closed form for a double sum

Question

How can I prove that $$\underset{k\geq1}{\sum}\left(\underset{m=-\infty}{\overset{\infty}{\sum}}\frac{\left(-1\right)^{m}}{\left(2k-1\right)^{2}+m^{2}}\right)=\frac{\pi\log\left(2\right)}{8}\,?$$I tried possion summation but it seems doesn't work.

Answer 1

We can exploit $$ \int_{0}^{+\infty}\frac{\sin(at)}{a}e^{-bt}\,dt=\frac{1}{a^2+b^2} $$ giving: $$\begin{eqnarray*}\sum_{m\in\mathbb{Z}}\frac{(-1)^m}{(2k-1)^2+m^2} &=& \frac{1}{(2k-1)^2}+2\sum_{m\geq 1}\int_{0}^{+\infty}\frac{\sin((2k-1)t)}{2k-1}(-1)^m e^{-mt}\,dt\\&=&\frac{1}{(2k-1)^2}-2\int_{0}^{+\infty}\frac{\sin((2k-1)t)}{2k-1}\frac{dt}{e^t+1}.\end{eqnarray*}$$ Since: $$g(t)=\sum_{k\geq 1}\frac{\sin((2k-1)t)}{2k-1}$$ is the rectangular wave whose value is $\frac{\pi}{4}$ on $(0,\pi)$, $-\frac{\pi}{4}$ on $(\pi,2\pi)$ and so on, the original series equals: $$\frac{\pi^2}{8}-2\int_{0}^{+\infty}\frac{g(t)}{e^t+1}\,dt.$$ Since: $$ \int_{k\pi}^{(k+1)\pi}\frac{dt}{e^t+1}=\log(1+e^{-k\pi})-\log(1+e^{-(k+1)\pi})$$ the final answer can be derived from the Weierstrass product for the hyperbolic cosine function.