In this Physics Stack Exchange post, I detailed my discovery of the following equation $$\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{(-1)^{i+j}}{(2i+1)(2j+1)\cosh\left(\frac{\sqrt{(2i+1)^2+(2j+1)^2}\pi}{2}\right)} = \frac{\pi^2}{48},$$
through solving a physics problem in two different ways.
Question: I was wondering if there exists any other proof of this equation, that is more 'mathematical'?
Here, I mean 'mathematical', in the sense that it does not delve into the physical context of a specific problem as done in the post, but rather relies on theorems of pure mathematics to prove the result.
$\DeclareMathOperator{\sech}{sech}$After a few days, I was able to construct a proof of this via the help of the Mittag-Leffer Expansion of $\sech(\pi z)$ which shows that:
$$\sech (\pi z) = \frac{4}{\pi} \sum_{k \mathop = 0}^\infty (-1)^k \frac {2 k + 1} { 4 z^2 + (2 k + 1)^2}$$
The double summation thus becomes,
$$ \begin{align*} \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{(-1)^{i+j}\sech\left(\frac{\sqrt{(2i+1)^2+(2j+1)^2}\pi}{2}\right)}{(2i+1)(2j+1)} &= \frac{4}{\pi} \sum_{i, j, k \geq 0} \frac{(-1)^{i+j+k}(2k+1)}{(2i+1)(2j+1)[(2i+1)^2+(2j+1)^2+(2k+1)^2]}\\ &= \frac{4}{\pi}\sum_{i, j, k \geq 0} \frac{(-1)^{i+j+k}(2k+1)^2}{(2i+1)(2j+1)(2k+1)[(2i+1)^2+(2j+1)^2+(2k+1)^2]} \\ \end{align*} $$
Notice that $i, j, k$ are interchangeable. Letting the sum be $S$, we get,
$$ \begin{align*} S &= \frac{4}{\pi}\sum_{i, j, k \geq 0} \frac{(-1)^{i+j+k}(2k+1)^2}{(2i+1)(2j+1)(2k+1)[(2i+1)^2+(2j+1)^2+(2k+1)^2]} \\ &= \frac{4}{\pi}\sum_{i, j, k \geq 0} \frac{(-1)^{i+j+k}(2i+1)^2}{(2i+1)(2j+1)(2k+1)[(2i+1)^2+(2j+1)^2+(2k+1)^2]} \\ &= \frac{4}{\pi}\sum_{i, j, k \geq 0} \frac{(-1)^{i+j+k}(2j+1)^2}{(2i+1)(2j+1)(2k+1)[(2i+1)^2+(2j+1)^2+(2k+1)^2]} \\ \end{align*}$$ Summing them, $$ \begin{align*} 3S &= \frac{4}{\pi}\sum_{i, j, k \geq 0} \frac{(-1)^{i+j+k}[(2i+1)^2 + (2j+1)^2 + (2k+1)^2]}{(2i+1)(2j+1)(2k+1)[(2i+1)^2+(2j+1)^2+(2k+1)^2]} \\ &= \frac{4}{\pi}\sum_{i, j, k \geq 0} \frac{(-1)^{i+j+k}}{(2i+1)(2j+1)(2k+1)} \\ \end{align*}$$
Again, since $i, j, k$ are interchangeable, we get,
$$ S = \frac{4}{3\pi}\left(\sum_{i = 0}^{\infty} \frac{(-1)^i}{2i+1}\right)^3$$
Note that $\sum_{i = 0}^{\infty} \frac{(-1)^i}{2i+1}$ is the famous Leibniz formula for $\pi$ which is exactly equal to $\frac{\pi}{4}$. We then get the final result:
$$ \begin{align*} S &= \frac{4}{3\pi}\left(\frac{\pi}{4}\right)^3 \\ &= \frac{\pi^2}{48}\\ \end{align*}$$