Introduction to Theta Functions I

Posted on May 12, 2021 by Shobhit Bhatnagar

The Jacobi theta functions are defined for all complex variables of $z$ and $q$ such that $|q| < 1$, as follows: $$ \begin{aligned} \vartheta_1 (z,q) &= -i \sum_{n=-\infty}^{\infty} (-1)^n q^{(n+1/2)^2} e^{i(2n+1)z} \\ \vartheta_2 (z,q) &= \sum_{n=-\infty}^{\infty} q^{(n+1/2)^2} e^{i(2n+1)z} \\ \vartheta_3 (z,q) &= \sum_{n=-\infty}^{\infty} q^{n^2} e^{2inz} \\ \vartheta_4 (z,q) &= \sum_{n=-\infty}^{\infty} (-1)^n q^{n^2} e^{2inz} \\ \end{aligned} $$ 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: $$ \begin{aligned} \vartheta_3(x,q)\vartheta_3(y,q) &= \sum_{r=-\infty}^{\infty} \sum_{s=-\infty}^{\infty} q^{2(r^2 +s^2)}e^{2i(r(x+y)+s(x-y))} \\ &\quad + \sum_{r=-\infty}^{\infty} \sum_{s=-\infty}^{\infty}q^{2\left[\left(r+\frac{1}{2} \right)^2 + \left(s+\frac{1}{2} \right)^2\right]} e^{i(2r+1)(x+y) + i(2s+1)(x-y)} \\ &= \vartheta_3(x+y,q^2)\vartheta_3(x-y,q^2) + \vartheta_2(x+y,q^2)\vartheta_2(x-y,q^2) \quad (1) \end{aligned} $$ Some similar identities that can be derived using this method are: $$ \begin{aligned} \vartheta_1(x,q) \vartheta_1(y,q) &= \vartheta_3(x+y,q^2)\vartheta_2(x-y,q^2) – \vartheta_2(x+y,q^2)\vartheta_3(x-y,q^2) \quad (2) \\ \vartheta_2(x,q) \vartheta_2(y,q) &= \vartheta_2(x+y,q^2)\vartheta_3(x-y,q^2) + \vartheta_3(x+y,q^2)\vartheta_2(x-y,q^2) \quad (3) \\ \vartheta_4(x,q) \vartheta_4(y,q) &= \vartheta_3(x+y,q^2)\vartheta_3(x-y,q^2) – \vartheta_2(x+y,q^2)\vartheta_2(x-y,q^2) \quad (4) \\ \vartheta_1(x,q) \vartheta_2(y,q) &= \vartheta_1(x+y,q^2)\vartheta_4(x-y,q^2) + \vartheta_4(x+y,q^2)\vartheta_1(x-y,q^2) \quad (5) \\ \vartheta_3(x,q) \vartheta_4(y,q) &= \vartheta_4(x+y,q^2)\vartheta_4(x-y,q^2) – \vartheta_1(x+y,q^2)\vartheta_1(x-y,q^2) \quad (6) \end{aligned} $$ Squaring and subtracting identities (4) and (2) gives us: $$ \begin{aligned} &\; \vartheta_4^2(x,q) \vartheta_4^2(y,q) – \vartheta_1^2(x,q) \vartheta_1^2(y,q)\\ &= \left[\vartheta_3^2(x+y,q^2)-\vartheta_2^2(x+y,q^2) \right]\times \left[\vartheta_3^2(x-y,q^2)-\vartheta_2^2(x-y,q^2) \right] \quad (7) \end{aligned} $$ 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 (7) gives us: $$ \vartheta_4(x+y,q)\vartheta_4(x-y,q)\vartheta_4^2(0,q) = \vartheta_4^2(x,q) \vartheta_4^2(y,q) – \vartheta_1^2(x,q) \vartheta_1^2(y,q) \quad (8) $$ 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 $\frac{\pi}{2},\frac{\pi \tau}{2},\frac{\pi+\pi\tau}{2}$: $$ \begin{aligned} \vartheta_1(x+y)\vartheta_1(x-y)\vartheta_4^2(0) &= \vartheta_3^2(x) \vartheta_2^2(y) – \vartheta_2^2(x) \vartheta_3^2(y) \\ &= \vartheta_1^2(x) \vartheta_4^2(y) – \vartheta_4^2(x) \vartheta_1^2(y)\quad (9) \\ \vartheta_2(x+y)\vartheta_2(x-y)\vartheta_4^2(0) &= \vartheta_4^2(x) \vartheta_2^2(y) – \vartheta_1^2(x) \vartheta_3^2(y) \\ &= \vartheta_2^2(x) \vartheta_4^2(y) – \vartheta_3^2(x) \vartheta_1^2(y)\quad (10) \\ \vartheta_3(x+y)\vartheta_3(x-y)\vartheta_4^2(0) &= \vartheta_4^2(x) \vartheta_3^2(y) – \vartheta_1^2(x) \vartheta_2^2(y) \\ &= \vartheta_3^2(x) \vartheta_4^2(y) – \vartheta_2^2(x) \vartheta_1^2(y)\quad (11) \\ \vartheta_4(x+y)\vartheta_4(x-y)\vartheta_4^2(0) &= \vartheta_3^2(x) \vartheta_3^2(y) – \vartheta_2^2(x) \vartheta_2^2(y) \\ &= \vartheta_4^2(x) \vartheta_4^2(y) – \vartheta_1^2(x) \vartheta_1^2(y)\quad (12) \\ \end{aligned} $$ One can keep on deriving similar identities by squaring and adding equations (1) and (2) and squaring and subtracting equations (3) and (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: $$ \vartheta_2^2(0) \vartheta_4^2(z) -\vartheta_3^2(0) \vartheta_1^2(z) = \vartheta_4^2(0) \vartheta_2^2(z) \quad (13) $$ A similar technique yields the identity: $$ \vartheta_3^2(0) \vartheta_4^2(z) – \vartheta_2^2(0) \vartheta_1^2(z) = \vartheta_4^2(0) \vartheta_3^2(z) \quad (14) $$ Incrementing $z$ by $\frac{\pi}{2}$ in (11) and (12) gives two additional identities: $$ \begin{aligned} \vartheta_4^2(0) \vartheta_1^2(z) &= \vartheta_2^2(0) \vartheta_3^2(z) – \vartheta_3^2(0) \vartheta_2^2(z) \quad (15) \\ \vartheta_4^2(0) \vartheta_4^2(z) &= \vartheta_3^2(0) \vartheta_3^2(z) – \vartheta_2^2(0) \vartheta_2^2(z) \quad (16) \end{aligned} $$ With these relations one can express any theta function in terms of any other pair of theta functions. Putting $z=0$ in (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 (5) with respect to $x$ and substituting $x=y=0$, we get: $$ \vartheta_1′(0,q) \vartheta_2(0,q) = 2 \vartheta_1′(0,q^2)\vartheta_4(0,q^2) \quad (17) $$ Substituting $x=y=0$ in equations (3) and (6) gives: $$ \begin{aligned} \vartheta_2^2(0,q) &= 2\vartheta_2(0,q^2) \vartheta_3(0,q^2) \quad (18) \\ \vartheta_3(0,q)\vartheta_4(0,q) &= \vartheta_4^2(0,q^2) \quad (19) \end{aligned} $$ Now, dividing (17) by both (18) and (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: $$ \frac{\vartheta_1′(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)} = \lim_{q\to 0}\frac{\vartheta_1′(0,q)}{\vartheta_2(0,q)\vartheta_3(0,q)\vartheta_4(0,q)} \quad (20) $$ 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 (20) equals 1 and $$\vartheta_1′(0)=\vartheta_2(0)\vartheta_3(0)\vartheta_4(0)$$ is established as desired.

