TITLE: An interesting identity involving Jacobi $\theta_4$ and $\zeta(2)$
QUESTION [16 upvotes]: 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:

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;
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;
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.

REPLY [8 votes]: 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)).