An interesting identity involving Jacobi $\theta_4$ and $\zeta(2)$

Question

A recent question mentioned an integral identity involving Dedekind $\eta$ function and a special value for the complete elliptic integral of the first kind. I refrained from providing a complete answer, I rather tried to guide the OP through some hints, but we apparently reached a dead spot, concerning the following simplified version of the original problem: $$\boxed{ \int_{0}^{+\infty}\left[\sum_{n\geq 1}(-1)^n e^{-n^2 x}\right]^2\,dx = \frac{\pi^2-3\pi\log 2}{12}.} \tag{A}$$ My solution goes as follows:

1. The LHS of $(A)$ can be written in terms of $\sum_{m,n\geq 1}\frac{(-1)^{m+n}}{m^2+n^2}$, to be dealt with care since it is not absolutely convergent;
2. We have $\frac{(-1)^{m+n}}{m^2+n^2}=\int_{0}^{+\infty}\frac{(-1)^n\sin(nx)}{n}(-1)^m e^{-mx}\,dx$, where $\sum_{m\geq 1}(-1)^m e^{-mx}$ is a simple geometric series and $\sum_{n\geq 1}\frac{(-1)^n \sin(nx)}{n}=-\arctan\tan\frac{x}{2}$ almost everywhere;
3. The problem boils down to integrating over $\mathbb{R}^+$ the product between a sawtooth wave and the function $\frac{1}{e^x+1}$. Through the dilogarithms machinery or the residue theorem, to reach the RHS of $(A)$ is not difficult.

I would use this question for collecting alternative/shorter/slicker proofs.

Answer 1

Using the Poisson sommation formula, one can show that \begin{equation} \sum_{n=1}^\infty (-1)^n e^{-n^2x}=-\frac{1}{2}+\sqrt{\frac{\pi}{x}}\sum_{n=1}^\infty e^{-\frac{(2n-1)^2\pi^2}{4x}} \end{equation} then, \begin{align} I&=\int_{0}^{+\infty}\left[\sum_{n\geq 1}(-1)^n e^{-n^2 x}\right]^2\,dx\\ &=-\frac{1}{2}\int_{0}^{+\infty}\sum_{p\geq 1}(-1)^p e^{-p^2 x}\,dx+ \sqrt{\pi}\sum_{n,p\geq 1}\int_{0}^{+\infty}(-1)^p e^{-p^2 x-\frac{(2n-1)^2\pi^2}{4x}}\frac{dx}{\sqrt{x}}\\ &=\frac{\pi^2}{24}+\pi\sum_{n,p\geq 1}\frac{(-1)^p}{p}e^{-(2n-1)p\pi}\\ &=\frac{\pi^2}{24}-\pi\sum_{n\geq 1}\ln\left(1+e^{-(2n-1)\pi} \right)\\ &=\frac{\pi^2}{24}+\frac \pi 2 \sum_{p\geq 1}\frac{(-1)^p}{p\sinh p\pi} \end{align} The integral representation of $K_{-1/2}(.)$ was used to evaluate the last integral.

The obtained results seem numerically correct, however, I couldn't succeed in expressing them as $I=\pi^2/12-\pi/4\ln(2)$, for example showing that $$\prod_{n\geq 1}\left(1+e^{-(2n-1)\pi} \right)\stackrel{?}{=}2^{1/4}e^{-\pi/24}$$ Edit: Finally, a proof of this last identity can be found in this article by Xu Ce (expression (6.3)).

Answer 2

$$ \sum_{m,n=1}^\infty (-1)^{m+n}/(m^2+n^2)= \frac{1}{4}(\frac{\pi^2}{3} - \pi\log(2) )$$ The proof will rely on a lattice sum identity, and series manipulation. Lattice sums are interesting because they generalize vastly, and the proper machinery is Dirichlet series. Below, the brackets will indicate the argument within the double sum.

$$ \sum_{m,n=1}^\infty \{ \ \} = \frac{1}{4}\big( \sum_{m,n=-\infty}^\infty'\{ \ \}-4 \sum_{m,n=1}^\infty (-1)^m/m^2 \big) $$ where the prime on the summation means that the term $m=0$ and $n=0$ is excluded. But from lattice sum theory (see Wolfram site for notation)

$$ \sum_{m,n=-\infty}^\infty' (-1)^{m+n}/(m^2+n^2)= \lim_{s\to 1} \big(-4\ \beta(s)\ \eta(s)\ \big). $$

It is then just a matter of looking up the closed forms, and likewise the sum over the single variable, to get the expression at the top of this answer.