References

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

Posted in Elliptic Functions, Elliptic Integrals | Tagged | Leave a comment

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

Posted on December 12, 2020 by Shobhit Bhatnagar

It is easy to see with the help of 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 it’s order is given by the Euler’s Totient function: $\varphi(n) = |(\mathbb{Z}/p\mathbb{Z})^*| $. 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 $s \nmid p$. 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.

The fact that $(\mathbb{Z}/n\mathbb{Z})^*$ can be cyclic only for the cases discussed above can also be verified easily using the Chinese remainder theorem.

References

  • M. Ram Murty. Problems in analytic number theory. Springer New York, 01-Nov-2008
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Evaluating very nasty logarithmic integrals: Part III

Posted on September 7, 2020 by Shobhit Bhatnagar

This is part 3 of our series on very nasty logarithmic integrals. Please have a look at part 1 and part 2 before reading this post.

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 anti-derivative 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 $1-\epsilon$.
  • $\gamma_{4,\epsilon}$ 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 $$ \begin{aligned} \lim_{\epsilon\to 0^+} \int_{\gamma_{1,\epsilon}} f(z)\; dz &= \int_0^1\frac{\arctan(x)}{1+x^2}\left(\text{arctanh}^2(x) + \frac{\pi^2}{16} \right)dx \\ &= \int_0^1\frac{\arctan(x) \text{arctanh}^2(x) }{1+x^2}dx + \frac{\pi^2}{16} \frac{\arctan^2(x)}{2}\Big|_0^1 \\ &= \frac{1}{4}\int_0^1 \frac{\arctan\left(\frac{1-x}{1+x} \right)\log^2(x)}{1+x^2}dx + \frac{\pi^4}{512} \quad \left(x\mapsto \frac{1-x}{1+x} \right) \\ &= \frac{1}{4}\int_0^1 \frac{\left(\frac{\pi}{4}-\arctan(x) \right)\log^2(x)}{1+x^2}dx + \frac{\pi^4}{512} \\ &= -\frac{I_1}{4} +\frac{\pi}{16}\int_0^1 \frac{\log^2(x)}{1+x^2}dx +\frac{\pi^4}{512} \\ &= -\frac{I_1}{4} + \frac{3\pi^4}{512}\quad \color{blue}{\cdots (1)} \end{aligned} $$ Here, we used the fact that $\int_0^1 \frac{\log^2(x)}{1+x^2}dx = \frac{\pi^3}{16}$. Similarly, on $\gamma_{4,\epsilon}$ we have $$ \begin{aligned} \lim_{\epsilon\to 0^+} \int_{\gamma_{5,\epsilon}} f(z) \; dz &= \int_i^0 \frac{\arctan(x)}{1+x^2}\left(\text{arctanh}^2(x) + \frac{\pi^2}{16} \right)dx \\ &= \int_0^1 \frac{\text{arctanh}(x)}{1-x^2}\left(\frac{\pi^2}{16}-\arctan^2(x) \right)dx \quad \left(x\mapsto i x \right) \\ &= \int_0^1\frac{\arctan(x) \text{arctanh}^2(x) }{1+x^2}dx \quad (\text{IBP}) \\ &= -\frac{I_1}{4}+\frac{\pi^4}{256} \quad \color{blue}{\cdots (2)} \end{aligned} $$

