Order of Zeros of the Riemann Zeta Function Is it true that all zeros of the Riemann Zeta Function are of order 1?
Let $h(z) = \frac{\zeta'(z)}{\zeta(z)}\frac{x^z}{z}$, where $x$ is a positive real number ($x > 1$, probably?) , and $\zeta$ is the Riemann zeta function. I'm computing the residues of $h$.
The only place I'm having problems computing the residue of $h$ are at the zeros of $\zeta$. I just need to know the order of the zeros of $\zeta$. It seems like for the trivial zeros (the negative even integers) the orders are 1.
But I don't know what to do about the non-trivial zeros. I'm probably confused about something. When I look in the derivation of the von Mangoldt formula shown here, it lists the residues of $h$, and from their computation, it seems that the orders of the zero at the non-trivial zeros are also 1.
I was under the impression that the order of the the zeros of $\zeta$ were not all known.
What am I missing?
EDIT: I found this. On page 2, it says that if $\rho$ is a zero of $\zeta$ contributes $\frac{x^\rho}{\rho}$ to the residue of $h$, counted with multiplicity. (I'm off by a negative sign because of the way I defined $h$).  Is there a special way to compute the residue without knowing the order of $\rho$?
 A: You might be interested in this recent arXiv paper: On simple zeros of the Riemann zeta-function

We show that at least 19/27 of the zeros of the Riemann zeta-function are simple, assuming the Riemann Hypothesis (RH). This was previously established by Conrey, Ghosh and Gonek [Proc. London Math. Soc. 76 (1998), 497--522] under the additional assumption of the Generalised Lindelöf Hypothesis (GLH). We are able to remove this hypothesis by careful use of the generalised Vaughan identity. 

A: Your Google Books link doesn't work for me; here's one that does.
I think there's nothing mysterious going on here, and no assumption about the multiplicities of the zeros of $\zeta$ is being made. The sum is written as a sum over the zeros, but it's implied that these contribute according to their multiplicity. The residue isn't computed in any special way (unless you count forming the residue of the logarithmic derivative as a special way); the residue from $\zeta'/\zeta$ is the multiplicity, and this gets multiplied by $x^\rho/\rho$.
