# Riemann zeta function at odd positive integers

Starting with the famous Basel problem, Euler evaluated the Riemann zeta function for all even positive integers and the result is a compact expression involving Bernoulli numbers. However, the evaluation of the zeta function at odd positive integers (in terms of getting a closed form sum) is still open. There has been some progress in the form of Apery's theorem and other results such as "infinitely many of $\zeta(2n+1)$ are irrational" or "at least one of $\zeta(5),\zeta(7),\zeta(9)$ or $\zeta(11)$ is irrational".

Question(s): Is there a high level understanding for this disparity between even and odd integers? Is it a case of there being a simple expression for $\zeta(3)$ that is out there waiting for an ingenious attack like Euler did with $\zeta(2)$? Or is the belief that such a closed form summation is unlikely? Where do the many many proofs powerful enough to evaluate $\zeta(2n)$ stumble when it comes to evaluating $\zeta(2n+1)$?

Motivation: The Basel problem and Euler's solution are my all-time favorites for the sheer surprise factor and ingenuity of proof (what do $\pi$ and $\frac{sin(x)}{x}$ have to do with $\zeta(2)$??). However, I currently lack the more advanced analytical tools to appreciate the deeper results of this area. I have wondered for a while about the questions above and Internet search hasn't helped much. I would greatly appreciate any answers/references. Thanks.

• If you don't know yet this list compiled by Wadim Zudilin, one of the researchers in this field, colects all the relevant papers concerning the subject wain.mi.ras.ru/zw "Several references and links devoted to the arithmetic study of values of the Riemann zeta function at positive integers and related constants are put on this page." – Américo Tavares Dec 3 '10 at 0:51

The zeta function is defined as a sum over the positive integers, but as far as actually evaluating it, it turns out to be more natural to think of it as a sum over all nonzero integers; thus we should really be thinking about $\sum_{n \neq 0} \frac{1}{n^k}$. For $k$ even this is just $2 \zeta(k)$ and there are various ways to evaluate this more symmetric sum, e.g. by writing down a meromorphic function with the right poles, or a Fourier series with the right coefficients, etc. But for $k$ odd this is equal to zero, since terms cancel with their negatives! Written in this way, the zeta function at even integers reveals its alter ego as an Eisenstein series in one dimension.

This cancellation phenomenon occurs in Euler's classic "proof," since the infinite product for $\sin z$ that he uses has zeroes at all integer multiples of $\pi$, not just the positive ones. It also occurs in the general proof that proceeds by considering the generating function $\frac{z}{e^z - 1}$ for the Bernoulli numbers. As you might know, the closed form of $\zeta(2k)$ involves Bernoulli numbers, and again $\frac{z}{e^z - 1}$ has poles at $2 \pi i n$ for all nonzero integers $n$, not just the positive ones. I describe how this works in slightly more detail here.

Another way to think about the difference between the even and odd cases is that one can think of the even cases as $L^2$ norms of appropriate Fourier series; this is precisely how a standard proof of the evaluation of $\zeta(2)$ works. But for the odd cases we don't get an $L^2$ norm; instead we get a mysterious inner product.

• Thanks for the wonderful explanation. Do you have any thoughts on how likely a closed form summation is for $\zeta(2n+1)$? – Dinesh Dec 3 '10 at 0:22
• @Dinesh: nope. The values of the zeta function at odd integers are related to some deep mathematics and I'm totally unqualified to comment on whether any of these deep approaches do or do not support the idea that there is a nice closed form. – Qiaochu Yuan Dec 3 '10 at 0:28
• Can you expand on the mysterious inner product? – zerosofthezeta Dec 3 '13 at 21:09

One expects that the numbers $\zeta(2n+1)$ are algebraically independent of one another, and of $\pi$, and so one should think of them as new'' numbers; you can't expect any closed form expression in terms of powers of $\pi$, say. Unfortunately, this conjecture seems very much out of reach at the moment.

For an explanation of why people believe this conjecture, one can see for example this answer to a related mathoverflow question.

• @ Matt E: Thanks for the explanation. Its fascinating that each of $\zeta(3),\zeta(5),\dots$ is likely to be a "new" irrational number. – Dinesh Dec 4 '10 at 0:14
• Why do you expect them to be algebraically independent of pi, when the even ones aren't? – Peter Webb Sep 12 '15 at 12:31
• Your question: " Is there a high level understanding for this disparity between even and odd integers?" is much deeper that you could think. It is at the heart of current advanced research on the "special values of L-functions of motives". I recommend the recent book of Proceedings of the Pune conference , "The Bloch-Kato conjecture for the Riemann Zeta function", ed. Coates & al., London Math. Soc. LNS 418 (2015). – nguyen quang do May 20 '16 at 6:24
• From the preface :"The values of Zeta(s) at even positive integers and odd negative integers have been known since the time of Euler. However, its values at odd positive integers remains mysterious to this day. The Bloch-Kato conjectures provide a conjectural framework to understand the special values of all motivic L-functions. The principal aim of this workshopwas to prove the B-K conjecture in the simplest "non critical" example, i.e. the values of Riemann Zeta(s) at odd positive integers." – nguyen quang do May 20 '16 at 6:27
• To make things short : 1) The "critical values" of Zeta(s) are its values at all even positive and (via the functional equation) odd negative integers are classically known (Euler, etc.) to be given by precise formulae involving Bernouilli numbers 2) The "non critical values" are the values at all odd positive and (again via the functional eq.) all even negative integers. However, the latter are null (the so called simple "trivial zeroes"), and the mystery actually lies in the first terms of the Taylor expansions . – nguyen quang do May 20 '16 at 6:39

Have you looked at this ICM paper before. Ramanujan has found out some wierd formula. The paper is not viable to read. But I guess, this will give some idea about the progress made regarding this problem.

