Sum of Reciprocals of Squares plus 1

Theorem

 * $\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2 + 1} = \frac 1 2 \paren {\pi \coth \pi - 1}$

Proof
Let $\map f x$ be the real function defined on $\hointl {-\pi} \pi$ as:


 * $\map f x = e^x$

From Fourier Series: $e^x$ over $\openint {-\pi} \pi$, we have:
 * $(1): \quad \displaystyle \map f x \sim \map S x = \frac {\sinh \pi} \pi \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\cos n x} {1 + n^2} - \frac {n \sin n x} {1 + n^2} } }$

Let $x = \pi$.

By Fourier's Theorem, we have:

Thus setting $x = \pi$ in $(1)$, we have: