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I have the function $y(x)=e^{iax}, a\in\mathbb{R}$. Where $0\le x\le 2\pi$, and y(x) is also $2\pi$-periodic.

They asked me to find the fourier coefficients of y(x), which I found out to be $\frac{e^{i2\pi a}-1}{2\pi i(a-n)}$.

But I'm also asked to show that $\sum_{-\infty}^\infty \frac{1}{(a-n)^{2}} = \frac{\pi^{2}}{sin^{2}\pi a} $

In my attempt for a solution, I used Parseval's formula: $\frac{(e^{i2\pi a}-1)^{2}}{4\pi ^{2}}\sum_{-\infty}^\infty \frac{1}{(a-n)^{2}} = \frac{1}{2\pi}\int_{0}^{2\pi} |e^{iax}|^{2} $

I concluded that the right hand side will equal to 1, then I get that:

$\sum_{-\infty}^\infty \frac{1}{(a-n)^{2}} = \frac{4\pi^{2}}{(e^{i2\pi a}-1)^{2}} $

I have tried to play around with the expression at the right hand side of the equality, but cant manage to obtain $\frac{\pi^{2}}{sin^{2}\pi a}$

Any help?

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If we expand out $e^{iax}/\sqrt{2\pi}$ defined on $(-\pi,\pi)$ in terms of the orthonormal sequence $e^{ikn}/\sqrt{2\pi}$, $n\in {\mathbb N}$ the coefficients are

$$ c_n=\frac 1{2\pi} \int_{-\pi}^{\pi} e^{iax}e^{-inx}=\frac{\sin(\pi(a-n))}{\pi(a-n)} $$

and from Parseval we get $$ \frac 1{2\pi} \int_{-\pi}^{\pi}dx= 1= \sum_{n=-\infty}^{\infty} \frac{\sin^2(\pi(a-n))}{\pi^2(a-n)^2}. $$ As the squared sine is $\pi$ periodic, this is the same as $$ 1= \sum_{n=-\infty}^{\infty} \frac{\sin^2(\pi a)}{\pi^2(a-n)^2}. $$
You problem is the same as this except that youy need to slide interval over by $\pi$.

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    I notice alao that you took the $n$ depended coefficiient outside the sum. I assume that you realize that this is not correct. --- AAhh I misread the rather small text incorrectly, you did not do this! – 2017-02-14