Riemann zeta function at odd positive integers

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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: -

• Bernoulli Numbers and The Riemann Zeta function by B.Sury.

• A note on Value of the Riemann Zeta function at Odd Positive Integers by Andrzej Dabrowski.

Answer 4

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.

Answer 5

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 $0$.

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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.

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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.

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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 $k$.

Answer 6

$$\DeclareMathOperator{sinc}{sinc}$$ To piggy-back off of the discussion from @Matt E above, I would like to add that there is probably good reason to believe that all odd integer values for $\zeta(s)$ are algebraically independent of each other, but not necessarily of $\pi$. Euler evaluated the $\sinc(x)$ function to obtain a closed form value for $\zeta(2)$. You can also see this function in the product formula:

$$f(x) = \prod_{k=1}^\infty 1 -\frac{x^2}{k^2 \pi^2} = \frac{\sin(x)}{x}$$

If we look at a similar function for $\zeta(4)$, we obtain:

$$f(x) = \prod_{k=1}^\infty 1 -\frac{x^4}{k^4 \pi^4} = \frac{\sin(x) \sinh(x)}{x^2}$$

There are other transcendental functions associated with the higher, even integer values for $\zeta(s)$. Given these transcendental functions, i.e. the $\sinc(x)$ function and the function involving the hyperbolic $\sin$ associated with $\zeta(4)$, you can simply take the absolute value of the second term in each of their Taylor series expansions and set $x = \pi$. You then obtain the closed form values for $\zeta(2)$ and $\zeta(4)$. You can similarly perform this for the higher, even values for $\zeta(s)$.

The question: Is there a transcendental function that we can similarly use to obtain the value for $\zeta(3)$? We do know this, using the gamma function:

$$f(x) = \prod_{k=1}^\infty 1 -\frac{x^3}{k^3 \pi^3} = \frac{1}{\Gamma(1 - \frac{x}{\pi}) \Gamma(1 + \frac{(-1)^{(1/3)} x}{\pi}) \Gamma(1 - \frac{(-1)^{(2/3)} x}{\pi})}$$

If you look at the second term of the series expansion for this inverted "multi-gamma" function above, we obtain:

$$\frac{x^3 ψ^2(1)}{2 \pi^3}$$ where $ψ^n(x)$ is $n^{\text{th}}$ derivative of the digamma function.

When you set $x = \pi$, you can see that it cancels out with the $\pi^3$ in the denominator and you're left with the digamma function value divided by 2, which is $\zeta(3)$. So it seems that $\pi^3$ is kind of related to $\zeta(3)$, but not in a way that's tangible. My opinion is that there is most likely another transcendental quantity involved along with $\pi$, one that may need to be discovered, or one that is in lieu of $\pi$. That is, $\pi$ might be a stepping stone, so to speak, to reach another transcendental which is then involved with a closed form value for $\zeta(3)$.

Answer 7

Relation of the Eisenstein series, zeta functions at even and odd numbers can see from 11.17 page 117 of "Automorphic forms on $Sl_2(R)$" of Borel gives a formula: $$c(s) = {\pi ^{1/2}}\frac{{\zeta (s)}}{{\zeta (s + 1)}}\frac{{\Gamma (s/2)}}{{\Gamma ((1 + s)/2)}}$$ where $c(s)$ comes from constant term of Eisenstein series $E(s,z)_P$: $$E{(s,z)_P} = {y^{(1 + s)/2}} + c(s){y^{(1 - s)/2}}$$ with $z=x+iy$.

Answer 8

For where $m∈ℤ^+$,

$$ζ(2m+1)=(-1)^{m}2^{2m+1}π^{2m}\left(∫_1^{∞}\frac{z^{-2m-1}}{e^{z}-1}dz+∑_{k=0}^{∞}\frac{B_{k}}{k!}\frac{1}{k-2m-1}\right)$$

where $B_{k}$ is the $k$-th Bernoulli number.
It is not "closed form" but I found this and I wonder what I found. I hope someone tell me if it has a meaning instead of saying rubbish.