Evaluating zeta function at 6, ζ(6)

Here is my answer to this Mathematics Stach Exchange question  by Chon  on how to compute \zeta(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.\qquad (1)

The computation is as follows:

f(x)=x^{2p}=\dfrac{a_{0,2p}}{2}+\displaystyle\sum_{n=1}^{\infty }\left( a_{n,2p}\cos nx+b_{n,2p}\sin nx\right),

where the coefficients are given by the following integrals

\begin{aligned}a_{0,2p}&=\dfrac{1}{\pi }\displaystyle\int_{-\pi }^{\pi }x^{2p}\;\mathrm{d}x=\dfrac{2\pi ^{2p}}{2p+1}\\a_{n,2p}&=\dfrac{1}{\pi }\displaystyle\int_{-\pi }^{\pi }x^{2p}\cos nx\;\mathrm{d}x=\dfrac{2}{\pi }\displaystyle\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x,\\b_{n,2p}&=\dfrac{1}{\pi }\displaystyle\int_{-\pi }^{\pi }x^{2p}\sin nx\;\mathrm{d}x=0.\end{aligned}

The series expansion is thus

x^{2p}=\dfrac{\pi ^{2p}}{2p+1}+\dfrac{2}{\pi }\sum_{n=1}^{\infty }\cos nx\displaystyle\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x.\qquad(2)

For f(\pi )=\pi ^{2p} we obtain

\begin{aligned}\pi ^{2p}=\dfrac{\pi ^{2p}}{2p+1}+\dfrac{2}{\pi }\displaystyle\sum_{n=1}^{\infty }\cos n\pi\displaystyle\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x,  \end{aligned}

where the integral

\begin{aligned}I_{n,2p}:=\displaystyle\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x  \end{aligned}

satisfies the following recurrence, as can be shown by integration by parts

\begin{aligned}  I_{n,2p}=\dfrac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\dfrac{2p(2p-1)}{n^{2}}  I_{n,2\left( p-1\right) },\qquad I_{n,0}=0.\qquad(3)  \end{aligned}

For p=1, we get

\begin{aligned}  I_{n,2}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi  \end{aligned}


\begin{aligned}  \pi ^{2} &=\dfrac{\pi ^{2}}{3}+\dfrac{2}{\pi }\displaystyle\sum_{n=1}^{\infty }\cos n\pi  \cdot I_{n,2}\\&=\dfrac{\pi ^{2}}{3}+\dfrac{2}{\pi }\displaystyle\sum_{n=1}^{\infty }\cos n\pi \left(\dfrac{2}{n^{2}}\pi \cos n\pi \right)\\&=\dfrac{\pi ^{2}}{3}+4\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}\\&\Rightarrow\zeta (2)=\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}=\dfrac{\pi ^{2}}{6}.\end{aligned}

For p=2, we get

\begin{aligned}I_{n,4}=\left( \dfrac{4\pi ^{3}}{n^{2}}-\dfrac{24\pi }{n^{4}}\right) \cos n\pi\end{aligned}


\begin{aligned}\pi ^{4}&=\dfrac{\pi ^{4}}{5}+\dfrac{2}{\pi }\displaystyle\sum_{n=1}^{\infty }\cos n\pi\cdot I_{n,4}=\dfrac{\pi ^{4}}{5}+\dfrac{4\pi ^{4}}{3}-48\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{4}}\\  &\Rightarrow\zeta (4)=\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{4}}=\dfrac{\pi ^{4}}{  90}.\end{aligned}

Finally for p=3, we get

\begin{aligned}I_{n,6}=\left( \dfrac{6\pi ^{5}}{n^{2}}-\dfrac{120\pi ^{3}}{n^{4}}+\dfrac{720}{n^{6}}\right) \cos n\pi\end{aligned}


\begin{aligned}\pi ^{6}=\dfrac{\pi ^{6}}{7}+2\displaystyle\sum_{n=1}^{\infty }\left( \dfrac{6\pi ^{4}}{n^{2}}-\dfrac{120\pi ^{2}}{n^{4}}+\dfrac{720}{n^{6}}\right),\end{aligned}

from which the result follows

\zeta(6)=\begin{aligned}\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{6}}=\dfrac{\pi ^{6}}{945}.\end{aligned}

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}.


Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Análise de Fourier, Matemática, Math, Mathematics Stack Exchange, Séries com as etiquetas , . ligação permanente.

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