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

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)).

$$ \sum_{m,n=1}^\infty (-1)^{m+n}/(m^2+n^2)= \frac{1}{4}(\frac{\pi^2}{3} - \pi\log(2) )$$ 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.