How to find the sum of the Series $\sum_{n \in \Bbb{Z}} \vert c_n \vert^2$ which terms are the special integral representations

Question

For $n \in \Bbb{Z}$, define $$ c_n=\frac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} e^{i(n-i)x} dx$$ where $i^2=-1$. Then $\displaystyle\sum_{n \in \Bbb{Z}} \vert c_n \vert^2$ equals........?

a) $\cosh(\pi)$ $\hspace{2cm}$ b) $\sinh(\pi)$$\hspace{2cm}$ c) $\cosh(2\pi)$ $\hspace{2cm}$ d) $\sinh(2\pi)$

How to approach this ?

I'm trying to find the integral value but it gives $\displaystyle\int_{-\pi}^{\pi} e^{i(n-i)x} dx=\frac{-2i\sinh(\pi+i\pi n)}{n-i}$

How to proceed further? or Is there any shortcut method to find this ?

Any help ?

Answer 1

Note that with straightforward arithmetic we have

$$\begin{align} \sum_{n=-\infty}^\infty|c_n|^2&=\sum_{n=-\infty}^\infty\left(\frac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi e^{inx}e^x\,dx\right)\left(\frac1{\sqrt{2\pi}}\int_{-\pi}^\pi e^{-inx'}e^{x'}\,dx'\right)\\\\ &=\frac1{2\pi}\sum_{n=-\infty}^\infty \int_{-\pi}^\pi\int_{-\pi}^\pi e^{x+x'}e^{in(x-x')}\,dx\,dx'\\\\ &=\frac1{2\pi}\lim_{N\to \infty}\int_{-\pi}^\pi\int_{-\pi}^\pi e^{x+x'}\underbrace{\sum_{n=-N}^N e^{in(x-x')}}_{\text{Dirichlet Kernel}}\,dx\,dx'\\\\ &=\int_{-\pi}^\pi e^{2x}\,dx\\\\ &=\sinh(2\pi) \end{align}$$

And we are done!

Answer 2

hint. use parseval theorem in wiki link. let $x=-u$ then $$c_n=\frac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} e^{i(n-i)x} dx=\frac{1}{2\pi} \int_{-\pi}^{\pi} \sqrt{2\pi}e^{-u}e^{-inu} du=\frac{1}{2\pi} \int_{-\pi}^{\pi} f(u)e^{-inu} du$$ where $f(u)=\sqrt{2\pi}e^{-u}$, then with the Fourier transform of $f$ in $(-\pi,\pi)$, $$\frac{1}{\pi}\int_{-\infty}^{\infty}|f|^2du=\frac{1}{2\pi}\int_{-\infty}^{\infty}|c_n|^2du$$