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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?

I am pi
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  • Here you go https://math.stackexchange.com/questions/736860/find-the-infinite-sum-of-the-series-sum-n-1-infty-frac1n2-1 – Sonal_sqrt Dec 21 '17 at 05:16

2 Answers2

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Hint: There is a much simpler way to go. Have you heard of Parseval? Try to write $$e^{i(n-i)x} = e^{inx} e^x $$

H. H. Rugh
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0

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

operatorerror
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Sonal_sqrt
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