# 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 $f$:

$$S(f)= \frac{\pi^2}{6}-\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n^2}$$

$$x=0, \zeta(2)=\pi^2/6$$

• What did you find for the Fourier series? Commented Mar 3, 2012 at 15:34
• Generally for $\zeta(6)$ you will need to do the Fourier series for some polynomial of higher degree. Commented Mar 3, 2012 at 17:01
• Replace $f$ with $f_0 = f - \bar{f}$, where $\bar{f}$ is the mean of $f$ over the interval in question. This is again a quadratic function and the Fourier series can be obtained from that of $f$. Now let $F_0$ be a periodic antiderivative of $f_0$. Its Fourier series can be obtained by integrating $S(f)$ and therefore will have coefficients like $n^{-3}$. Parseval's Theorem should allow you to arrive at $\zeta(6)$. Commented Apr 5, 2012 at 12:08
• The following MSE link shows a standard technique that can be used to find $\zeta(6).$ Commented Feb 19, 2014 at 23:29

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}

• Totally spiffy! Commented Mar 27, 2022 at 18:14
• How do you draw the conclusion after $(3)$ that $a_n = -2\zeta(2n)/\pi^{2n}$? Commented Apr 30 at 16:32
• The equations \begin{align}\sum_{k=1}^\infty\zeta(2k)\,x^{2k}&=\frac12(1-\pi x\cot(\pi x))\tag{5a}\\&=\frac12\left(1-\sum_{k=0}^\infty a_k\pi^{2k}x^{2k}\right)\tag{5b}\end{align} follow from $(1)$ and the first equation of $(2)$. Then, by equating coefficients of $x^{2k}$ we get $a_k=-2\frac{\zeta(2k)}{\pi^{2k}}$.
– robjohn
Commented May 1 at 2:46

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 $$(\zeta(2p))_{p\ge 1}$$.

• I think you have a typo: $\pi^6/945$, not $\pi^4/945$. Commented Mar 3, 2012 at 23:45
• @alex.jordan: Thanks! Fixed. Commented Mar 3, 2012 at 23:48
– Chon
Commented Mar 4, 2012 at 17:22
• @Chon: You are welcome! Commented Mar 5, 2012 at 14:32
• @clathratus tanks for changing integration variable in integrals Commented Jun 2, 2022 at 9:09

You may start with (for $x \in [0,1]$)
$$B_1(x)=x-\frac 12=2\sum_{n=1}^{\infty} \frac{(-1)^n}{2\pi n}\sin\left(2\pi n\left(x-\frac 12\right)\right)$$

and integrate multiple times $x-\frac 12$ (adding the necessary constant when required!) to get :

$$B_{2k}(x)=2(-1)^k\sum_{n=1}^{\infty} \frac{(-1)^n}{(2\pi n)^{2k}}\cos\left(2\pi n\left(x-\frac 12\right)\right)$$ (this is detailed in this paper page 6)

You could too use following formula $$\cot(\pi z)=\frac1{\pi}\left[\frac1{z}-\sum_{k=1}^{\infty}\frac{2z}{k^2-z^2}\right]$$ (obtained from computation of Fourier series of $\cos(zx)$ for $-\pi \le x \le \pi$ with the result :

$$\cos(zx)=\frac{2z\sin(\pi z)}{\pi}\left[\frac1{2z^2}+\frac{\cos(1x)}{1^2-z^2}-\frac{\cos(2x)}{2^2-z^2}+\frac{\cos(3x)}{3^2-z^2}-\cdots\right]$$ applied to $x=\pi$)

and deduce an interesting expansion of $\frac z2\left(\cot\left(\frac z2\right)\right)$ in powers of $z^2$.

