Lobachevsky-type logarithmic integrals via partial-fraction series

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

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