Infinite Sum of 1/n^2 Variant

Let $f(x) = \cosh x$, and let's try to find the Fourier Series of $f(x)$ on $[-\pi,\pi]$.

$c_n = \dfrac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}e^{-inx}\cosh x\,dx$ $= \dfrac{1}{4\pi}\displaystyle\int_{-\pi}^{\pi}e^{-inx}(e^x+e^{-x})\,dx$ $= \dfrac{1}{4\pi}\displaystyle\int_{-\pi}^{\pi}(e^{(1-in)x}+e^{(-1-in)x})\,dx$

$= \dfrac{1}{4\pi}\left[\dfrac{e^{(1-in)x}}{1-in} + \dfrac{e^{(-1-in)x}}{-1-in}\right]_{-\pi}^{\pi}$ $= \dfrac{1}{4\pi}\left[\dfrac{e^{(1-in)\pi} - e^{-(1-in)\pi}}{1-in} + \dfrac{e^{(-1-in)\pi} - e^{-(-1-in)\pi}}{-1-in}\right]$

$= \dfrac{(-1)^n}{4\pi}\left[\dfrac{e^{\pi} - e^{-\pi}}{1-in} + \dfrac{e^{-\pi} - e^{\pi}}{-1-in}\right]$ $= \dfrac{(-1)^n\sinh \pi}{2\pi}\left[\dfrac{1}{1-in} - \dfrac{1}{1+in}\right]$ $= \dfrac{(-1)^n\sinh \pi}{\pi(n^2+1)}$.

Therefore, $f(x) = \cosh x = \displaystyle\sum_{n = -\infty}^{\infty}\dfrac{(-1)^n\sinh \pi}{\pi(n^2+1)}e^{inx}$.

Thus, $f(\pi) = \cosh \pi = \displaystyle\sum_{n = -\infty}^{\infty}\dfrac{\sinh \pi}{\pi(n^2+1)}$, and so, $\displaystyle\sum_{n = -\infty}^{\infty}\dfrac{1}{n^2+1} = \dfrac{\pi \cosh \pi}{\sinh \pi}$.

Subtracting the $n = 0$ term and dividing by $2$ gives: $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n^2+1} = \dfrac{\pi \cosh \pi}{2\sinh \pi} - \dfrac{1}{2} = \dfrac{\pi}{2}\coth \pi - \dfrac{1}{2}$.

EDIT: This is essentially what C L. did, except with complex Fourier coefficients instead.

Bob's right: the answer reminded me of some Fourier transform computations I did some years back.

Let $f(x) = \cosh(x)$ on the interval $(-\pi, +\pi]$ and repeated periodically through the real line via $f(x+2\pi) = f(x)$. It's continuous and nice enough that its Fourier series converges pointwise.

Convergence of Fourier Transform

Specifically, at the point $x = \pi$, the one-sided derivatives exist:

$\lim_{x \to \pi^-} \frac{f(x) - \cosh\pi} {x - \pi} = \sinh\pi, \quad \lim_{x\to \pi^+} \frac{f(x) - \cosh\pi} {x - \pi} = \sinh(-\pi) = -\sinh \pi$

where the second equality follows from $f(x) = \cosh(x - 2\pi)$ when x is slightly more than $2 \pi$. Furthermore, the one-sided derivatives of $f(x)$ have only discrete discontinuities. One can then apply Dini criterion to show that the Fourier series converges.

Computing the Fourier Transform

Omitting the details, the Fourier coefficients of $f(x)$ are given by:

$\begin{aligned} a_n &= \frac 1 \pi \int_{-\pi}^{\pi} f(x)\cos nx dx = \frac 1 \pi \int_{-\pi}^\pi \cosh x \cos nx dx = 2(-1)^n \frac{\sinh \pi}{\pi},\\ b_n &= \frac 1 \pi \int_{-\pi}^{\pi} f(x)\sin nx dx = \frac 1 \pi \int_{-\pi}^\pi \cosh x \sin nx dx = 0.\end{aligned}$

Thus, the Fourier expansion gives:

$\begin{aligned} f(x) &= \frac 1 2 a_0 + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx). \\ &= \frac{\sinh \pi}\pi \left( 1 + 2\sum_{n=1}^\infty (-1)^n \frac {\cos nx} {n^2 + 1}\right) \end{aligned}$

Substituting $x = \pi$ then gives:

$\cosh \pi = \frac{\sinh \pi}\pi \left( 1 + 2\sum_{n=1}^\infty \frac 1 {n^2 + 1}\right)$

which gives Bob's answer.

Here's a shorter way.

Note that $\cot( x)=\frac{1}{x}+ \sum_{k=1}^{\infty} \frac{2x}{x^2-k^2 \pi^2}$ (Proof: use infinite product for sine, take natural logs, differentiate, rearrange). Thus,

$\large \frac{1}{\pi^2} \sum_{k=1}^{\infty} \frac{2x}{\frac{x^2}{\pi^2}-k^2 }=\cot( x)-\frac{1}{x}$

$\large \sum_{k=1}^{\infty} \frac{x}{k^2-\frac{x^2}{\pi^2} }=\frac{-\pi^2}{2} \left(\cot( x)-\frac{1}{x}\right)$

Let $x=i\pi$,

$\large \sum_{k=1}^{\infty} \frac{i \pi}{k^2+1 }=\frac{-\pi^2}{2} \left(\cot( i \pi)-\frac{1}{i \pi}\right)$

$\large \sum_{k=1}^{\infty} \frac{1}{k^2+1 }=\frac{i\pi}{2} \left(\cot( i \pi)-\frac{1}{i \pi}\right)$

$\large \sum_{k=1}^{\infty} \frac{1}{k^2+1 }=\frac{i\pi}{2} \frac{\cos(i \pi)}{\sin(i \pi)}-\frac{1}{2}$

$\large \sum_{k=1}^{\infty} \frac{1}{k^2+1 }=\frac{i\pi}{2} \frac{\cosh( \pi)}{i\sinh( \pi)}-\frac{1}{2}$

$\large \sum_{k=1}^{\infty} \frac{1}{k^2+1 }=\frac{\pi}{2} \frac{\cosh( \pi)}{\sinh( \pi)}-\frac{1}{2}$

Mathematica gave $\frac{-1}{2} + \frac{\pi}{2}\coth \pi$. I know this isn't exactly the point, but sometimes the answer is the best place to start a question like this.

Is there a systematic way to approach these questions? I've seen Fourier Series solutions before, but they always struck me as most likely discovered by accident. For a simpler example, why would you immediately think to consider the Fourier Series of $f(x) = |x|$ if you wanted to find $\zeta(2)$?

Excellent work, everybody!