Closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$. What is the closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$? We can use Fourier series of $e^{-bx}$ ($|x|<\pi$) to evaluate $\sum_{n=-\infty}^{\infty}\frac{1}{n^2+b^2}$. But this one, it seems to me it is tough to get the closed form.

This one can also be done with the standard technique of using the $\pi \cot(\pi z)$ multiplier, integrating $$f(z) = \frac{\pi \cot(\pi z)}{(z-a)^2+b^2}$$ along a circle using the same technique as here.
We integrate $f(z)$ along a circle of radius $R$ with $R$ going to infinity and the integral disappears in the limit so that the residues sum to zero (actually a square with vertices $$(\pm(N+1/2),\pm(N+1/2))$$ $N$ a positive integer is easier to handle computationally). The poles of $f(z)$ other than at the integers are at $$z_{0,1} = a\pm ib$$ and the residues are
$$\operatorname{Res}(f(z); z=z_0) =
\left.\frac{\pi \cot(\pi z)}{2(z-a)}\right|_{z=z_0}
= \frac{\pi\cot(\pi(a+bi))}{2bi}
= \frac{\pi}{2bi} i \frac{e^{i\pi(a+bi)}+ e^{-i\pi(a+bi)}}{e^{i\pi(a+bi)}- e^{-i\pi(a+bi)}}$$
and
$$\operatorname{Res}(f(z); z=z_1) =
\left.\frac{\pi \cot(\pi z)}{2(z-a)}\right|_{z=z_1}
= \frac{\pi\cot(\pi(a-bi))}{-2bi}
= \frac{\pi}{-2bi} i \frac{e^{i\pi(a-bi)}+ e^{-i\pi(a-bi)}}{e^{i\pi(a-bi)}- e^{-i\pi(a-bi)}}.$$
Now put $x=e^{i\pi a}$ and $y=e^{-\pi b}.$
The first residue becomes
$$\operatorname{Res}(f(z); z=z_0) =
\frac{\pi}{2b}\frac{xy+1/x/y}{xy-1/x/y}$$
and the second residue is
$$\operatorname{Res}(f(z); z=z_1) =
-\frac{\pi}{2b}\frac{x/y+y/x}{x/y-y/x}.$$
Adding the two contributions and simplifying with Euler's formula yields
$$-\frac{\pi x^2 (y^4-1)}{b (x^2y^2-1)(x^2-y^2)} =
-\frac{\pi (y^2-1/y^2)}{b (x^2-1/y^2)(1-y^2/x^2)} \\=
-\frac{\pi (y^2-1/y^2)}{b (x^2+1/x^2-y^2-1/y^2)} =
\frac{\pi}{b} \frac{\sinh(2\pi b)}{\cos(2\pi a)-\cosh(2\pi b)}.$$
Now with $S$ being our sum we have by the Cauchy Residue Theorem that
$$ S + \frac{\pi}{b} \frac{\sinh(2\pi b)}{\cos(2\pi a)-\cosh(2\pi b)} = 0$$
so that finally
$$ S = \frac{\pi}{b} \frac{\sinh(2\pi b)}{\cosh(2\pi b)-\cos(2\pi a)}.$$
This MSE link contains a similar computation, this one including the computation of the relevant bounds. 
