# Fourier series method for zeta functions

In a previous post we discussed some methods for showing that

$\displaystyle \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$

where the zeta function is defined to be

$\displaystyle \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$

for $s \in \mathbb{C}$.

Apéry’s constant is the value of the zeta function at $s = 3$

$\displaystyle \zeta(3) = \sum_{n=1}^{\infty} \frac{1}{n^3} = 1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{64} + \frac{1}{125} + \dots$

named for Roger Apéry who proved this number irrational in 1978 [1].

Prior to Apéry’s result, it was known that the zeta function, evaluated at the positive even numbers, was irrational with closed form

$\displaystyle \zeta(2k) = \sum_{n=1}^{\infty} \frac{1}{n^{2k}} = (-1)^{k-1}\frac{(2\pi)^{2k}}{2\cdot (2k)!} B_{2k}$

for $k \in \mathbb{N}$ where $B_{2k}$ are (rational) Bernoulli numbers (see [3] for proof).

For example $\displaystyle \zeta(2) = \frac{\pi^2}{6}$, $\displaystyle \zeta(4) = \frac{\pi^4}{90}$, $\displaystyle \zeta(6) = \frac{\pi^6}{945}$, etc.

Yet no simple form is known for Apéry’s constant $\zeta(3)$ or indeed any of the other odd zeta values, with the exception of $\zeta(1)$. Indeed, it has been postulated that no such simple form is even possible. Nevertheless, there are various closed form solutions involving infinite sums or integrals.

So why is finding a simple expression for the odd values of the zeta function so much more difficult than the even case? We do not know the answer to this question, but what we can do is try to generalise the proofs previously given for $\zeta(2k)$ and see where they fail. This is our focus for the remainder of this post.

Consider the proof that $\zeta(2) = \frac{\pi}{6}$ based on Fourier series given here for example. To construct an equivalent statement for $\zeta(3)$ it seems natural to consider the Fourier series of $x^3$ over the domain $-\pi \leq x \leq \pi$ so that

$\displaystyle f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx) + \sum_{n=1}^{\infty} b_n \sin(nx)$

is convergent to $x^3$ on $[-\pi,\pi]$ where

$a_n = \displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} x^3 \cos(nx) \,dx$

for $n = 0,1,2,3, \dots$ and

$b_n = \displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} x^3 \sin(nx) \,dx = \frac{2}{\pi} \int_0^{\pi} x^3 \sin(nx) \,dx$

for $n = 1,2,3,\dots$ Thus $a_n = 0$ for $n \in \mathbb{N}$.

Integrating by parts thrice we have (up to an additive constant)

$\displaystyle \int x^3 \sin(nx) \,dx = - \frac{2\sin(nx)}{n^4} + \frac{6x \cos(nx)}{n^3} + \frac{3x^2 \sin(nx)}{n^2} - \frac{x^3 \cos(nx)}{n}$

And therefore

\begin{aligned} b_n & = \displaystyle \frac{2}{\pi} \Big[- \frac{2\sin(nx)}{n^4} + \frac{6x \cos(nx)}{n^3} + \frac{3x^2 \sin(nx)}{n^2} - \frac{x^3 \cos(nx)}{n} \Big]_0^{\pi} \\ & = (-1)^n \Big(\frac{12}{n^3} - \frac{2 \pi^2}{n} \Big) \end{aligned}

Thus we have the Fourier series expansion

$f(x) = \displaystyle \sum_{n=1}^{\infty} (-1)^n \Big(\frac{12}{n^3} - \frac{2 \pi^2}{n} \Big) \sin(nx)$

We see now a clear difference in the situations, as we have inherited a sum over $\frac{1}{n}$ (the harmonic series) which would seem to complicate the analysis. Considering $x = \pi$ is no longer effective as $\sin(nx) = 0$ whenever $x = \pi$. Perhaps then we might take $x = \frac{\pi}{2}$ so that

\begin{aligned} \displaystyle \Big( \frac{\pi}{2} \Big)^3 & = \sum_{n=1}^{\infty} (-1)^n \sin(n \frac{\pi}{2}) \Big( \frac{12}{n^3} - \frac{2 \pi^2}{n} \Big) \\ & = \sum_{k=1}^{\infty} (-1)^k \Big( \frac{12}{(2k-1)^3} - \frac{2 \pi^2}{2k-1} \Big) \end{aligned}

We know that $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^k}{2k-1} = - \pi/4$ so then

\begin{aligned} \displaystyle \Big( \frac{\pi^3}{8} - \frac{2\pi^3}{4} \Big) / 12 & = \sum_{k=1}^{\infty} \frac{ (-1)^k}{(2k-1)^3} \end{aligned}

or equivalently

$\displaystyle \sum_{k=0}^{\infty} \frac{ (-1)^k}{(2k+1)^3} = \frac{\pi^3}{32}$

Thus, in spite of our original goal, we are lead to a well-known value for the Dirichlet beta function.

