p-s_n \]

are all positive and less than \(\frac{p}{2}\). Now, we will prove that no two of these integers are equal. Suppose that for some choice of \(i\) and \(j\) we have \(p-s_i = r_j\). We can choose integers \(u\) and \(v\), with \(1 \leq u,v \leq \frac{p-1}{2}\) and satisfying

\[ \begin{aligned} s_i &\equiv u a \; (\text{mod }p) \\ r_j &\equiv v a \; (\text{mod }p) \end{aligned} \]

Now, we have

\[ s_i+r_j \equiv a(u+v) \equiv p \equiv 0 \; (\text{mod }p) \]

This implies that \(u+v \equiv 0 \; (\text{mod }p)\). However, this is not possible because \(1\leq u+v \leq p-1\). Thus, we have proven that the numbers \(r_1,\cdots, r_m, p-s_1, \cdots p-s_n\) are simply a rearrangement of the integers \(1,2,\cdots, \frac{p-1}{2}\). Their product is equal to \(\left(\frac{p-1}{2} \right)!\). Therefore,

\[ \begin{aligned} \left(\frac{p-1}{2}\right)! &= r_1 \cdots r_m (p-s_1)\cdots (p-s_n) \\ &\equiv (-1)^n r_1 \cdots r_m s_1\cdots s_n \; (\text{mod }p) \\ &\equiv (-1)^n a\cdot 2a\cdots \left(\frac{p-1}{2}\right)a \; (\text{mod }p) \\ &\equiv (-1)^n a^{\frac{p-1}{2}} \left( \frac{p-1}{2}\right)! \; (\text{mod }p) \end{aligned} \]

The \(\left( \frac{p-1}{2}\right)!\) term can be cancelled from both sides as \(p\not| \left( \frac{p-1}{2}\right)!\). In other words, we have \(a^{\frac{p-1}{2}}\equiv (-1)^n \; (\text{mod }p)\). This completes the proof of Gauss's lemma.

We are now ready to prove the Quadratic reciprocity theorem.

Proof of the Quadratic Reciprocity Theorem:

Using the periodicity properties of \(\sin\) and Gauss's lemma, it is easy to verify the following result:

Lemma: Let \(p\) and \(q\) be distinct odd primes and let \(A = \left\{ \alpha\in \mathbb{Z} | 1\leq \alpha \leq \frac{p-1}{2}\right\}\) be a half system modulo \(p\). Then,

\[ \begin{equation}\label{eisensteins-proof-of-quadratic-reciprocity-eq-1} \left(\frac{q}{p}\right) = \prod_{\alpha\in A}\frac{\sin\left(\frac{2\pi}{p}q\alpha\right)}{\sin\left(\frac{2\pi}{p}\alpha\right)} \end{equation} \]

We start by examining the right hand side of equation \(\eqref{eisensteins-proof-of-quadratic-reciprocity-eq-1}\). The addition theorem for trigonometric functions yields \(\sin 2\alpha = 2\sin\alpha\cos\alpha\) and \(\sin 3\alpha = \sin\alpha(3-4\sin^2\alpha)\). Induction shows that \(\sin q\alpha = \sin\alpha P(\sin\alpha)\) for all odd \(q\geq 1\), where \(P\in \mathbb{Z}[X]\) is a polynomial of degree \(q-1\) and highest coefficient \((-4)^{\frac{q-1}{2}}\). After factoring out the leading coefficient, there are rational coefficients \(a_i\) such that

\[ \begin{aligned}\frac{\sin qz}{\sin z} &= (-4)^{\frac{q-1}{2}} \left( (\sin z)^{q-1}+a_{q-2} (\sin z)^{q-2}+\cdots + a_0 \right) \\ &= (-4)^{\frac{q-1}{2}} \psi(X), \quad \text{where }X=\sin z\end{aligned} \]

Since \(\phi(z)=\frac{\sin qz}{\sin z}\) is an even function, so is \(\psi(X)\), hence all coefficients of odd powers of \(X\) vanish. Now \(\phi(z)\) has zeros \(\left\{ \pm \frac{2\pi}{q}\beta, \ 1\leq \beta \leq \frac{q-1}{2}\right\}\). Since \(\psi\) is monic of degree \(q-1\), we may write

\[ \psi(X) = \prod_{\beta\in B}\left(X^2-\sin^2\frac{2\pi\beta}{q} \right) \]

where \(B=\left\{1,\cdots,\frac{q-1}{2} \right\}\) is a half system modulo \(q\). Replacing \(X\) by \(\sin z\), we get

\[ \begin{equation}\label{eisensteins-proof-of-quadratic-reciprocity-eq-2} \frac{\sin qz}{\sin z} = (-4)^{\frac{q-1}{2}} \prod_{\beta\in B }\left( \sin^2z -\sin^2\frac{2\pi\beta}{q}\right) \end{equation} \]

