TITLE: Summation containing $\pi$ gives an expression containing $e$
QUESTION [5 upvotes]: This is an interesting relation between $\pi$ and $e$.
$$\sum_{k=1}^{\infty }\frac{1}{(k\pi)^2+1}=\frac{1}{\pi^2+1}+\frac{1}{4\pi^2+1}+\frac{1}{9\pi^2+1}+\frac{1}{16\pi^2+1}+\dots=\frac{1}{e^2-1}$$
I tried to prove it using Fourier series expansion, but I could not. Any hint will be appreciated.

REPLY [6 votes]: Using the result that
$$
\mathrm{PV}\sum_{k\in\mathbb{Z}}\frac1{x+k}=\pi\cot{\pi x}
$$
we get
$$
\begin{align}
\sum_{k=1}^\infty\frac1{1+k^2\pi^2}
&=\frac{i}{2\pi}\mathrm{PV}\sum_{k\in\mathbb{Z}}\frac1{i/\pi+k}-\frac12\\
&=\frac{i}{2\pi}\pi\cot\left(\pi\frac{i}\pi\right)-\frac12\\[3pt]
&=\frac12\coth(1)-\frac12\\[3pt]
&=\frac1{e^2-1}
\end{align}
$$