TITLE: Compute Double Sum $\sum_{n,m=1}^{\infty}\frac{(-1)^{n-1}}{n^2+m^2}=\frac{\pi^2}{24}+\frac{\pi \ln(2)}{8}$
QUESTION [20 upvotes]: EDIT
After a long search, it seems that the solution is related to Jacobi elliptical functions.
For instance, if we try using this itegral representation
$$\boxed{\frac{1}{m^2+ n^2}=\int_{0}^{\infty}e^{-(m^2+n^2)t}dt}$$
We end up with sums of the following form, which as far as I found, are related to Jacobi elliptical functions.
$$\sum_{m=1}^{\infty}e^{-m^2t}$$
Another strategy would be starting with the following identity
$$\boxed{\sum_{n=1}^{\infty}\frac{(-1)^n}{m^2+n^2}=\frac{\pi \mathrm{csch}(\pi m)}{2m}-\frac{1}{2m^2}}$$
Also leads to evaluate a sum of the form below, which also belongs to the elliptical functions family
$$\sum_{m=1}^{\infty}\frac{1}{m \sinh(\pi m)}$$
Even the infinite product
$$\prod_{k=1}^{\infty} \Big(1+e^{-\pi k2} \Big)$$
seems to be related to these special functions!

First attempt
$$\boxed{\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{(-1)^{n-1}}{n^2+m^2}=\frac{\pi^2}{24}+\frac{\pi \ln(2)}{8}}$$
Consider
$$S=\sum_{n,m=1}^{\infty}\frac{(-1)^{n-1}}{n^2+m^2}$$
$$S=-\sum_{n=1}^{\infty}(-1)^n\sum_{m=1}^{\infty}\frac{1}{n^2+m^2}$$
Recall
$$\boxed{\sum_{m=1}^{\infty}\frac{1}{n^2+m^2}=\frac{\pi \coth(\pi n)}{2n}-\frac{1}{2n^2}}$$
$$S=-\sum_{n=1}^{\infty}(-1)^n\frac{\pi \coth(\pi n)}{2n}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}$$
$$S=-\frac{\pi}{2}\sum_{n=1}^{\infty}(-1)^n\frac{ \coth(\pi n)}{n}-\frac{\pi^2}{24}$$
Also
$$\boxed{\coth(\pi n)-1=\frac{2}{e^{2 \pi n}-1}}$$
$$S=-\frac{\pi}{2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n} \bigg\{\frac{2}{e^{2 \pi n}-1}+1 \bigg\}-\frac{\pi^2}{24}$$
$$S=-{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n}{n} \bigg\{\frac{1}{e^{2 \pi n}-1}\bigg\}-\frac{\pi}{2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n}-\frac{\pi^2}{24}$$
$$S=-{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n}{n} \bigg\{\frac{1}{e^{2 \pi n}-1}\bigg\}+\frac{\pi \ln(2)}{2}-\frac{\pi^2}{24}$$
$$S=-{\pi}\sum_{n=1}^{\infty}\frac{(-1)^ne^{-2 \pi n}}{n}\sum_{k=0}^{\infty}e^{-2 \pi n k} +\frac{\pi \ln(2)}{2}-\frac{\pi^2}{24}$$
$$S=-\pi\sum_{k=0}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^ne^{-2 \pi n(k+1)}}{n}+\frac{\pi \ln(2)}{2}-\frac{\pi^2}{24}$$
$$S=\pi\sum_{k=1}^{\infty}\ln(1+e^{-2\pi k})+\frac{\pi \ln(2)}{2}-\frac{\pi^2}{24}$$
$$S=\pi\ln(\prod_{k=1}^{\infty}1+\big(e^{-\pi k}\big)^2)+\frac{\pi \ln(2)}{2}-\frac{\pi^2}{24}$$
From this point on, I can´t finish. Any help is welcome.

