Let $ f$ be a function such that $ f\in C_{2\pi}^{0}(\mathbb{R},\mathbb{R}) $ (f is $2\pi$-periodic) such that $ \forall x \in [0,\pi]$: $$f(x)=x(\pi-x)$$
Computing the Fourier series of $f$ and using Parseval's identity, I have computed $\zeta(2)$ and $\zeta(4)$.
How can I compute $ \zeta(6) $ now?
Fourier series of $ f $:
$$ S(f)= \frac{\pi^2}{6}-\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n^2}$$
$$ x=0, \zeta(2)=\pi^2/6$$
Best Answer
One method is to consider the generating function of $\zeta(2k)$: $$ \begin{align} f(x) &=\sum_{k=1}^\infty\zeta(2k)\,x^{2k}\\ &=\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{x^{2k}}{n^{2k}}\\ &=\sum_{n=1}^\infty\frac{x^2/n^2}{1-x^2/n^2}\\ &=\sum_{n=1}^\infty\frac{x^2}{n^2-x^2}\\ &=-\frac{x}{2}\sum_{n=1}^\infty\left(\frac{1}{x-n}+\frac{1}{x+n}\right)\\ &=-\frac{x}{2}\left(\pi\cot(\pi x)-\frac1x\right)\\ &=\frac12(1-\pi x\cot(\pi x))\tag{1} \end{align} $$ In light of equation $(1)$, find the power series of $$ x\cot(x)=\sum_{k=0}^\infty a_kx^{2k} $$ $$ \cos(x)=\frac{\sin(x)}{x}\sum_{k=0}^\infty a_kx^{2k} $$ $$ \begin{align} \sum_{n=0}^\infty(-1)^n\!\frac{x^{2n}}{(2n)!} &=\sum_{n=0}^\infty(-1)^n\!\frac{x^{2n}}{(2n+1)!}\;\;\sum_{k=0}^\infty a_kx^{2k}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n(-1)^k\!\frac{a_{n-k}}{(2k+1)!}\right)x^{2n}\tag{2} \end{align} $$ Comparing the coefficients of the powers of $x$ in $(2)$ yields $$ \begin{align} a_n &=\frac{(-1)^n}{(2n)!}-\sum_{k=1}^n(-1)^k\!\frac{a_{n-k}}{(2k+1)!}\\ &=\frac{(-1)^n2n}{(2n+1)!}-\sum_{k=1}^{n-1}(-1)^k\!\frac{a_{n-k}}{(2k+1)!}\tag{3} \end{align} $$ Since $a_n=-2\dfrac{\zeta(2n)}{\pi^{2n}}$ for positive $n$, $(3)$ becomes $$ \zeta(2n)=\frac{(-1)^{n-1}\pi^{2n}}{(2n+1)!}n+\sum_{k=1}^{n-1}\!\frac{(-1)^{k-1}\pi^{2k}}{(2k+1)!}\zeta(2n-2k)\tag{4} $$ Equation $(4)$ gives $\zeta(2n)$ recursively for positive $n$: $$ \begin{align} \zeta(2)&=\frac{\pi^2}{3!}=\frac{\pi^2}{6}\\ \zeta(4)&=-\frac{\pi^4}{5!}2+\frac{\pi^2}{3!}\zeta(2)=\frac{\pi^4}{90}\\ \zeta(6)&=\frac{\pi^6}{7!}3-\frac{\pi^4}{5!}\zeta(2)+\frac{\pi^2}{3!}\zeta(4)=\frac{\pi^6}{945}\\ \zeta(8)&=-\frac{\pi^8}{9!}4+\frac{\pi^6}{7!}\zeta(2)-\frac{\pi^4}{5!}\zeta(4)+\frac{\pi^2}{3!}\zeta(6)=\frac{\pi^8}{9450} \end{align} $$