Summation containing $\pi$ gives an expression containing $e$

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

Hint: Use the expansion $$\coth x=\sum_{n=-\infty}^{\infty}\dfrac{x}{n^2\pi^2+x^2}$$