Prove $\sum\limits_{n\mathop=0}^{\infty}\frac{1}{n^2+1}=\frac{1}{2}+\frac{\pi}{2}\left(\frac{e^{2\pi}+1}{e^{2\pi}-1}\right)$

Question

How do I show that

$$\sum\limits_{n\mathop=0}^{\infty}\frac{1}{n^2+1}=\frac{1}{2}+\frac{\pi}{2}\left(\frac{e^{2\pi}+1}{e^{2\pi}-1}\right)$$

Answer 1

For example, using 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 $f(x)=\frac{1}{x^2+1}$, we obtain \begin{align} \hat{f}(\nu)=\int_{-\infty}^{\infty}\frac{e^{-2\pi i \nu x}dx}{x^2+1}=\int_{-\infty}^{\infty}\frac{e^{-2\pi i \nu x}dx}{x^2+1}= \pi e^{-2\pi |\nu|}, \end{align} where we calculated the last integral by residues. Therefore, we can write $$\sum_{n\in\mathbb{Z}}\frac{1}{n^2+1}=\pi\sum_{k\in\mathbb{Z}}e^{-2\pi |k|}=\frac{\pi\sinh2\pi}{\cosh2\pi-1}=\pi \coth \pi.$$ This is obviously equivalent to the formula you want to prove.