Closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$.

In my opinion, this is a rather nice example of application of the Poisson summation formula: $$ \sum_{n\in\mathbb{Z}}f(n)=\sum_{k\in\mathbb{Z}}\hat{f}(k),\qquad \hat{f}(\nu)=\int_{-\infty}^{\infty}f(x)e^{-2\pi i \nu x}dx.$$ Namely, setting $f(x)=\frac{1}{(x-a)^2+b^2}$, we find \begin{align} \hat{f}(\nu)=\int_{-\infty}^{\infty}\frac{e^{-2\pi i \nu x}dx}{(x-a)^2+b^2}=e^{-2\pi i \nu a}\int_{-\infty}^{\infty}\frac{e^{-2\pi i \nu x}dx}{x^2+b^2}= \frac{\pi}{b}e^{-2\pi i \nu a-2\pi |\nu| b}, \end{align} where we have assumed that $b>0$ and calculated the last integral using residues. Therefore, the sum we are trying to calculate reduces to geometric series: $$\sum_{n\in\mathbb{Z}}\frac{1}{(n-a)^2+b^2}=\frac{\pi}{b}\sum_{k\in\mathbb{Z}}e^{-2\pi i k a-2\pi |k| b}=\frac{\pi\sinh2\pi b}{b\left(\cosh2\pi b-\cos2\pi a\right)}.$$

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.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{#c00000}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \sum_{n\ =\ -\infty}^{\infty}{1 \over \pars{n - a}^{2} + b^{2}}} \\[5mm] = &\ \sum_{n\ =\ -\infty}^{-1}{1 \over \pars{n - a}^{2} + b^{2}} +{1 \over a^{2} + b^{2}} +\sum_{n\ =\ 1}^{\infty}{1 \over \pars{n - a}^{2} + b^{2}} \\[5mm]&=\sum_{n\ =\ 1}^{\infty}{1 \over \pars{n + a}^{2} + b^{2}} +{1 \over a^{2} + b^{2}} +\sum_{n\ =\ 1}^{\infty}{1 \over \pars{n - a}^{2} + b^{2}} \\[5mm]&=-\,{1 \over a^{2} + b^{2}} +\bracks{\color{#c00000}{% \sum_{n\ =\ 0}^{\infty}{1 \over \pars{n - a}^{2} + b^{2}}} + \pars{a \to -a}} \end{align}

\begin{align}&\color{#c00000}{% \sum_{n\ =\ 0}^{\infty}{1 \over \pars{n - a}^{2} + b^{2}}} =\sum_{n\ =\ 0}^{\infty} {1 \over \bracks{n - \pars{a + b\ic}}\bracks{n - \pars{a - b\ic}}} \\[5mm]&={\Psi\pars{a + b\ic} - \Psi\pars{a - b\ic}\over \pars{a + b\ic} - \pars{a - b\ic}} ={2\ic\,\Im\Psi\pars{a + b\ic} \over 2b\ic}={\Im\Psi\pars{a + b\ic} \over b} \end{align}

\begin{align}&\color{#66f}{\large% \sum_{n\ =\ -\infty}^{\infty}{1 \over \pars{n - a}^{2} + b^{2}}} \\[5mm] = &\ \color{#66f}{\large -\,{1 \over a^{2} + b^{2}} + {\Im\Psi\pars{a + b\ic} + \Im\Psi\pars{-a + b\ic} \over b}} \end{align}

$\ds{\Psi\pars{z}}$ is the Digamma Function .