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?
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 $\pi$.