Infinite product representation of a function in terms of its non-trivial zeroes? 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
 A: (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.
A: 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.
