In the introduction to Akshay Venkatesh's thesis "Limiting Forms of the Trace Formula" we have the following statement : 



> 
> "For, in summing over primes, the limit
>  $\lim\_{X\to\infty}\frac{1}{X}\sum\_{p<X}\log(p)\lambda(p,\pi,\rho)$ is
>  a relatively harmless constant: the multiplicty $m(\pi,\rho)$. In summing over integers, we will be considering a
>  sum of the form
>  $\lim\_{X\to\infty}\frac{1}{X}\sum\_{n<X}\lambda(p,\pi,\rho)$, which
>  essentially evaluates the residue of $L(s,\pi,\rho)$ at $s=1$."
> 
> 
> 


In the above $\lambda(n,\pi,\rho)$ is the $n$-th coefficient in the Dirichlet series $L(s,\pi,r)$, and $m(\pi,\rho)$ is the multiplicity of the pole at $s=1$.


From a note of Arthur this should follow from the Wiener-Ikehara theorem, but as I understand it the setting of said theorem is a series (the logarithmic derivative of $L(s,\pi,\rho)$) with a *simple* pole at $s=1$, and the sum over the integers gives the residue.


So how should this give the *multiplicity,* or the order of the pole instead, by summing over primes? 



Wiener-Ikehara tauberian theorem and order of pole at s=1
Here's a more detailed answer - fleshing out Daniel Loughran's comment. For
$\lim\_{X \to \infty} \frac{1}{X} \sum\_{n < X} \lambda(p,\pi,\rho)$, one applies the Wiener-Ikehara theorem to $L(s,\pi,\rho)$, but for $\lim\_{X \to \infty} \frac{1}{X} \sum\_{p < X} \log(p) \lambda(p,\pi,\rho)$, one applies the Wiener-Ikehara theorem to $-\frac{L'(s,\pi,\rho)}{L(s,\pi,\rho)}$.


If $L(s,\pi,\rho) = (s-1)^{-m} g(s)$, where $g(s)$ is holomorphic and non-vanishing at $s = 1$, then a simple calculation (one you should do if you haven't seen it before!) gives that the residue of $-\frac{L'(s,\pi,\rho)}{L(s,\pi,\rho)}$ as $s = 1$ is equal to $m$. Now the Dirichlet series 
$$-\frac{L'(s,\pi,\rho)}{L(s,\pi,\rho)} = \sum\_{n} \frac{\Lambda\_{\pi,\rho}(n)}{n^{s}}
$$ 
(which comes from taking the logarithmic derivative of the Euler product) satisfies
$\Lambda\_{\pi,\rho}(p) = \log(p) \lambda(p,\pi,\rho)$ if $p$ is prime. If $\Lambda\_{\pi,\rho}(n)$ is nonzero, then $n$ will be a prime power, but the contribution to
$\frac{1}{X} \sum\_{n \leq X} \Lambda\_{\pi,\rho}(n)$ from all $n = p^{k}$ with $k > 1$ will be negligible.


