introduction

Say \(f(t)\) is an \(2L\)-periodic even function. The cosine series for \(f\) is defined as \[\frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(\pi nt / L)\] where you compute the coefficients via inner-products: \[a_i = \frac{1}{L}\int_{-L}^{L} f(t) \cos(\pi nt / L) dt.\] Because \(\{\cos(\pi n t / L) : n\in \mathbb{N}\}\) forms an orthonormal basis [nasty integral / trig identity exercise: check this] for nice \(2L\)-periodic even functions (note: I’m not totally sure what the space spanned by these cosines is, ofc it is not all functions, maybe continuous is the right word?) it is then a basic fact of linear algebra that if we project onto these basis vectors we get another way of representing the function.

computing the ingtegral

Well, ok. So one function you could compute the fourier transform of is the even \(2\pi\)-periodic extension of the parabola \(t\mapsto t^2.\)

If you do so you get, for all \(t\in [-\pi,\pi]\) \[t^2 = \sum_{n=1}^{\infty} \frac{4(-1)^n}{n^{2}}\cos(nt).\] Evaluating at \(\pi\) we find \[\sum_{n\ge 1}\frac{1}{n^{2}} = \frac{\pi^2}{6}.\] Nice.