Put \(z=\frac{2\pi\alpha}{p}\) in equation \(\eqref{eisensteins-proof-of-quadratic-reciprocity-eq-2}\) and plug the result into equation \(\eqref{eisensteins-proof-of-quadratic-reciprocity-eq-1}\).

\[ \begin{equation}\label{eisensteins-proof-of-quadratic-reciprocity-eq-3} \begin{aligned} \left(\frac{q}{p}\right) &= \prod_{\alpha\in A} (-4)^{\frac{q-1}{2}} \prod_{\beta\in B} \left( \sin^2\frac{2\pi\alpha}{p} -\sin^2\frac{2\pi\beta}{q}\right)\\ &= (-4)^{\frac{q-1}{2} \frac{p-1}{2}} \prod_{\alpha\in A}\prod_{\beta\in B} \left( \sin^2\frac{2\pi\alpha}{p} -\sin^2\frac{2\pi\beta}{q}\right) \end{aligned} \end{equation} \]

Exchanging \(p\) and \(q\) on the right side of \(\eqref{eisensteins-proof-of-quadratic-reciprocity-eq-3}\) give rise to factor of \((-1)^{(p-1)(q-1)/4}\). Therefore,

\[ \begin{equation}\label{eisensteins-proof-of-quadratic-reciprocity-eq-4} \left(\frac{q}{p}\right) = (-1)^{(p-1)(q-1)/4}\left(\frac{p}{q}\right) \end{equation} \]

which is the quadratic reciprocity law.

References

Lemmermeyer, Franz. "Reciprocity Laws: From Euler to Eisenstein". New York, Springer, 2000
Burton, David M. "Elementary Number Theory". New Delhi, Tata McGraw-Hill Publishing Company Limited, May 1 2006

Simplification of Incomplete Beta Functions

2020-01-24T07:47:14+00:00

We will evaluate two algebraic integrals by rewriting them as incomplete beta functions. The first yields to a pair of substitutions; the second leads to the addition and duplication formulas for the Weierstrass elliptic function.

\[ \begin{align} \int_0^1\frac{dx}{\sqrt{1-x}\sqrt[6]{9-x}\sqrt[3]x}&=\frac\pi{\sqrt3} \label{hello-world-eq-1}\\ \int_0^1\frac{dx}{\sqrt{1-x}\sqrt[4]x\sqrt[4]{2-x\sqrt3}} &= \frac{2\sqrt{2}\pi}{3\sqrt[8]{3}} \label{hello-world-eq-2} \end{align} \]

These problems are from a math.stackexchange.com question posted by Vladimir Reshetnikov.

Alternate forms

The equations \(\eqref{hello-world-eq-1}\) and \(\eqref{hello-world-eq-2}\) can be expressed in terms of the incomplete beta function using some elementary hypergeometric identities. This makes our problems equivalent to proving that

\[ \begin{align} B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) &= \frac{\sqrt{3}}{2^{\frac{4}{3}}\pi}\Gamma\left(\frac{1}{3}\right)^3 \label{hello-world-eq-3}\\ B\left(\frac{\sqrt{3}}{2};\frac{1}{4},\frac{1}{4}\right) &= \frac{2}{3\sqrt{\pi}}\Gamma\left(\frac{1}{4}\right)^2 \label{hello-world-eq-4} \end{align} \]

Proof of \(\eqref{hello-world-eq-3}\)

\[ B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) = \int_0^{\frac{1}{9}}x^{-\frac{5}{6}}\left( 1-x\right)^{-\frac{2}{3}} dx \]

Substitute \(x=\frac{y^3}{1+y^3}\) in the above integral to get

\[ B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) =3\int_0^{\frac{1}{2}}\frac{1}{\sqrt{y+y^4}}dy \]

We can now transform this integral into a beta integral using the substitution \(y=\frac{1-z}{2+z}\).

\[ B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) =3\int_0^1\frac{1}{\sqrt{1-z^3}}dz = B\left( \frac{1}{3},\frac{1}{2}\right)=\frac{\sqrt{3}}{2^{\frac{4}{3}}\pi}\Gamma\left(\frac{1}{3}\right)^3 \]

This proves equations \(\eqref{hello-world-eq-1}\) and \(\eqref{hello-world-eq-3}\).

