Fourier series to calculate infinite series

Question

I try to show that $\sum_{i=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ using Fourier series and $f(x) = x$ on $L^2_{\mathbb{C}}[-\pi, \pi]$, with basis $e_n(x) = \frac{1}{\sqrt{2\pi}}e^{inx}$. I found $\langle f, f \rangle = \int_{-\pi}^{\pi} x^2 \, dx = \frac{2\pi^3}{3}$, but I am having difficulties calculating the inner product $\langle f, e_n \rangle$. Could somebody give me a hint?

Answer 1

Use integration by parts as follows$$\langle f, e_n \rangle=\frac{1}{\sqrt{2\pi}}\int\limits_{-\pi}^{\pi} x e^{inx}\, dx \\= \frac{1}{\sqrt{2\pi}}\left[\left.x\cdot \frac{e^{inx}}{in}\right|_{-\pi}^{\pi}-\int\limits_{-\pi}^{\pi} \frac{e^{inx}}{in}\cdot 1\, dx\right]$$

Added

The integral will be $$c_n=\frac{1}{\sqrt{2\pi}}\cdot \frac{2\pi\cos n\pi}{in} \\ ||c_n||^2= \frac{2\pi}{n^2} $$ and so $$\frac{1}{2\pi}\cdot \frac{2\pi^3}{3}=\sum \frac{2\pi}{n^2}$$