# Thread: Analytic continuation of zeta function

1. ## Analytic continuation of zeta function

I realise that there are various ways to prove the analytic continuation of Riemann's zeta function to the complex plane, but could anybody explain it in plain english?

I'm trying to get my head around various proofs, but am failing miserably. Are there any straight-forward proofs, or proofs that can be explained intuitively?

2. ## Re: Analytic continuation of zeta function

Let'sw start from the 'standard' definition of the Riemann zeta function...

$\displaystyle \zeta(s)= \sum_{n=1}^{\infty} \frac{1}{n^{s}}$ (1)

It is well known that (1) converges if $\displaystyle \text{Re} (s)>1$. In particular $\displaystyle \zeta(s)$ has a singularity in $\displaystyle s=1$ and we suppose preliminarly that in all remaining complex plane it is analytic. In that case we can express the function 'somewhere around' a point $\displaystyle s_{0} \ne 1$ in Taylor series...

$\displaystyle \zeta (s) = \sum_{k=0}^{\infty} a_{k}\ (s-s_{0})^{k}$ (2)

... where...

$\displaystyle a_{k}= \frac{1}{k!}\ \frac{d^{k}}{d s^{k}} \zeta(s)_{s=s_{0}}$ (3)

... and the derivatives in (3) are given by ...

$\displaystyle \frac{d^{k}}{d s^{k}} \zeta(s)= (-1)^{k}\ \sum_{n=1}^{\infty} \frac{\ln^{k} n}{n^{s}}$ (4)

Now we start the analysis setting [why not?...] $\displaystyle s_{0}=2$. Because we have supposed that the only singularity is in $\displaystyle s=1$, the Taylor series around $\displaystyle s_{0}=2$ computed with (2), (3) and (4) converges in a circle centered in $\displaystyle s_{0}=2$ with radious $\displaystyle r=1$, the 'red circle' in the figure...

The Taylor series converges in any interior point of the 'red circle' so that nobody forbids to expand $\displaystyle \zeta(s)$ in Taylor series in any point internal to it, using (2) to compute the derivatives of the function in the new 'central point' that call $\displaystyle s_{1}$. Let suppose to determine $\displaystyle s_{1}$ by a '$\displaystyle \frac{\pi}{4}$ counterclockwise rotation' of $\displaystyle s_{0}$ taking $\displaystyle s=1$ as 'pivot' [see figure...], we obtain a new Taylor expansion in a circle centered in $\displaystyle s=s_{1}$ with radious $\displaystyle r=1$, i.e. the 'black circle' of figure. In that way we have in some way 'extended' the region in which $\displaystyle \zeta(s)$ is analytic. Proceeding we can hop to a new 'central pivot point' $\displaystyle s_{2}$ with a '$\displaystyle \frac{\pi}{4}$ counterclockwise rotation' of $\displaystyle s_{1}$' and the same for $\displaystyle s_{3}$ and $\displaystyle s_{4}$ [the 'blue circle' in the figure...]. Of course is $\displaystyle s_{4}=0$ and that means that we have 'extended' the domain of $\displaystyle \zeta(s)$ on the real axis till to $\displaystyle s=-1$. At this point with four more '$\displaystyle \frac{\pi}{4}$ counterclockwise rotation' we return to $\displaystyle s_{0}$. At the end of the work we have considerably extended the domain of $\displaystyle \zeta(s)$, in particular in part of the region where is $\displaystyle \text{Re}(s)<1$. A spontaneous question: how to obtain a larger domain?... an obvious possibility is tho choose a larger value of $\displaystyle s_{0}$ like 3,4, ... or, why not, $\displaystyle e^{10000}$... 'theoretically' we can extend the domain of $\displaystyle \zeta(s)$ to the whole complex plane, with the only exception of $\displaystyle s=1$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: Analytic continuation of zeta function

Thanks for this.

I'm going to see if I can get my head around some other proofs today.