Here we present an application of Parseval's equality to number theory. Let $\zeta(s)$ denote the Riemann zeta function. We will compute the value $$\zeta(2)$$ with the help of Fourier analysis.

Example:

Let $f\colon \Reals \to \Reals$ be the ``identity'' function, defined by $$f(x)=x, \text{ for all }x\in\Reals.$$

The Fourier series of this function has been computed in the entry example of Fourier series.

Thus \begin{eqnarray*} f(x)=\ x&=& a_0^f + \sum_{n=1}^{\infty}(a_n^f\cos(nx)+b_n^f\sin(nx)) \\ &=& \sum_{n=1}^{\infty}(-1)^{n+1}\frac{2}{n} \sin(nx), \quad \forall x\in (-\pi,\pi). \end{eqnarray*} Parseval's theorem asserts that:

$$\frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)dx = 2(a_0^f)^2 + \sum_{k=1}^{\infty}[(a_k^f)^2+(b_k^f)^2].$$

So we apply this to the function $f(x)= x $ $$2(a_0^f)^2 + \sum_{k=1}^{\infty}[(a_k^f)^2+(b_k^f)^2]= \sum_{n=1}^{\infty} \frac{4}{n^2}= 4\sum_{n=1}^{\infty}\frac{1}{n^2}$$ and $$\frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)dx = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2dx= \frac{2\pi^2}{3}.$$

Hence by Parseval's equality $$4\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{2\pi^2}{3}$$ and hence $$\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.$$