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By applying Parseval's identity (Lyapunov equation) 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:
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 $$\frac{1}{2\pi}\int_{-\pi}^\pi(f(x))^2\,dx = \frac{a_0^2}{4}+\frac{1}{2}\sum_{n=1}^\infty(a_n^2+b_n^2)$$ reads now $$\frac{1}{\pi}\int_0^\pi(x^2)^2\,dx = \frac{1}{\pi}\!\sijoitus{0}{\quad\pi}\frac{x^5}{5} = \frac{\pi^4}{5}$$ and its right hand side $$\frac{1}{4}\!\left(\frac{2\pi^2}{3}\right)^2+\frac{1}{2}\sum_{n=1}^\infty\left(\frac{4}{n^2}\right)^2 = \frac{\pi^4}{9}+8\sum_{n=1}^\infty\frac{1}{n^4} = \frac{\pi^4}{9}+8\zeta(4).$$ Accordingly, we obtain the result
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