I've been playing with lots of Poisson sums lately, and I thought this one to be interesting: $$\sum_{k\in\mathbb{Z}}\left(\frac{1}{(k+x)\sinh{(k+x)\pi q}}-\frac{1}{\pi q (k+x)^2}\right)$$I want to find a closed form for this sum and its derivatives over $x$ when $x=0$ and $q=1$. Since its poles are of the form $k+\frac{in}{q}\,(k,n\text { integers})$ with double-order poles at the integers, I figure its expression may include trigonometric and theta functions...but I can't figure anything beyond its singularities. Any help would be appreciated.
I've managed to turn the sum into a Fourier series $\left(-4\sum_{k\ge 1}\ln(1+e^{-2k\pi / q})\cos{2k\pi x}\right)\quad\;$, but even with its simplicity, I haven't been able to crack it.
(Edit) I think I have a way to evaluate the Fourier series: if I expand the cosines into Taylor series, then I just have to sum series of the form $$\sum_{k \ge 1}k^{2n}\ln(1+e^{-2k\pi/q})$$ which I can rewrite as $$\sum_{m\ge 1}\frac{(-1)^{m-1}}m\sum_{k\ge 1}k^{2n}e^{-2km\pi/q}$$and since $\displaystyle{\sum_{k\ge 1}e^{-2km\pi/q}=\frac1{e^{2m\pi/q}-1}}$, $\displaystyle{\sum_{k\ge 1}k^2e^{-2km\pi/q}=\frac14\frac{\cosh{\frac{m\pi}q}}{\sinh^3{\frac{m\pi}q}}}$ and subsequent sums consist of a hyperbolic cosine times an odd reciprocal polynomial in the hyperbolic sine, I've reduced my problem to evaluating sums of the form $$\sum_{m \ge 1}\frac{(-1)^{m-1}}m \frac{\cosh{\frac{m\pi}q}}{\sinh^{2n+1}{\frac{m\pi}q}}$$which is also $\displaystyle{\int{\sum_{k\ge 1}\frac{(-1)^{k-1}\sinh{kz}}{\sinh^{2n+1}{\frac{k\pi}q}}\, dz}}$ with $z=\frac{i\pi}q$. But here I am stuck.