# Convergence of Euler Product on the line Re(s) = 1

Consider the Euler product for the Riemann Zeta function: $$\zeta(s) =? \prod_p (1-p^{-s})^{-1}$$ When I started studying this product, I read that for $Re(s) > 1$, the Euler product converges, and that for $Re(s) < 1$, the product diverges. Those make sense. However, I wondered how the Euler product behaves on $Re(s) = 1$. This is probably ignorance on my part, but so far, I have not found any complete discussion of this on either math SE, or the Stein Shakarchi complex analysis book. So I want to take a poke at this question and seek help.

What I am thinking is that I can evoke a theorem from complex analysis. Generally, supposing that $\sum_n |a_n^2|$ converges absolutes, then $\prod_n (1 + a_n)$ converges iff $\sum_n a_n$ converges. Now define $a_n = 0$ when $n$ is not a prime, and $a_n = - p^{-s}$ otherwise. Along the line $Re(s) = 1$, $\sum_n |a_n|^2$ converges because the summand decays as $n^{-2}$ which converges by comparison test. Thus the convergence of the Euler product hinges upon the convergence of $\sum_p p^{-s}$.

Now we focus on $s = 1 + ix$. If $x = 0$, then this sum diverges as $\log \log x$. This is the pole at $\zeta(1)$. However, I want to say that for $x \in \mathbb{R}$ and $x = 0$, this in fact converges. Here is a sketch of my reasoning.

Consider the sum $\sum_p p^{-1} e^{-ix \log p}$ as a finite sum up to a large prime $p^*$ and an infinite tail. If $p^*$ is sufficiently large, then the sum can be well approximated by an integral after $p^*$ and we just need to check the convergence of the integral. Using prime number theorem, we can then estimate the sum as: $$\sum_{p=p^*}^{N} p^{-1} e^{-ix\log p} \sim \int_{p^*}^N \frac{dp}{\log p} p^{-1} e^{-ix\log p}$$ Now I make a change of variables to $u = \log p$: $$du = \frac{1}{p} dp \quad \rightarrow \quad \frac{dp}{\log p} = \frac{e^u}{u} du$$

We can simplify the integral to: $$\sum_{p=p^*}^{N} p^{-1} e^{-ix\log p} \sim \int_{\log p^*}^{\log N} du \frac{e^u}{u} e^{-u} e^{-ixu} = \int_{\log p^*}^{\log N} du \frac{e^{-ixu}}{u}$$ If we now take the limit as $N \rightarrow \infty$, this final integral indeed converges for all $x \neq 0$ to the incomplete gamma function $\Gamma(0,i x \log p^* )$. This leads me to think that the Euler product for zeta function converges everywhere along Re(s) = 1 except at s = 1.

Is my conclusion correct? If so, how can I make this sketch rigorous? If not, what are the critical flaws in the argument?