Consider the Euler product for the Riemann Zeta function: \begin{equation} \zeta(s) =? \prod_p (1-p^{-s})^{-1} \end{equation} 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: \begin{equation} \sum_{p=p^*}^{N} p^{-1} e^{-ix\log p} \sim \int_{p^*}^N \frac{dp}{\log p} p^{-1} e^{-ix\log p} \end{equation} Now I make a change of variables to $ u = \log p$: \begin{equation} du = \frac{1}{p} dp \quad \rightarrow \quad \frac{dp}{\log p} = \frac{e^u}{u} du \end{equation}
We can simplify the integral to: \begin{equation} \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} \end{equation} 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?
You would probably have to estimate the sum differently... The Idea is correct, but you would probably do partial integration to switch to the summatory function, and then use the prime number theorem to approximate $\pi(x)$, then maybe if you like switch back using partial integration, or just solve the integral.
So we have for p^* large enough by partial integration:
$\sum \limits_{p>p^*~prime} p^{-s}=\int \limits_\mathbb{R} \delta_{p>p^*} p^{-s}dp=\int \limits_\mathbb{R}\pi(p)sp^{-s-1}=s\int \limits_\mathbb{R}\pi(p)p^{-s-1}$
Due to the prime number theorem we furthermore have:
$\pi(x)=Li(x)+R(x)$ with $R(x)$ some error term not fully understood yet.
If the bounds for the error term ensure convergence of the "error term part" of the integral (under the Riemann Hypothesis this would certainly be the case), the (not absolute) convergence of the integral depends only on the (not absolute) convergence of:
$\int \limits_{p>p^*} Li(p)p^{-s-1}dp=c(s)-\frac{1}{s}\int \limits_{p>p^*} \frac{1}{ln(p)}p^{-s} dp$
Using Wolfram Alpha, one quickly realizes that this actually converges iff the exponential Integral converges for purely imaginary arguments going to "$i\infty$", which it does, as can be simply checked by using the definition and partial integration.
So at least under the Riemann Hypothesis, you are correct, without the Riemann Hypothesis tough im afraid you may not be able to prove that the $R(x)$ integral converges.
Very interesting question and very correct tought tough.
Also, it appears that it is actually known that it converges on that line, see the paper "THE EULER PRODUCT FOR THE RIEMANN ZETA-FUNCTION IN THE CRITICAL STRIP".