Let me ask some basic question on the zeta function.


As you know well, the Riemann zeta function $\sum\_{n=1}^{\infty}\frac{1}{n^s}$ for $Re(s)>1$ has a meromorphic continuation to the whole complex plane and has a simple pole at $s=1$. We denote it by $\zeta(s)$. On the other hand, the zeta function has an Euler product whose local factor at $p$ is $\frac{1}{1-p^{-s}}$ that we call $\zeta\_{p}(s)$. In contrast to the global zeta function $\zeta(s)$, this local zeta function is defined in itself on the whole complex plane except where $p^s=1$. Then I am wondering whether there is a relation between the zeta function $\zeta(s)$ and the product of all these local zeta functions, especially on the half-plane $Re(s)<1$. It is obvious these two are not equal, because while the former one equals $-\frac{1}{2}$ at s=0, the latter has an infinite order pole at $s=0$. If these are not equal, then what would could be said about the local factors of $\zeta(s)$?


Could you give me some explanation on this?



On the analytic continuation of zeta function
The product $\prod\_p \zeta\_p(s)$ diverges at every point $s$ with $\Re(s)<1$, meaning that the partial products do not converge to a nonzero complex number. This is relatively straightforward to prove for $\Re(s)\leq 0$, while for $0<\Re(s)<1$ a proof can be found [here](https://mathoverflow.net/questions/63714/is-the-euler-product-formula-always-divergent-for-0res1).


For $\Re(s)\geq 1$ and $s\neq 1$ the product converges to $\zeta(s)$, see Section 3.15 in Titchmarsh: The Theory of the Riemann Zeta-function, while for $s=1$ it clearly diverges.


