Riemann zeta at even integers - MathOverflow

Riemann zeta at even integers

2012-04-12T16:14:26Z

I am talking about this in a course I am teaching, and hence am wondering: what are the various derivations of the values of Riemann zeta function at even integers? There are two incredibly cool proofs in Don Zagier's paper (section 1), but there must several other proofs floating around. Also, I recall reading that Euler originally proved the formula for $\zeta(2)$ by thinking of $\sin(x)$ as a polynomial -- has this argument been made rigorous since?

EDIT I did not realize that this was known as the "Basel Problem", so did not find @Yemon's answer myself. I conjecture, however, that the Robin Chapman list is incomplete, since I have found yet another proof, not contained in Robin's list, so maybe there are more yet out there...

Answer by Yemon Choi for Riemann zeta at even integers

2012-04-12T16:52:32Z

The Wikipedia page http://en.wikipedia.org/wiki/Basel_problem has a link to several proofs for zeta(2), compiled by Robin Chapman.

Answer by i707107 for Riemann zeta at even integers

2012-04-12T17:37:12Z

These are the proofs that I have seen:

The proof using Fourier series: Reference: Stein Shakarchi "Fourier Analysis, the introduction" p 97 Exercise 4 The key ingredient of this proof is the following identity $$\sum_{n=1}^{\infty} \frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan{\alpha\pi}}$$ which can be proved by using Fourier series of $\cos(\alpha x)$. This allows an expression of $\zeta(2n)$ by Bernoulli numbers.

The proof using the functional equation of $\zeta(s)$: Reference: E.Titchmarsh "The Theory of the Riemann zeta function" p18 (2.4 Second method)

Chapter 2 of this book is entirely devoted to the proofs of the functional equation $$\pi^{-s/2}\Gamma(s/2)\zeta(s)=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s).$$ Section 2.4 is one of the proofs, it uses te residue theorem of complex analysis, and derives the formula $$\zeta(1-2m)=\frac{(-1)^mB_{2m}}{2m}$$ for $m=1,2,\cdots$ The formula for $\zeta(2n)$ is now followd by the functional equation.

The result is: $$2\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{(2n)!}B_{2n}$$ where $$\frac{z}{e^z-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}z^n.$$

Answer by Américo Tavares for Riemann zeta at even integers

2012-04-28T12:58:03Z

If we evaluate 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 $2\pi,$ we get

$$\begin{equation*}x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos nx\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x.\tag{1}\end{equation*}$$

So, for $f(\pi )=\pi ^{2p}$ we have

$$\begin{equation*}\pi ^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x,\tag{2}\end{equation*}$$

where the integral

$$\begin{equation*}I_{2p}:=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x\tag{3}\end{equation*}$$

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

$$\begin{equation*}I_{2p}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\frac{2p(2p-1)}{n^{2}}I_{2\left( p-1\right) },\qquad I_{0}=0.\tag{4}\end{equation*}$$

For $p=1$, we obtain $$\begin{equation*}I_{2}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi.\end{equation*}\tag{5}$$ and

$$\pi ^{2}=\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\left(\frac{2}{n^{2}}\pi \cos n\pi \right)=\frac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\frac{1}{n^{2}}.\tag{6}$$ Consequently, $$\zeta (2)=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6}\tag{7}$$

By using $(1)$ to $(4)$ we can generate recursively the sequence $(\zeta(2p))_{p\ge 1}$. For instance, I evaluated $\zeta(4)$ and $\zeta(6)$ in this math.SE answer.