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

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!

• I think $L=1$ would be a good choice. At least it would give you a concrete number. – Joel Nov 4 '14 at 16:33
• Also, note that your $A_n$ seems to be from $0$ to $L$ rather than $-L$ to $L$. – Joel Nov 4 '14 at 16:35
• possible duplicate of Different methods to compute $\sum\limits_{n=1}^\infty \frac{1}{n^2}$ – dustin Nov 7 '14 at 3:34

$L$ 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).
• I am trying to do it on $[1, \infty)$ – Amad27 Nov 4 '14 at 17:52
• @Amad27: You cannot do fourier series expansions for functions whose period is not finite. If you want to do your problem on $[1,\infty)$, then do the periodic extension above. The point is that if you solve for the fourier series in any window whose length is the period of the function, then that solution will work in any other window. Once again, the hint is to assume $f(x)=x$ is defined on $(-1,1)$ and then extended periodically to all other windows. – Alex R. Nov 4 '14 at 18:47