value of Riemann zeta function at $s=4$

By applying Parseval’s identity (Lyapunov equation (http://planetmath.org/PersevalEquality)) to the Fourier series

$$\frac{a_0}{2}+(a_1\cos x+b_1\sin x)+(a_2\cos 2x+b_2\sin 2x)+\mathrm{\dots }$$

of $x^2$ on the interval  $[-\pi ,\pi ]$,  one may derive the value of Riemann zeta function at  $s=4$.

Let us first find the needed Fourier coefficients $a_n$ and $b_n$.  Since $x^2$ defines an even function, we have

$$b_n=0\mathit{\hspace{1em}}\forall n=1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots }.$$

Then

$$a_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^2𝑑x=\frac{2}{\pi }{\int }_0^{\pi }{x}^2𝑑x=\frac{2{\pi }^2}{3}.$$

For other coefficients $a_n$, we must perform twice integrations by parts:

$a_n={\displaystyle \frac{1}{\pi }}{\displaystyle {\int }_{-\pi }^{\pi }}{x}^2\cos nxdx$$={\displaystyle \frac{2}{\pi }}{\displaystyle {\int }_0^{\pi }}{x}^2\cos nxdx$

$={\displaystyle \frac{2}{\pi }}\left(\underset{0}{\overset{\pi }{/}}{x}^2\cdot {\displaystyle \frac{\sin nx}{n}}-{\displaystyle {\int }_0^{\pi }}2x\cdot {\displaystyle \frac{\sin nx}{n}}𝑑x\right)$

$=-{\displaystyle \frac{4}{n\pi }}{\displaystyle {\int }_o^{\pi }}x\sin nxdx$

$=-{\displaystyle \frac{4}{n\pi }}\left(\underset{0}{\overset{\pi }{/}}x\cdot {\displaystyle \frac{-\cos nx}{n}}-{\displaystyle {\int }_0^{\pi }}1\cdot {\displaystyle \frac{-\cos nx}{n}}𝑑x\right)$

$=-{\displaystyle \frac{4}{n\pi }}\underset{0}{\overset{\pi }{/}}\left({\displaystyle \frac{-x\cos nx}{n}}-{\displaystyle \frac{\sin nx}{n^2}}\right)$

$={\displaystyle \frac{4\cos n\pi }{n^2}}={\displaystyle \frac{4{(-1)}^n}{n^2}}\mathit{\hspace{1em}}\forall n=1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots }$

Thus

$$x^2=\frac{{\pi }^2}{3}+\sum _{n=1}^{\mathrm{\infty }}\frac{4{(-1)}^n}{n^2}\cos nx\mathit{\hspace{1em}}\text{for}-\pi \leqq x\leqq \pi .$$

The left hand side of Parseval’s identity

$$\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{(f(x))}^2𝑑x=\frac{a_0^2}{4}+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}(a_n^2+b_n^2)$$

reads now

$$\frac{1}{\pi }{\int }_0^{\pi }{(x^2)}^2𝑑x=\frac{1}{\pi }\underset{0}{\overset{\pi }{/}}\frac{x^5}{5}=\frac{{\pi }^4}{5}$$

and its right hand side

$$\frac{1}{4}{\left(\frac{2{\pi }^2}{3}\right)}^2+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{4}{n^2}\right)}^2=\frac{{\pi }^4}{9}+8\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^4}=\frac{{\pi }^4}{9}+8\zeta (4).$$

Accordingly, we obtain the result

$\zeta (4)=\mathrm{\hspace{0.33em}1}+{\displaystyle \frac{1}{2^4}}+{\displaystyle \frac{1}{3^4}}+\mathrm{\dots }={\displaystyle \frac{{\pi }^4}{90}}.$(1)