• Thank you for your help!
– Chon
Commented Mar 3, 2012 at 22:09

Let $f(t):=t^3\ \ (-\pi\leq t\leq \pi)$, extended to all of ${\mathbb R}$ periodically with period $2\pi$. The Fourier series of this function is $$t^3=\sum_{k=1}^\infty {2(-1)^{k-1}(k^2\pi^2-6)\over k^3}\sin(kt)\qquad(-\pi\leq t\leq \pi).$$ We now use Parseval's formula $$\|f\|^2=\sum_{k=0}^\infty |c_k|^2.$$ But $$\|f\|^2={1\over\pi}\int_{-\pi}^\pi x^6\>dt={2\pi^6\over7}$$ and
$$c_k^2={4(k^2\pi^2-6)^2\over k^6}=4\left(\frac{36}{k^6}-\frac{12\pi^2}{k^4}+\frac{\pi^4}{k^2}\right),k\geq1.$$ Noting that $$\sum_{k=1}^\infty \frac{12\pi^2}{k^4}=12\pi^2\frac{\pi^4}{90}=\frac{2\pi^6}{15}, \sum_{k=1}^\infty\frac{\pi^4}{k^2}=\pi^4\frac{\pi^2}{6}=\frac{\pi^6}{6}$$ we have $$\sum_{k=1}^\infty \frac{1}{k^6}=\left({\pi^6\over14}+\frac{2\pi^2}{15}-\frac{\pi^4}{6}\right)/36=\frac{\pi^6}{945}.$$

Similar to @RobJohn's solution: Setting $$z=\pi$$ in the Fourier series of $$\cos(z x)$$:

$$\cos(zx)=\frac{2x\sin(\pi x)}{\pi}\left[\frac1{2x^2}-\sum_{k=1}^\infty\frac{(-1)^k\cos(kz)}{k^2-x^2}\right],\quad x\notin\mathbb{Z}$$

gives

\begin{align} \cot(\pi x)&=\frac1{\pi x}-\frac2{\pi x}\sum_{k=1}^\infty\frac{x^2}{k^2-x^2}\\ &=\frac1{\pi x}-\frac2{\pi x}\sum_{k=1}^\infty\frac{\frac{x^2}{k^2}}{1-\frac{x^2}{k^2}}\\ &=\frac1{\pi x}-\frac2{\pi x}\sum_{k=1}^\infty\sum_{n=1}^\infty\left(\frac{x^2}{k^2}\right)^n\\ &=\frac1{\pi x}-\frac2{\pi x}\sum_{n=1}^\infty x^{2n}\sum_{k=1}^\infty\frac{1}{k^{2n}}\\ &=\frac1{\pi x}-\frac2{\pi x}\color{red}{\sum_{n=1}^\infty \zeta(2n)x^{2n}}\\ \end{align}

Expanding $$\cot x$$ in Taylor series:

$$\cot x=\frac1{x}\sum_{n=0}^\infty\!\frac{(-1)^{n} B_{2n}}{(2n)!}(2x)^{2n}$$

or \begin{align} \cot(\pi x)&=\frac1{\pi x}\sum_{n=0}^\infty\!\frac{(-1)^{n} B_{2n}}{(2n)!}(2\pi x)^{2n}\\ &=\frac1{\pi x}-\frac2{\pi x}\color{red}{\sum_{n=1}^\infty\!\frac{(-1)^{n+1} B_{2n}(2\pi)^{2n}}{2(2n)!}x^{2n}} \end{align}

yields

$$\sum_{n=1}^\infty\!\zeta(2n)x^{2n} =\sum_{n=1}^\infty\!\frac{(-1)^{n+1} B_{2n}(2\pi)^{2n}}{2(2n)!}x^{2n}$$ Compare the coefficients of $$x^{2n}$$ to get

$$\zeta(2n)=\frac{(-1)^{n+1} B_{2n}(2\pi)^{2n}}{2(2n)!}.$$

Let $$g(x) = \sum_{n=1}^{\infty} \frac{\cos(2nx)}{n^2}$$

Then

$$g^3(x)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\sum_{k=1}^{\infty} \frac{\cos(2nx)\cos(2mx)\cos(2kx)}{n^2m^2k^2}$$

Since $\cos(a)\cos(b)=\frac{1}{2}\left(\cos(a+b)+\cos(a-b)\right)$ then

$$g^3(x)=\frac{1}{2}\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\sum_{k=1}^{\infty} \frac{\cos(2nx)\cos(2mx+2kx)+\cos(2nx)\cos(2mx-2kx)}{n^2m^2k^2}$$

Due to orthogonality of cosine function we get after integration :

$$\int_0^\pi g^3(x)\;\mathrm{d}x=\frac{\pi}{4}\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\sum_{k=1}^{\infty} \frac{\delta_{n,m+k}+\delta_{n,m-k}}{n^2m^2k^2}= \frac{\pi}{2}\sum_{m=1}^{\infty}\sum_{k=1}^{\infty} \frac{1}{(m+k)^2m^2k^2}$$