value of Riemann zeta function at $s=4$

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

 $\frac{a_{0}}{2}+(a_{1}\cos{x}+b_{1}\sin{x})+(a_{2}\cos{2x}+b_{2}\sin{2x})+\ldots$

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

 $b_{n}=0\quad\forall\,n=1,\,2,\,3,\,\ldots.$

Then

 $a_{0}=\frac{1}{\pi}\int_{-\pi}^{\pi}x^{2}\,dx=\frac{2}{\pi}\int_{0}^{\pi}x^{2}% \,dx\,=\,\frac{2\pi^{2}}{3}.$

For other coefficients $a_{n}$, we must perform twice integrations by parts:

 $\displaystyle a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}x^{2}\cos{nx}\,dx$ $\displaystyle=\frac{2}{\pi}\int_{0}^{\pi}x^{2}\cos{nx}\,dx$ $\displaystyle=\frac{2}{\pi}\left(\!\operatornamewithlimits{\Big{/}}_{\!\!\!0}^% {\,\quad\pi}x^{2}\cdot\frac{\sin{nx}}{n}-\int_{0}^{\pi}2x\cdot\frac{\sin{nx}}{% n}\,dx\right)$ $\displaystyle=-\frac{4}{n\pi}\int_{o}^{\pi}x\sin{nx}\,dx$ $\displaystyle=-\frac{4}{n\pi}\left(\operatornamewithlimits{\Big{/}}_{\!\!\!0}^% {\,\quad\pi}x\cdot\frac{-\cos{nx}}{n}-\int_{0}^{\pi}1\cdot\frac{-\cos{nx}}{n}% \,dx\right)$ $\displaystyle=-\frac{4}{n\pi}\operatornamewithlimits{\Big{/}}_{\!\!\!0}^{\,% \quad\pi}\left(\frac{-x\cos{nx}}{n}-\frac{\sin{nx}}{n^{2}}\right)$ $\displaystyle=\frac{4\cos{n\pi}}{n^{2}}\;=\;\frac{4(-1)^{n}}{n^{2}}\quad% \forall\,n=1,\,2,\,3,\,\ldots$

Thus

 $x^{2}\;=\;\frac{\pi^{2}}{3}+\sum_{n=1}^{\infty}\frac{4(-1)^{n}}{n^{2}}\cos{nx}% \quad\mbox{for}\;\;-\pi\leqq x\leqq\pi.$

The left hand side of Parseval’s identity

 $\frac{1}{2\pi}\int_{-\pi}^{\pi}(f(x))^{2}\,dx=\frac{a_{0}^{2}}{4}+\frac{1}{2}% \sum_{n=1}^{\infty}(a_{n}^{2}+b_{n}^{2})$

 $\frac{1}{\pi}\int_{0}^{\pi}(x^{2})^{2}\,dx=\frac{1}{\pi}\!% \operatornamewithlimits{\Big{/}}_{\!\!\!0}^{\,\quad\pi}\frac{x^{5}}{5}=\frac{% \pi^{4}}{5}$
 $\frac{1}{4}\!\left(\frac{2\pi^{2}}{3}\right)^{2}+\frac{1}{2}\sum_{n=1}^{\infty% }\left(\frac{4}{n^{2}}\right)^{2}=\frac{\pi^{4}}{9}+8\sum_{n=1}^{\infty}\frac{% 1}{n^{4}}=\frac{\pi^{4}}{9}+8\zeta(4).$
 $\displaystyle\zeta(4)\;=\;1+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots\;=\;\frac{% \pi^{4}}{90}.$ (1)
Title value of Riemann zeta function at $s=4$ ValueOfRiemannZetaFunctionAtS4 2013-03-22 18:22:06 2013-03-22 18:22:06 pahio (2872) pahio (2872) 7 pahio (2872) Example msc 11M06 SubstitutionNotation CosineAtMultiplesOfStraightAngle ValueOfTheRiemannZetaFunctionAtS2