Solving $\sum_{n=1}^{\infty} \frac{1}{n^2}$ using the fourier series.

Question

Please do NOT solve the problem, I just need some help, not a full solution. I would like to try this myself.

Find $\zeta(2) = \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$

The fourier series for $f(x)$ on the interval $-L \le f(x) \le L$ is:

$\displaystyle f(x) = \sum_{n=0}^{\infty} A_n\cos(\frac{n\pi x}{L}) + \sum_{n=1}^{\infty} B_n\sin(\frac{n\pi x}{L})$ Where $A_n, B_n$ are Fourier coefficients.

$A_n = \displaystyle \frac{2}{L} \int_{0}^{L} f(x)\cos(\frac{n\pi x}{L})$

I found for $f(x) = x$, $A_n = \displaystyle \frac{-Lx}{n \pi} \sin(\frac{mx \pi}{L}) + (\frac{L}{\pi n})^2\cos(\frac{n \pi x}{L})$

Any thoughts?

The Problem: $f(x) = x$ is not $2\pi$ periodic so how do we fix the $L$ so that is for $n \to \infty$ because from $n = 1$ to $\infty$ is the goal? What should $L$ be?

Thanks!

Answer 1

$L$ shouldn't worry you. You may as well set it equal to 1,and define $f(x)=x$ on $(-1,1)$ and then extend it everywhere else periodically $f(x\pm 2)=f(x)$. The fourier series will work everywhere in that case. You're not trying to do a fourier series expansion on $(-\infty,\infty)$ afterall (which is impossible).

Then you might want to take a look at Parseval's identity.