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 $$ \begin{aligned} \text{Re}\left[\lim_{\epsilon\to 0^+} \int_{\gamma_{3,\epsilon}}f(z) \; dz \right] &= \int_0^\frac{\pi}{2} \frac{\frac{1}{16}\log^2\left(\tan\frac{\theta}{2} \right)\log\left(\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}} \right)+\frac{\pi^2}{32}\log\left(\tan\frac{\theta}{2} \right)}{\cos \theta} d\theta \\ &= \int_0^1 \frac{\frac{1}{8}\log^2\left(u \right)\log\left(\frac{1-u}{1+u} \right)+\frac{\pi^2}{16}\log\left(u \right)}{1-u^2} du \quad \left(u= \tan\frac{\theta}{2} \right) \\ &= -\frac{1}{8} \int_0^1 \frac{\log^2(u)\log\left(\frac{1+u}{1-u}\right)}{1-u^2} + \frac{\pi^2}{16}\left(-\frac{\pi^2}{8} \right) \\ &= \frac{1}{16}\int_0^1 \frac{\log(u) \log^2\left(\frac{1-u}{1+u} \right)}{u}du – \frac{\pi^4}{128} \quad \color{blue}{\cdots (3)} \end{aligned} $$ 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 (1), (2) and (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 $$ I_2 = \int_0^1 \frac{\log^3(x) \arctan(x)}{1+x^2}dx \quad \color{blue}{\cdots (4)} $$ Using the transformation $x\mapsto \frac{1}{x}$, we write the integral as: $$ \begin{aligned} I_2 &= -\int_1^\infty\frac{\log^3(x)\left(\frac{\pi}{2}-\arctan(x) \right)}{1+x^2}dx \quad \color{blue}{\cdots (5)} \end{aligned} $$ Adding up equations (4) and (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 $$ I_2 = – \frac{3\pi}{2}\beta(4) + \frac{1}{2}\int_0^\infty \frac{\log^3(x) \arctan(x)}{1+x^2}dx \quad \color{blue}{\cdots (6)} $$ 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: $$ \begin{aligned} \int_0^\infty \frac{x\left((\log (x)-i\pi)^2 – (\log (x)+i\pi)^2 \right)}{(1+x^2)(1+s^2 x^2)}dx &= 2i\pi \Big(\mathop{\text{Res}}\limits_{z=i} \; g(z)+ \mathop{\text{Res}}\limits_{z=-i}\; g(z) + \mathop{\text{Res}}\limits_{z=i/s}\; g(z) \\ &\quad + \mathop{\text{Res}}\limits_{z=-i/s}\; g(z) \Big) \\ \implies -4i\pi \int_0^\infty \frac{x\log(x)}{(1+x^2)(1+s^2 x^2)}dx &= -2i\pi \frac{\log^2(s)}{1-s^2} \\ \implies \int_0^\infty \frac{x\log(x)}{(1+x^2)(1+s^2 x^2)}dx &= \frac{\log^2(s)}{2(1-s^2)} \quad \color{blue}{\cdots (7)} \end{aligned} $$ 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: $$ \begin{aligned} \int_0^\infty \frac{x\left((\log (x)-i\pi)^4 – (\log (x)+i\pi)^4 \right)}{(1+x^2)(1+s^2 x^2)}dx &= 2i\pi \Big(\mathop{\text{Res}}\limits_{z=i} \; h(z)+ \mathop{\text{Res}}\limits_{z=-i}\; h(z) + \mathop{\text{Res}}\limits_{z=i/s}\; h(z) \\ &\quad + \mathop{\text{Res}}\limits_{z=-i/s}\; h(z) \Big) \\ \implies -8i\pi \int_0^\infty \frac{x\left(\log^3(x) -\pi^2 \log(x) \right)}{(1+x^2)(1+s^2 x^2)}dx &= 2i\pi \frac{\frac{3\pi^2}{2}\log^2(s)-\log^4(s)}{1-s^2} \\ \implies \int_0^\infty \frac{x\log^3(x)}{(1+x^2)(1+s^2 x^2)}dx &= \frac{\pi^2\log^2(s)+2\log^4(s)}{8(1-s^2)} \quad \color{blue}{\cdots (8)} \end{aligned} $$ Note that we used equation (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.

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