Computing $\zeta(6)=\sum\limits_{k=1}^\infty \frac1{k^6}$ with Fourier series. 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 $:
$$ S(f)= \frac{\pi^2}{6}-\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n^2}$$
$$ x=0,  \zeta(2)=\pi^2/6$$

I have posted here in Portuguese a recursive method based on the computation of the Fourier trigonometric series expansion for the function defined in $\left[ -\pi ,\pi \right] $ by  $f(x)=x^{2p}$ and extended to all of ${\mathbb R}$ periodically with period $2\pi.$ This is a shorter description than the original. In this reply I outline the case $\zeta(4)$. For $p=3$ the expansion is
$$x^{6}=\dfrac{\pi ^{6}}{7}+2\displaystyle\sum_{n\ge 1}^{}\left( \left( \dfrac{6}{n^{2}}\pi ^{4}-\dfrac{120}{n^{4}}\pi ^{2}+\dfrac{720 }{n^{6}}\right)\cos n\pi \right) \cos nx.\tag{1}$$
The computation is as follows:
$$\begin{equation*}
f(x)=x^{2p}=\frac{a_{0,2p}}{2}+\sum_{n=1}^{\infty }\left( a_{n,2p}\cos
nx+b_{n,2p}\sin nx\right) ,
\end{equation*}$$
where the coefficients are given by the following integrals
$$\begin{eqnarray*}
a_{0,2p} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }x^{2p}\;\mathrm{d}x=\frac{2\pi ^{2p}}{2p+1},
\\
a_{n,2p} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }x^{2p}\cos nx\;\mathrm{d}x=\frac{2}{\pi }
\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x, \\
b_{n,2p} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }x^{2p}\sin nx\;\mathrm{d}x=0.
\end{eqnarray*}$$
The series expansion is thus
$$\begin{equation*}
x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos
nx\int_{0}^{\pi }t^{2p}\cos nt\;\mathrm{d}t.\tag{2}
\end{equation*}$$
For $f(\pi )=\pi ^{2p}$ we obtain
$$
\begin{equation*}
\pi ^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi
\int_{0}^{\pi }t^{2p}\cos nt\;\mathrm{d}t,
\end{equation*}$$
where the integral
$$
\begin{equation*}
I_{n,2p}:=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x
\end{equation*}$$
satisfies the following recurrence, as can be shown by integration by parts
$$\begin{equation*}
I_{n,2p}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\frac{2p(2p-1)}{n^{2}}
I_{n,2\left( p-1\right) },\qquad I_{n,0}=0.\tag{3}
\end{equation*}$$

*

*For $p=1$, we get
$$\begin{equation*}
I_{n,2}=\frac{2}{n^{2}}\pi\cos n\pi. 
\end{equation*}$$
and
$$\begin{eqnarray*}
\pi ^{2} &=&\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi
\cdot I_{n,2} \\
&=&\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi \left( 
\frac{2}{n^{2}}\pi \cos n\pi \right)  \\
&=&\frac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\frac{1}{n^{2}} \\
&\Rightarrow &\zeta (2)=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6
}
\end{eqnarray*}$$

*For $p=2$, we get
$$
\begin{equation*}
I_{n,4}=\left( \frac{4\pi ^{3}}{n^{2}}-\frac{24\pi }{n^{4}}\right) \cos n\pi 
\end{equation*}$$
and
$$
\begin{eqnarray*}
\pi ^{4} &=&\frac{\pi ^{4}}{5}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi
\cdot I_{n,4}=\frac{\pi ^{4}}{5}+\frac{4\pi ^{4}}{3}-48\sum_{n=1}^{\infty }
\frac{1}{n^{4}} \\
&\Rightarrow &\zeta (4)=\sum_{n=1}^{\infty }\frac{1}{n^{4}}=\frac{\pi ^{4}}{
90}.
\end{eqnarray*}$$

*Finally for $p=3$, we get
$$\begin{equation*}
I_{n,6}=\left( \frac{6\pi ^{5}}{n^{2}}-\frac{120\pi ^{3}}{n^{4}}+\frac{720}{
n^{6}}\right) \cos n\pi 
\end{equation*}$$
and
$$
\begin{equation*}
\pi ^{6}=\frac{\pi ^{6}}{7}+2\sum_{n=1}^{\infty }\left( \frac{6\pi ^{4}}{
n^{2}}-\frac{120\pi ^{2}}{n^{4}}+\frac{720}{n^{6}}\right), 
\end{equation*}$$
from which the result follows
$$\zeta(6)=
\begin{equation*}
\sum_{n=1}^{\infty }\frac{1}{n^{6}}=\frac{\pi ^{6}}{945}.
\end{equation*}$$
Plots of the periodic function defined in $\left[ -\pi ,\pi \right] $ by $f(x)=x^{6}$ (blue curve) and of the partial sum with the first 10 terms of its Fourier trigonometric series (red curve).

This method generates recursively the sequence $(\zeta(2p))_{p\ge 1}$.
