Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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...

share|improve this question
5  
One of the gaps in Euler's proof was the problem of possible nonreal zeroes of the sine function. Once you have rigorously proved the product expansion of the sine, everything follows easily. –  Franz Lemmermeyer Apr 12 '12 at 17:10
    
I've long wondered about the following. Fix a disc $D$ in with large radius $R$. $D$ contains about $\pi R^2$ points, so about $\pi^2 R^4$. About $6/\pi^2$ of these pairs determine a line segment that does not pass through any other lattice point, so about $6R^4$ total. Can one prove an asymptotic result along these lines without appeal to prior determination of $\zeta(2)$? Ideally one would give an approximately bijective proof": $6R^4$ counts ordered pairs of points in 6 squares of side $R$. (Perhaps I should promote this comment to a full fledged question.) –  David Feldman Apr 13 '12 at 3:13
add comment

3 Answers

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

share|improve this answer
    
Thanks! That's great, though see the edit :) –  Igor Rivin Apr 12 '12 at 17:10
add comment

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 $$\frac{z}{e^z-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}z^n.$$

share|improve this answer
1  
In the functional equation for ζ(s), the term Γ(s) should be Γ(s/2), and the term Γ(1-s) should be Γ((1-s)/2). –  Daniel Asimov Sep 14 '12 at 2:51
    
@Daniel Asimov: You are right!! –  i707107 Sep 19 '12 at 6:31
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.