# Prove $\sum_{n\geq1}\frac1{n^2+1}=\frac{\pi\coth\pi-1}2$ [duplicate]

I am trying to prove $$\sum_{n\geq1}\frac1{n^2+1}=\frac{\pi\coth\pi-1}2$$ Letting $$S=\sum_{n\geq1}\frac1{n^2+1}$$ we recall the Fourier series for the exponential function $$e^x=\frac{\sinh\pi}\pi+\frac{2\sinh\pi}\pi\sum_{n\geq1}\frac{(-1)^n}{n^2+1}(\cos nx-n\sin nx)$$ Plugging in $$x=\pi$$ $$e^\pi=\frac{\sinh\pi}\pi+\frac{2\sinh\pi}\pi\sum_{n\geq1}\frac{(-1)^n}{n^2+1}(\cos n\pi-n\sin n\pi)$$ $$e^\pi=\frac{\sinh\pi}\pi+\frac{2\sinh\pi}\pi\sum_{n\geq1}\frac{(-1)^n}{n^2+1}((-1)^n-n\cdot0)$$ $$e^\pi=\frac{\sinh\pi}\pi+\frac{2\sinh\pi}\pi S$$ $$S=\frac{\pi e^\pi}{2\sinh\pi}-\frac12$$ But that is nowhere near to correct. What did I do wrong, and how do can I prove the identity? Thanks.

The formula $$e^x=\frac{\sinh\pi}\pi+\frac{2\sinh\pi}\pi\sum_{n\geq1}\frac{(-1)^n}{n^2+1}(\cos nx-n\sin nx)$$ is valid only for $$|x|<\pi$$ since $$e^x$$ is regarded as $$2\pi$$-periodic function extended from $$(-\pi, \pi).$$ Since $$2\pi$$-periodic function $$x\mapsto e^x$$ is of bounded variation, its Fourier series converges to the mean of its left and right limit at every point. So if you evaluate the RHS at $$x=\pi$$, then you get $$\frac{e^{\pi}+e^{-\pi}}{2} = \cosh\pi = \frac{\sinh\pi}\pi+\frac{2\sinh\pi}\pi S,$$ and hence $$S = \sum_{n\geq1}\frac1{n^2+1}=\frac{\pi\coth\pi-1}2,$$as desired.
Recalling that $$\frac{\psi(z)-\psi(s)}{z-s}=\sum_{n\ge 0} \frac{1}{(n+z)(n+s)}$$, your sum is equal to $$\sum_{n\ge 1} \frac{1}{n^2+1}=\frac{\psi(i)-\psi(-i)}{2i}-1$$ Now from two identities of the digamma function, \begin{align} \psi(1-z)-\psi(z)&=\pi\cot(\pi x) \\ \psi(1+z)-\psi(z)&=\frac{1}{z} \\ \end{align} We may find that $$\psi(i)-\psi(-i)=-\frac{1}{i}-\pi\cot(\pi i)$$. Thus, we can conclude that$$\sum_{n\ge 1} \frac{1}{n^2+1}=\frac{\pi}{2}\coth(\pi)-\frac{1}{2}$$