value of the Riemann zeta function at $s=2$

Here we present an application of Parseval’s equality to numbertheory. Let $\zeta (s)$ denote the Riemann zeta function. We willcompute the value

$$\zeta (2)$$

with the help of Fourier analysis.

Example:

Let $f:ℝ\to ℝ$ be the “identity” function,defined by

$$f(x)=x,\text{ for all }x\in ℝ.$$

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

Thus

$f(x)=x$$=$$a_0^f+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}(a_n^f\cos (nx)+b_n^f\sin (nx))$

$=$${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{(-1)}^{n+1}{\displaystyle \frac{2}{n}}\sin (nx),\forall x\in (-\pi ,\pi ).$

Parseval’s theorem asserts that:

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

So we apply this to the function $f(x)=x$:

$$2{(a_0^f)}^2+\sum _{k=1}^{\mathrm{\infty }}[{(a_k^f)}^2+{(b_k^f)}^2]=\sum _{n=1}^{\mathrm{\infty }}\frac{4}{n^2}=4\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}$$

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

$$4\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{2{\pi }^2}{3}$$

and hence

$$\zeta (2)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}.$$