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From Wikipedia's Weierstrass Factorization Theorem, I learned that every entire function can be represented as a product involving its zeroes. Examples are the sine and cosine function. The Riemann zeta function, however, is not entire.

Let us assume the Riemann Hypothesis. Can $\zeta(s)$ be represented by an infinite product involving both its trivial zeroes at $\zeta(s)=-2n$ (for $n \in \mathbb{N}$) and its non-trivial zeroes at $\zeta(s)=\frac{1}{2} + i t$?

Thanks, Max

Max Muller
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  • To give a bit of historical context, the Weierstrass factorization is not enough to pin down the factors involved; one has to use Hadamard's factorization instead which relies on information about the growth rate of the function. Actually it is said that Hadamard developed his theory in order to get at the prime number theorem. First, the Riemann xi function was factorized by von Mangoldt by using Hadamard's theory. A few years later Hadamard and de la Vallee-Pousin proved the prime number theorem continuing the same line of development. – timur Dec 29 '10 at 16:45
  • @timur: thank you for the historical background. I like it when mathematics is put into (historical) perspective. – Max Muller Dec 29 '10 at 17:00
  • I have to say that the factorization formula was already stated by Riemann in his famous paper, without rigorous proof. He gave some reasons as to why the formula is valid, and this reasoning was shown to be valid provided that Hadamard's theory is known. The point I wanted to make in my earlier comment but forgot was that the factorization formulas have been from the beginning, and still are one of the most powerful tools to study Riemann's zeta function. – timur Dec 29 '10 at 17:31

2 Answers2

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The standard approach to this problem is to write $\xi(s) = \dfrac{s(s-1)}{2}\pi^{-s/2}\Gamma(s/2)\zeta(s).$ This function is entire, and has zeroes precisely at the non-trivial zeroes of $\zeta(s)$. It also has a slow enough rate of growth that it can be written as a product over its zeroes: $$\xi(s) = \xi(0) \prod_{\rho}(1-\dfrac{s}{\rho}),$$ where the product is over zeroes of $\xi(s)$, i.e. over non-trivial zeroes of $\zeta(s)$. (Here I am following more-or-less the notation in Edward's book Riemann's zeta function, which is a good reference for these sort of things.) [Edit: Also, to ensure convergence, the product should be taken over "matching pairs" of zeroes, i.e. the factors for a pair of zeroes of the form $\rho$ and $1-\rho$ should be combined.] We can then write $$\zeta(s) = \dfrac{2\xi(0)}{s(s-1)}\pi^{s/2}\dfrac{1}{\Gamma(s/2)}\prod_{\rho}(1-\dfrac{s}{\rho}).$$ If we now replace $\dfrac{1}{\Gamma(s/2)}$ by its Weierstrass product, we get a product formula for $\zeta(s)$, namely $$\zeta(s) = \dfrac{\xi(0)}{s-1} (\pi e^{\gamma})^{s/2}\prod_{n=1}^{\infty}(1 +\dfrac{s}{2n})e^{-s/2n} \prod_{\rho}(1-\dfrac{s}{\rho}).$$ (Here $\gamma$ is Euler's constant.)

Note that we can now compute $\xi(0)$, because we know that the value of $\zeta(s)$ at $s = 0$ is equal to $-1/2$. We find that $\xi(0) = 1/2$, and so $$\zeta(s) = \dfrac{1}{2(s-1)}(\pi e^{\gamma})^{s/2}\prod_{n=1}^{\infty}(1+\dfrac{s}{2n})e^{-s/2n}\prod_{\rho}(1-\dfrac{s}{\rho}).$$

An added cultural remark: Riemann's explicit formula for the prime counting function is obtained by taking a Fourier transform of the logarithm of this formula. Combined with the fact that all the non-trivial zeroes $\rho$ have real part $< 1$ (proved by Hadamard and de la Vallee Poussin), this gives the prime number theorem.

Matt E
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  • Here's the MathWorld entry for $\xi(s)$. – J. M. ain't a mathematician Dec 29 '10 at 15:38
  • @ J.M. and especially Matt: thanks a lot! Is p for prime here or does it stand for the non-trivial zeroes of $\zeta(s)$? – Max Muller Dec 29 '10 at 15:46
  • Never mind I see it's the zeroes. Can we also do something about the $e^{-s/2n}$ part, as in incorporating it somehow in the part between the brackets (of the infinite product of (1+s/2n))? Something similar was done with the sine function (see the link I mentioned, you'll see it at the bottom). – Max Muller Dec 29 '10 at 15:56
  • To nitpick, the factors in the factorization of xi must be paired, as in $\xi(s)=\prod_{\Im\rho>0}(1-\frac{s}{\rho})(1-\frac{s}{1-\rho})$. – timur Dec 29 '10 at 16:37
  • @timur: Dear Timur, Thanks for pointing this out. I'll add this as a remark into the answer. Regards, – Matt E Dec 29 '10 at 17:08
  • @Max: Dear Max, These exponential factors are needed to ensure convergence of the product. In the case of $\sin(x)$, we can pair the factors corresponding to matching positive and negative zeroes so as to cancel out the exponential factors, but that trick isn't available here, and so I don't think they can be removed. Regards, – Matt E Dec 29 '10 at 17:11
  • @Max: They come from the Gamma function so they are harmless and I think no reason to try to remove them. The reason you cannot treat them as in sine is that $1/\Gamma$ is an entire function having essentially (negative) half of the zeroes of sine, and the other (positive) half is needed to pair them up and cancel the exponential factor. This is also the reason why you get $\sin(z)$ when you multiply $1/\Gamma(z)$ with something like $1/\Gamma(-z)$. – timur Dec 29 '10 at 17:40
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(Would be a comment, but I don't have the reputation.)

The Riemann zeta function is certainly not entire: it has a simple pole at s = 1. It is, however, meromorphic.

Zhen Lin
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  • Ok so when a function has a point at which the value of the function is 'infinity' the function is not entire, no? Thanks, Zhen. – Max Muller Dec 29 '10 at 14:45
  • Not entire, but such singularity (called poles) is better than the other case which is called essential singularity. – timur Dec 29 '10 at 16:38