TITLE: Order of Zeros of the Riemann Zeta Function
QUESTION [10 upvotes]: 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 $\rho$?

REPLY [3 votes]: 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.