Computing zeta(k), for k odd, using Fourier coefficients

Question

I'm not really sure what topics exactly this falls under, so I apologize if I've misclassified this question.

There is a neat way of computing $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ using Fourier analysis: Compute the Fourier series of $t^2$ extended $2\pi$-periodically, which turns out to be

$$\frac{\pi^{2}}{3}+4\sum_{n=1}^{\infty}\frac{1}{n^{2}}$$

By Fejer's theorem (I think), the Fourier series around $\pi$ converges, so we get an equation that can be solved for the $\zeta(2)$.

I think a similar approach can be taken for $\zeta(2k)$ by taking $t^{2k}$ extended $2\pi$-periodically, but all my attempts to do something like this for odd integers fail.

On the other hand, since $1/n^k$ for $k$ odd is in $\ell^2$, there should be an $L^2$ function that has the sequence as its Fourier coefficients. Can one be explicitly constructed? What if we allow the entries in the sequence to alternate, or let finitely many of them deviate from $1/n^k$?

Basically, I want to find an $L^2$, $2\pi$-periodic function whose Fourier coefficients would give a relatively straightforward computation of $\zeta(k)$ when $k$ is odd.

Answer 1

Well, I wish you the best. But I don't think you are going to succeed for two reasons:

(1) Proving almost anything about $\zeta(2k+1)$ is hard.

(2) There are nice formulas for $\sum_{n=1}^{\infty} \cos (n \theta)/n^{2k}$ and for $\sum_{n=1}^{\infty} \sin (n \theta)/n^{2k+1}$. There are not particularly nice formulas for $\sin$ with even powers or $\cos$ with odd. One way to think about this is that the imaginary part of $\log (1-e^{i \theta})$ has a simple formula but the real part does not.

Answer 2

Qiaochu, Parseval does apply, it just doesn't yield anything elementary. :) Still, it is easy to deduce (for example) the identity

$$\int_{0}^{2\pi}x(2\pi-x)\log(1-\cos{x})dx=8\pi \zeta(3)-\frac{4}{3}\pi^3\log{2}$$ but its not clear what good such an identity is, relative to the original problem.

Answer 3

The difficulty becomes apparent if you play around a little bit: the function whose $n$th Fourier coefficient is $n^{-3}$ is a twice-iterated indefinite integral of

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

which will be some highly mysterious transcendental function (a "trilogarithm") whose values are just as obscure as those of $\zeta(2k+1)$.

Answer 4

What you easily get for any integer $k \ge 2$ is $\sum 1/n^k$ summed over all nonzero integers $n$, both positive and negative. When $k$ is even, we can use this to evaluate $\zeta(k)$. When $k$ is odd, we get the (correct, but uninteresting) result $0$.

Answer 5

A point that hasn't been made yet is that one reason the sums for k even are easy to evaluate is that they occur, not only as values of periodic functions, but as $L^2$ norms of those functions. One can then apply Parseval's identity and evaluate the sum of the squares of 1/n^k as a definite integral, as is done in the Wikipedia article. Parseval's identity does not, on the other hand, apply to the odd values.

Answer 6

This is very open, from what I know. I think there was a recent (last 15 years) result where it was proved that zeta(3) is transcendental (or maybe just irrational?) but my understanding is that virtually nothing is known about odd zeta values, and someone needs to do something new in order to get any useful information about them.