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Show using Parseval's Theorem that: $$ \sum_{n=-\infty}^{\infty}\frac{1}{(n+a)^2}=\frac{\pi^2}{\sin^2(\pi a)} $$ I've tried to think about ways to solve this but haven't got anywhere. I must be fundamentally misunderstanding Parseval's theorem. The only thing I can think of doing is expanding the summation and showing it in terms of series relating to $\pi$ and $\sin$ but I don't think that can work because the sum starts at $- \infty$.

This is the function $f(x)$ given if needed: $$ f(x)=\frac{\pi e^{-iax}}{\sin(\pi a)},\ x\in[-\pi,\pi] $$

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Hint: compute the Fourier series of

$g(x):=f(x) \rightarrow [-\pi,\pi]; g(x+2\pi)=g(x)$

And use Parseval theorem

$\sum_n |a_n|^2=\frac{1}{2\pi}\int_{-\pi}^{\pi}|g(x)|^2dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left(\frac{\pi}{\sin(\pi a)}\right)^2dx=\left(\frac{\pi}{\sin(\pi a)}\right)^2$

Doing computations:

$a_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\pi}{\sin(\pi a)}e^{-i(n+a)x}dx= \frac{1}{2\sin(\pi a)}\frac{-1}{i(n+a)}e^{-i(n+a)x}\rvert^{\pi}_{-\pi}= \frac{1}{\sin(\pi a)(n+a)}\sin(\pi(n+a))=\frac{(-1)^{n}}{(n+a)}$

Thus:

$\sum_{n=-\infty}^{n=\infty}|a_n|^2=\sum_{n=-\infty}^{n=\infty}\frac{1}{(n+a)^2}=\left(\frac{\pi}{\sin(\pi a)}\right)^2$

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  • $\begingroup$ Thank you for your answer. Could you please explain what you mean by your '𝑔(𝑥):' section? I have seen the f(x) --> g(x +2𝜋) section before but not what it means. Also, I'm guessing from context that you say that the modulus of g(x) gets rid of the imaginary exponential. How is that treated with? Thank you. $\endgroup$ – Mathslearner.2000 Feb 22 at 23:02
  • $\begingroup$ @Mathslearner.2000 knowing a little of complex algebra: $|e^{ix}|=|\cos(x)+i\sin(x)|=\sqrt{\cos^2(x)+\sin^2(x)}=1$ $\endgroup$ – Gabriele Cassese Feb 23 at 6:26
  • $\begingroup$ @Mathslearner.2000 For $g$ I meant the periodic extension of $f$, I improved my formatting $\endgroup$ – Gabriele Cassese Feb 24 at 17:15

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