TITLE: Closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$.
QUESTION [36 upvotes]: 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.

REPLY [10 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\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 .