Find the sum of the series $ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \ $ Find the sum of the series $ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \ $ , where $ \ l=constant \ $ 
Answer
The given series is convergent clearly . 
$ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \\ = \frac{1}{1^2+l^2}+\frac{1}{2^2+l^2}+......+\frac{1}{n^2+l^2}+........ \\ = \frac{1}{l}  [ d \tan^{-1}(\frac{1}{l})+ d \tan^{-1}(\frac{2}{l})+........+ d \tan^{-1}(\frac{n}{l})+.....] \\ = \frac{1}{l} \lim_{n \to \infty}  \sum_{k=1}^{n}[d \tan^{-1}(\frac{1}{l})+ d \tan^{-1}(\frac{2}{l})+........+ d \tan^{-1}(\frac{k}{l})]    \\ = \frac{1}{l} \ d (\lim_{n \to \infty}  \sum_{k=1}^{n}[ \tan^{-1}(\frac{1}{l})+  \tan^{-1}(\frac{2}{l})+........+  \tan^{-1}(\frac{k}{l})]) $
but now i  can't proceed to find the sum of the series. 
Help me out.
 A: Let $f(z)=\frac{\cot(\pi z)}{z^2+\ell^2}$.  Note that $f$ has simple poles at $z=n$ and $z=\pm i\ell$.
Next, let $R_N=(N+1/2)$ in $(1)$.  Then, the residue theorem guarantees that
$$\oint_{|z|=R_N}f(z)\,dz=2\pi i \sum_{n=-N}^N\text{Res}\left(\frac{\cot(\pi z)}{z^2+\ell^2}, z=n\right)+2\pi i \text{Res}\left(\frac{\cot(\pi z)}{z^2+\ell^2}, z=\pm i\ell\right)\tag 1$$
In THIS ANSWER, I showed that $|\cot(\pi z)|$ is bounded on the circle $|z|=N+1/2$, where $N\in \mathbb{N}$.
So, letting $N\to \infty$, the contour integral in $(1)$ tends to $0$ and we find
$$\begin{align}
\sum_{n=-\infty }^\infty \frac{1}{\pi(n^2+\ell^2)}&=-\left(\frac{\cot(i\pi\ell)}{i2\ell}+\frac{\cot(-i\pi\ell)}{-i2\ell}\right)\\\\
&=\frac{\coth(\pi\ell)}{\ell}\tag2
\end{align}$$
Therefore, multiplying $(2)$ by $\pi$ reveals
$$\sum_{n=-\infty }^\infty \frac{1}{n^2+\ell^2}=\frac{\pi \coth(\pi\ell)}{\ell}$$
Finally, exploiting the evenness of the summands yields
$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1 }^\infty \frac{1}{\pi(n^2+\ell^2)}=\frac{\pi\coth(\pi\ell)}{2\ell}-\frac{1}{2\ell^2}}$$
A: Hint: Use the fact that
\begin{align}
\sum^\infty_{n=-\infty} \frac{1}{x^2+n^2} = \frac{\pi\coth(\pi x) }{x}.
\end{align}
A: At equation $(7)$ of this answer, it is concluded that, for all $z\in\mathbb{C}\setminus\mathbb{Z}$,
$$
\sum_{k=-\infty}^\infty\frac{1}{z+k}=\pi\cot(\pi z)
$$
Therefore,
$$
\begin{align}
\sum_{k=1}^\infty\frac1{k^2+z^2}
&=-\frac1{2iz}\sum_{k=1}^\infty\left(\frac1{iz-k}+\frac1{iz+k}\right)\\
&=-\frac1{2z^2}-\frac1{2iz}\sum_{k=-\infty}^\infty\frac1{iz+k}\\
&=-\frac1{2z^2}-\frac1{2iz}\pi\cot(\pi iz)\\
&=\bbox[5px,border:2px solid #C0A000]{-\frac1{2z^2}+\frac1{2z}\pi\coth(\pi z)}
\end{align}
$$
