If $c_n=\frac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi e^{i(n-i)x}dx $ ,then find $\sum_{n\in \mathbb{Z}}|c_n|^2$ Let $$c_n=\frac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi e^{i(n-i)x}dx .$$ Then what will be the value of $$\sum_{n\in \mathbb{Z}}|c_n|^2.$$
My attempt:

\begin{align*}
c_n & =\frac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi e^{i(n-i)x}dx\\
    & = \frac{1}{\sqrt{2\pi}}\cdot\frac{1}{i(n-i)} \left(e^{i(n-i)x}\right) \Big|_{x=-\pi}^{x=\pi}\\
& =  \frac{1}{\sqrt{2\pi}}\cdot\frac{1}{i(n-i)} \cos (n\pi) (2\sinh \pi)\\
\implies |c_n|^2 &= \frac{1}{2\pi}\cdot \frac{1}{n^2+1} \cos^2n\pi (4\sinh^2\pi)\\
&= \frac{2\sinh^2\pi}{\pi(n^2+1)}
\end{align*}
So, $$\sum_{n\in \mathbb{Z}}\frac{2\sinh^2\pi}{\pi(n^2+1)}=\frac{2\sinh^2\pi}{\pi}\sum_{n\in \mathbb{Z}}\frac{1}{n^2+1}$$
After that I stuck, what to do?


from here David Cardon
$$\sum_{k=-\infty}^{\infty}\frac1{1+k^2}=\pi\coth (\pi)$$
