I have just got this paper: http://people.mpim-bonn.mpg.de/zagier/files/doi/10.2307/2975232/fulltext.pdf and I have a serious doubt:
When proving that soft Tauberian theorem he explicitly uses analyticity of $g(s)$ in a small region on the left of the semiplane $Re(s)>0$.
On the other hand, the function $ \Phi (s) $ has an analytic extension to the semiplane $Re(s) \geq 1$, but for the Tauberian Theorem to be of any use here, we would want the function $\Phi(s)$ to have a holomorphic extension to a certain open set containing the semiplane $Re \geq 1$; am I not right?
Therefore, I would say that we need to prove that there exist an $\epsilon >0$ such that $\zeta(s)$ has no zeros on $Re(s) \geq 1-\epsilon$... :S
I don't believe Zagier can be wrong, so I guess either I am missing basic stuff or there is way to apply the theorem as it is.
Whatever it is, I will appreciate any help.
Thanks a lot
In Part (IV) it is proven that $\zeta(s) \neq 0$ for $\operatorname{Re}(s) = 1$. Therefore $\Phi(z)$ is holomorphic on the closed half-plane $\operatorname{Re}(z) \geq 0$. Therefore we can analytically continue $\Phi(z)$ to some open set containing $\operatorname{Re}(z) \geq 0$, but as you say, we certainly do not know that $\Phi(z)$ can be analytically continued to $\operatorname{Re}(z) \geq -\delta$, no matter how small we make $\delta$.
However, in the proof on the "Analytic Theorem," all that is asked for is that $\Phi(z)$ is holomorphic in the region $\{z \in \mathbb{C} \;|\; |z| \leq R, \operatorname{Re}(z) \geq -\delta \}$, where, critically, $\delta$ is allowed to depend on $R$. This is certainly possible.