# On the analytic continuation of zeta function

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?

• If you like my answer, please accept it officially (so that it turns green). Thanks in advance! Aug 19 '18 at 23:07
• @GH from MO, Sorry. I forgot the acceptance your answer. Your answer always helped me very much! Thank you very much!:) Aug 23 '18 at 14:16

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.
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.