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

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!

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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.

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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

$$\frac{1}{n^2+m^2}=\frac{1}{m}\int_{0}^{\infty}\sin(mt)e^{-nt}dt$$

$$S=\sum_{n=1}^{\infty}(-1)^{n+1}\sum_{m=1}^{\infty}\frac{1}{m}\int_{0}^{\infty}\sin(mt)e^{-nt}dt$$

$$S=\int_{0}^{\infty}\sum_{n=1}^{\infty}(-1)^{n+1}e^{-nt} \Big\{\sum_{m=1}^{\infty}\frac{1}{m}\sin(mt) \Big \}dt$$

$$S=\int_{0}^{\infty}\sum_{n=1}^{\infty}(-1)^{n+1}e^{-nt} \Big\{ \frac{\pi}{2}-\frac{t}{2} \Big \}dt$$

$$S=\sum_{n=1}^{\infty}(-1)^{n+1}\int_{0}^{\infty}e^{-nt} \Big\{ \frac{\pi}{2}-\frac{t}{2} \Big \}dt$$

$$S=\sum_{n=1}^{\infty}(-1)^{n+1} \Big\{ \frac{\pi}{2n}-\frac{1}{2n^2}\Big \}$$

$$S=\frac{\pi}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}-\frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$$

$$S=\frac{\pi \log(2)}{2}+ \frac{\pi^2}{24}$$

Answer 1

Show that $$\lim_{s\to 1^+} (s-1)\sum_{n,m\ne 0,0} |n+im|^{-2s}=\lim_{s\to 1^+} (s-1)\sum_{n,m\ne 0,0} (n^2+m^2)^{-s}$$ $$= \lim_{s\to 1^+} (s-1) \int_{|x|>1,|y|>1} (x^2+y^2)^{-s}dxdy=\pi$$ From $(1-|1+i|^{-2s}) \sum_{n,m\ne 0,0} |n+im|^{-2s}$ $=\sum_{n,m, 2\ \nmid \ |n+im|^2} |n+im|^{-2s}$ $= 2\sum_{n\ne 0,m} |2n+i(2m+1)|^{-2s}$ you'll get for $s >1$

$$ \begin{eqnarray}F(s)&=&\sum_{n\ge 1,m\ge 1} (-1)^{n-1} (n^2+m^2)^{-s}\\ &=&\frac14\left(\sum_{n,m\ne 0,0} (-1)^{n-1} |n+im|^{-2s} -\sum_{n\ne 0}(-1)^{n-1} |n|^{-2s}+\sum_{m\ne 0}|m|^{-2s}\right) \\&=& \frac14(1- (1-2^{-s}) -2^{1-2s}) \sum_{n,m\ne 0,0} |n+im|^{-2s}+ 2^{-2s} \zeta(2s) \end{eqnarray}$$

$\zeta(2)=\pi^2/6$ will give $$\sum_{n\ge 1}\sum_{m\ge 1} (-1)^{n-1}(n^2+m^2)^{-1} =\lim_{s\to 1^+}F(s)=\frac{\pi}8\log 2 + \frac{\pi^2}{24}$$ where I'm using that $\sum_{n\ge 1,m\ge 1} ( (2n-1)^2+m^2)^{-1}- ((2n)^2+m^2)^{-1})$ is absolutely convergent to write the double series as $\lim_{s\to 1^+}F(s)$.

Answer 2

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 $$S=\frac{\pi^2}{24}+\frac{\pi\log 2}{8}$$