It is natural then to enquire about the value of the sum

$G = \displaystyle \sum_{k=0}^{\infty} \frac{ (-1)^k}{(2k+1)^2}$

This sum is known as Catalan’s constant and it is unknown whether it is irrational.

The even values of the Dirichlet eta function are known, for example

$\displaystyle \eta(2) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} = \frac{\zeta(2)}{2} = \frac{\pi^2}{12}$

It is interesting to take $x = \pi/4$ in the Fourier series given above, as we know $\sin(\pi/4) = 1/\sqrt{2}$, and so

\begin{aligned} \displaystyle \Big( \frac{\pi}{4} \Big)^3 & = \sum_{n=1}^{\infty} (-1)^n \sin(n \frac{\pi}{4}) \Big( \frac{12}{n^3} - \frac{2 \pi^2}{n} \Big) \\ \end{aligned}

where $\sin(\frac{n\pi}{4})$ cycles through the values $(1/ \sqrt{2},1,1/ \sqrt{2},0,-1/ \sqrt{2},-1,-1/ \sqrt{2},0)$ so that

\begin{aligned} \displaystyle \Big( \frac{\pi}{4} \Big)^3 & = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{\sqrt{2}} \Big( \frac{12}{(4k-3)^3} - \frac{2 \pi^2}{4k-3} \Big) + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{\sqrt{2}} \Big( \frac{12}{(4k-1)^3} - \frac{2 \pi^2}{4k-1} \Big) \\ & + \sum_{k=1}^{\infty} (-1)^{k+1} \Big( \frac{12}{(4k-2)^3} - \frac{2 \pi^2}{4k-2} \Big) \end{aligned}

Wolfram Alpha gives

$\displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k}{4k+1} = \frac{\pi + 2\ln(1+\sqrt{2})}{4\sqrt{2}}$

$\displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k}{4k+2} = \frac{\pi}{8}$

and a monstrosity for

$\displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k}{4k+3}$

Next, we consider the Fourier series expansion of $|x^3|$ , in which case

$\displaystyle \int x^3 \cos(nx) \,dx = - \frac{6\cos(nx)}{n^4} - \frac{6x\sin(nx)}{n^3}+ \frac{3x^2\cos(nx)}{n^2} + \frac{x^3\sin(nx)}{n}$

so that

\begin{aligned} \displaystyle \int_0^{\pi} x^3 \cos(nx) \,dx & = (-1)^{n+1} \frac{6}{n^4} + (-1)^n \frac{3x^2}{n^2} - \frac{6}{n^4} \end{aligned}

Thus $a_n = \frac{2}{\pi} \Big( (-1)^{n+1} \frac{6}{n^4} + (-1)^n \frac{3x^2}{n^2} - \frac{6}{n^4} \Big)$ for $n = 1,2,3,\dots$ and

$a_0 = \displaystyle \frac{2}{\pi} \int_0^{\pi} x^3 \,dx = \frac{2}{\pi} \cdot \frac{\pi^4}{4} = \frac{\pi^3}{2}$

Then evaluating the Fourier series for $|x^3|$ at $x = \pi$ gives

\begin{aligned} \displaystyle \pi^3 & = \frac{\pi^3}{4} + \frac{2}{\pi} \sum_{n=1}^{\infty} \Big( (-1)^{n+1} \frac{6}{n^4} + (-1)^n \frac{3\pi^2}{n^2} - \frac{6}{n^4} \Big) \cos(n\pi) \\ & = \frac{\pi^3}{4} +\frac{2}{\pi} \sum_{n=1}^{\infty} \Big( - \frac{6}{n^4} + \frac{3\pi^2}{n^2} + (-1)^{n+1} \frac{6}{n^4} \Big) \\ & =\frac{\pi^3}{4} - \frac{12}{\pi} \sum_{n=1}^{\infty} \frac{1}{n^4} + 6 \pi \sum_{n=1}^{\infty} \frac{1}{n^2} + \frac{12}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^4} \\ & = \frac{\pi^3}{4} - \frac{12 \pi^3}{90} + \pi^3 + \frac{12}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^4} \end{aligned}

And thus

$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^4} = \frac{\pi^4}{90} - \frac{\pi^4}{48} = - \frac{7 \pi^4}{700}$

up to the sign error… that’s not bad.

References

[1] Apéry, R. “Irrationalité de ζ(2) et ζ(3)”, Astérisque, 61: 11–13.

[2] Euler, L. “Exercitationes Analyticae” http://eulerarchive.maa.org//docs/originals/E432.pdf

[3] van der Poorten, A. “Apéry’s proof of the irrationality of $\zeta(3)$https://web.archive.org/web/20110706114957/http://www.maths.mq.edu.au/~alf/45.pdf

Advertisements