Why is $\sum_{k=1}^\infty (-1)^{k+1}\frac{ \text{csch}(\pi k)}{k} = \frac{1}{2}(\ln(64) - \pi)$? While answering this question I found through Wolfram Alpha that
$$\sum_{k=1}^\infty (-1)^{k+1}\frac{ \text{csch}(\pi k)}{k} = \frac{1}{12}(\ln(64) - \pi).$$
Sadly Wolfram Alpha just spits the summation value out without any justification or references.
Is there an explanation or reference for this identity?
 A: The meromorphic function $f(z)=\frac{1}{z\sinh(\pi z)}$ has a double pole at the origin and simple poles at $z\in\{\pm i,\pm 2i,\pm 3i,\ldots\}$. By computing the residues at such points it follows that:
$$ \frac{\text{csch}(\pi z)}{z} = \frac{1}{\pi z^2}+\frac{i}{\pi}\sum_{m\geq 1}\frac{(-1)^{m+1}}{m}\left(\frac{1}{z-mi}-\frac{1}{z+mi}\right)\tag{1} $$
and now we need to evaluate both sides at $z=k$, multiply by $(-1)^{k+1}$ and sum over $k\geq 1$.
The contribute provided by the double pole at the origin is simple to compute:
$$ \sum_{k\geq 1}\frac{(-1)^{k+1}}{\pi k^2} = \frac{\eta(2)}{\pi} = \frac{\pi}{12}\tag{2} $$
It remains to evaluate:
$$ \frac{2}{\pi}\sum_{k\geq 1}(-1)^{k+1}\sum_{m\geq 1}\frac{(-1)^{m+1}}{m^2+k^2}\tag{3}$$
and that is possible through the identities
$$ \frac{1}{m^2+k^2} = \int_{0}^{+\infty}\frac{\sin(mx)}{m}e^{-kx}\,dx = \int_{0}^{+\infty}\cos(mx)\frac{e^{-kx}}{k}\,dx.\tag{4} $$
and Fourier series, or by recognizing in $(3)$ a Jacobi theta function. The final outcome is
$$ \sum_{k\geq 1}(-1)^{k+1}\frac{\text{csch}(\pi k)}{k}=\color{red}{\frac{\log 2}{2}-\frac{\pi}{12}}.\tag{5}$$