Added: Try emailing Prof. Bruce Berndt. Since this is related to Ramanujan, I am sure he might be knowing something along these lines. Another good source of information which I am sure you would like to read are: -

Where do the many many proofs powerful enough to evaluate ζ(2n) stumble when it comes to evaluating ζ(2n+1)?

Another method of evaluating $\zeta(2n) \$ not mentioned yet is by applying Parseval's Theorem to the Fourier series of the Bernoulli Polynomials. The Fourier coefficients of $B_n\$ have the form $c_k = (constants)\cdot \frac{1}{k}$. By Parseval's Theorem, $$\int_0^1 |B_n(t)|^2 dt = \sum_{k\neq0} |c_k|^2$$ $$(constant) = (more\ constants) \sum_{k\neq0} \frac{1}{k^{2n}}$$ (Note that the constants depend upon n)
The reason this method falls apart for odd integers is that, since Parseval's Theorem involves squaring the coefficients, so there is no way to get odd powers into the sum.

• This is equivalent to the proof using the generating function for the Bernoulli numbers, and I mentioned that one could use Parseval's theorem in my answer; the Fourier series you mention comes from the poles of that function. – Qiaochu Yuan Jan 4 '11 at 13:56
• Sorry about that. I only noticed the reference to the generating function and not the reference to the Fourier series. – Riley E Jan 4 '11 at 17:39

Riemann Zeta at Even Integers

One reason why $$\zeta(2k)$$ has a nice closed form is because $$\require{enclose} \newcommand{\Res}{\operatorname*{Res}} \lim_{R\to\infty}\int_{|z|=R}\frac{\pi\cot(\pi z)}{z^{2k}}\,\mathrm{d}z=0\tag1$$ and the residue of $$\pi\cot(\pi z)$$ is $$1$$ at each integer. Since the integral in $$(1)$$ is $$2\pi i$$ times the sum of the residues inside $$|z|=R$$, we get $$\zeta(2k)=-\tfrac12\Res_{z=0}\frac{\pi\cot(\pi z)}{z^{2k}}\tag2$$ As shown in this answer, we can find a recurrence for the coefficients of $$z\cot(z)$$ which leads to a recurrence for $$\zeta(2k)$$.

This approach fails if we use $$2k+1$$ in place of $$2k$$. Summing over the reciprocals of all non-zero integers with odd exponent yields a sum of $$0$$. This agrees with the fact that the residue of an even function at $$z=0$$ is $$0$$.

An Alternating Approach

The function $$\pi\sec(\pi z)$$ has a residue $$(-1)^n$$ at $$z=n-\frac12$$ for $$n\in\mathbb{Z}$$. This gives opposite residues at $$z=n-\frac12$$ and $$z=-n+\frac12$$. This is good for summing odd functions. \begin{align} 0 &=\frac1{2\pi i}\lim_{R\to\infty}\int_{|z|=R}\frac{\pi\sec(\pi z)}{z^{2k+1}}\,\mathrm{d}z\\ &=\Res_{z=0}\frac{\pi\sec(\pi z)}{z^{2k+1}}+2\sum_{n=1}^\infty\frac{(-1)^n}{\left(\small{n-\frac12}\right)^{2k+1}}\\ &=\Res_{z=0}\frac{\pi\sec(\pi z)}{z^{2k+1}}+4^{k+1}\sum_{n=1}^\infty\frac{(-1)^n}{(2n-1)^{2k+1}}\\ &=\Res_{z=0}\frac{\pi\sec(\pi z)}{z^{2k+1}}-4^{k+1}\beta(2k+1) \end{align} where $$\beta(s)$$ is the Dirichlet beta Function. Some values and a recursion for all even arguments can be found in this answer.

A Unilateral Attack on Zeta

The function $$\pi\tan(\pi z)$$ has a residue $$-1$$ at $$z=n-\frac12$$ for $$n\in\mathbb{Z}$$. This gives identical residues at $$z=n-\frac12$$ and $$z=-n+\frac12$$. This is good for summing even functions. \begin{align} 0 &=\frac1{2\pi i}\lim_{R\to\infty}\int_{|z|=R}\frac{\pi\tan(\pi z)}{z^{2k}}\,\mathrm{d}z\\ &=\Res_{z=0}\frac{\pi\tan(\pi z)}{z^{2k}}+2\sum_{n=1}^\infty\frac{-1}{\left(\small{n-\frac12}\right)^{2k}}\\ &=\Res_{z=0}\frac{\pi\tan(\pi z)}{z^{2k}}-2^{2k+1}\sum_{n=1}^\infty\frac1{(2n-1)^{2k}}\\ &=\Res_{z=0}\frac{\pi\tan(\pi z)}{z^{2k}}-\left(2^{2k+1}-2\right)\zeta(2k) \end{align} where $$\zeta(s)$$ is the Riemann zeta Function.

A Unified Result

The following is a rational multiple of $$\pi^k$$ $$\sum_{n=0}^\infty\frac{(-1)^{nk}}{(2n+1)^k}$$ This evaluates to $$\beta(k)$$ for odd $$k$$ and $$\left(1-2^{-k}\right)\zeta(k)$$ for even $$k$$.

## protected by Zev ChonolesMay 22 '16 at 3:07

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