Proof of \(\eqref{hello-world-eq-4}\)

The proof of integrals \(\eqref{hello-world-eq-2}\) and \(\eqref{hello-world-eq-4}\) is more involved. Using the substitution \(x=\frac{4t^2}{1+4t^2}\), the integral transforms into the Weierstrass form:

\[ B\left(\frac{\sqrt{3}}{2};\frac{1}{4},\frac{1}{4}\right) =\int_0^{\frac{\sqrt{3}}{2}}x^{-\frac{3}{4}}(1-x)^{-\frac{3}{4}}dx=2\sqrt{2} \int_0^\alpha \frac{dt}{\sqrt{4t^3+t}} \]

where \(\alpha=\frac{\sqrt{3+2\sqrt{3}}}{2}\). Let \(\wp(z)\) denote the Weierstrass Elliptic function with invariants \(g_2=-1\) and \(g_3=0\). \(\wp(z)\) satisfies the differential equation:

\[ \wp'(z)^2=4\wp(z)^3+\wp(z) \]

The half periods are \(\omega_1=\frac{1+i}{4}L\) and \(\omega_2=\frac{-1+i}{4}L\) where \(L=\frac{1}{\sqrt{2\pi}}\Gamma\left(\frac{1}{4}\right)^2\) is the Lemniscate constant. Now, we can express our integral as follows:

\[ B\left(\frac{\sqrt{3}}{2};\frac{1}{4},\frac{1}{4}\right) =2\sqrt{2}\left(\wp^{-1}(0)-\wp^{-1}(\alpha) \right) \]

Note that the following identity is easy to prove using the beta function:

\[ \int_0^\infty \frac{dt}{\sqrt{4t^3+t}}=\frac{1}{2\sqrt{2\pi}}\Gamma\left(\frac{1}{4}\right)^2=\frac{L}{2} \]

Therefore, \(\wp^{-1}(0)=\frac{L}{2}\). To prove the claim we only need to prove that \(\wp\left(\frac{L}{6}\right)=\alpha\). Using the addition theorem of \(\wp(z)\) and the fact that \(\wp\left(\frac{L}{2}\right)=0\) we get:

\[ \begin{equation}\label{hello-world-eq-5} \begin{aligned} \wp\left(\frac{L}{6}\right) &= \wp\left(\frac{L}{2}-\frac{L}{3}\right)\\ &= \frac{1}{4}\left(\frac{\wp'\left(\frac{L}{3}\right)}{\wp\left(\frac{L}{3}\right)} \right)^2-\wp\left(\frac{L}{3}\right) \\ &= \frac{4\wp\left( \frac{L}{3}\right)^3+\wp\left(\frac{L}{3}\right)}{4\wp\left(\frac{L}{3}\right)^2}-\wp\left(\frac{L}{3}\right)\\ &= \frac{1}{4\wp\left(\frac{L}{3}\right)} \end{aligned} \end{equation} \]

On the other hand, by the duplication theorem:

\[ \begin{equation}\label{hello-world-eq-6} \begin{aligned} \wp\left(\frac{L}{3}\right)&= \frac{1}{4}\frac{\left(6\wp\left(\frac{L}{6}\right)^2+\frac{1}{2}\right)^2}{4\wp\left(\frac{L}{6}\right)^3+\wp\left(\frac{L}{6}\right)}-2\wp\left(\frac{L}{6}\right) \end{aligned} \end{equation} \]

Combine equations \(\eqref{hello-world-eq-5}\) and \(\eqref{hello-world-eq-6}\) to get:

\[ \frac{1+4\wp\left(\frac{L}{6}\right)^2}{\left(6\wp\left(\frac{L}{6}\right)^2+\frac{1}{2} \right)^2-32\wp\left(\frac{L}{6}\right)^4-8\wp\left(\frac{L}{6}\right)^2}=1 \\ \implies 16\wp\left(\frac{L}{6}\right)^4-24\wp\left(\frac{L}{6}\right)^2-3=0 \]

Since \(\wp\left( \frac{L}{6}\right)\) is positive, we conclude that

\[ \wp\left(\frac{L}{6}\right)=\frac{\sqrt{3+2\sqrt{3}}}{2} \]

This completes our proof of equation \(\eqref{hello-world-eq-4}\) and \(\eqref{hello-world-eq-2}\).