Second attempt
I tried the method suggest bellow by @NoName, but I still missing one factor $\frac{1}{4}$ before the $\log(2)$. I suspect that this method fails because $\sum_{m=1}^{\infty}\frac{1}{m}\sin(mt)=\frac{\pi}{2}-\frac{t}{2}$ is only valid betweem $0$ and $2 \pi$.
$$S=\sum_{n=1}^{\infty}(-1)^{n+1}\sum_{m=1}^{\infty}\frac{1}{n^2+m^2}$$
rewrite
$$\frac{1}{n^2+m^2}=\frac{1}{m}\int_{0}^{\infty}\sin(mt)e^{-nt}dt$$
$$S=\sum_{n=1}^{\infty}(-1)^{n+1}\sum_{m=1}^{\infty}\frac{1}{m}\int_{0}^{\infty}\sin(mt)e^{-nt}dt$$
$$S=\int_{0}^{\infty}\sum_{n=1}^{\infty}(-1)^{n+1}e^{-nt} \Big\{\sum_{m=1}^{\infty}\frac{1}{m}\sin(mt) \Big \}dt$$
$$S=\int_{0}^{\infty}\sum_{n=1}^{\infty}(-1)^{n+1}e^{-nt} \Big\{ \frac{\pi}{2}-\frac{t}{2} \Big \}dt$$
$$S=\sum_{n=1}^{\infty}(-1)^{n+1}\int_{0}^{\infty}e^{-nt} \Big\{ \frac{\pi}{2}-\frac{t}{2} \Big \}dt$$
$$S=\sum_{n=1}^{\infty}(-1)^{n+1} \Big\{ \frac{\pi}{2n}-\frac{1}{2n^2}\Big \}$$
$$S=\frac{\pi}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}-\frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$$
$$S=\frac{\pi \log(2)}{2}+ \frac{\pi^2}{24}$$

REPLY [4 votes]: I liked your first approach and the following answer completes it.
Let $q=e^{-\pi} $ and you wish to evaluate the product $$F(q)=\prod_{n=1}^{\infty} (1+q^{2n})\tag{1}$$ We can rewrite the above product as $$F(q) =\prod_{n\geq 1}\frac{1+q^n}{1+q^{2n-1}}=\prod_{n\geq 1}\frac{1-q^{2n}}{(1-q^n)(1+q^{2n-1})}=\prod_{n\geq 1}\frac{1}{(1-q^{2n-1})(1+q^{2n-1})} \tag{2}$$ Now we bring some elliptic integrals and a bit of Ramanujan into picture. Let $k$ be the elliptic modulus corresponding to nome $q$ and $K$ be the corresponding complete elliptic integral of first kind and $k'=\sqrt{1-k^2}$.
We have by definition of Ramanujan class invariant $$G_m=G(q) =2^{-1/4}q^{-1/24}\prod_{n\geq 1}(1+q^{2n-1})\tag{3}$$ and $$g_m=g(q) =2^{-1/4}q^{-1/24}\prod_{n\geq 1}(1-q^{2n-1}) \tag{4}$$ where $q=e^{-\pi\sqrt{m}} $. Here we have $q=e^{-\pi} $ so that $m=1$. And the class invariants defined above are also linked to the elliptic moduli $k, k'$ via $$G(q) =(2kk') ^{-1/12}, g_m=g(q)=(2k/k'^2)^{-1/12} \tag{5}$$ For $q=e^{-\pi} $ we have $k=k'=1/\sqrt{2}$ and thus  $$G_1=G(q)=1, g_1=g(q)=2^{-1/8}\tag{6}$$ Now $$G(q) g(q) =2^{-1/2}q^{-1/12}/F(q)$$ and putting $q=e^{-\pi} $ and using $(6)$ we get $$F(q) = 2^{-3/8}e^{\pi/12}\tag{7}$$ Based on this your sum in question is $$S=\pi\log F(q) +\frac{\pi\log 2}{2}-\frac{\pi^2}{24}$$ which via $(7)$ equals $$S=\frac{\pi^2}{24}+\frac{\pi\log 2}{8}$$