9
$\begingroup$

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

  • 0
    I have to say that the factorizatio$n$ 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.2010-12-29

2 Answers 2

14

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.

  • 0
    @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 $1/\Gamma(-z)$.2010-12-29
3

(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.

  • 0
    Not entire, but such singularity (called poles) is better than the other case which is called essential singularity.2010-12-29