2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$ – GH from MO Aug 19 '18 at 23:07
  • 1
    $\begingroup$ @GH from MO, Sorry. I forgot the acceptance your answer. Your answer always helped me very much! Thank you very much!:) $\endgroup$ – Monty Aug 23 '18 at 14:16
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.