# Thread: The domain of the Riemann zeta function

1. ## The domain of the Riemann zeta function

Hi,

I'm currently looking at the Riemann zeta function (r.z.f), particularly it's domain. I know we can show it is holomorphic in Re(z) > 1 and that we can see it has a singularity at 1 by rewriting its formula however, after this I get a bit confused.

Before the analytic continuation of the r.z.f, is it well defined for all the complex plane apart from 1?

Do we consider the analytic continuation of the r.z.f to make it well defined for all the complex plane or to show that it is meromorphic in Re(z) > 0 or both or none of these?

Thanks

2. Originally Posted by TheFinalPush
Hi,

I'm currently looking at the Riemann zeta function (r.z.f), particularly it's domain. I know we can show it is holomorphic in Re(z) > 1 and that we can see it has a singularity at 1 by rewriting its formula however, after this I get a bit confused.

Before the analytic continuation of the r.z.f, is it well defined for all the complex plane apart from 1?

Do we consider the analytic continuation of the r.z.f to make it well defined for all the complex plane or to show that it is meromorphic in Re(z) > 0 or both or none of these?

Thanks

Before its analytic continuation The RZF only exists for Re(z) > 1. Only after the anal. cont. is done we can look at this function as analytic everywhere except as s = 1.

It's like $f(z):=\sum^\infty_{n=0}z^n$ . The function $f(z)$ is defined, and analytic, only for $|z|<1$ , but we can anal. continue it to the whole $\mathbb{C}-\{1\}$ by definining $f(z):=\frac{1}{1-z}$ ...

Tonio

3. A suggestive [though not very pratical...] approach to the analytic extension of the function $\zeta(*)$ is illustrated in the figure...

$\zeta (s) =\sum_{n=1}^{\infty} \frac{1}{n^{s}}$ , $\Re (s) >1$ (1)

... and suppose to expand $\zeta (*)$ in Taylor series around $s_{0} = 2$. With not difficult steps from (1) we derive...

$\frac{d^{k}}{d s^{k}} \zeta (s) _{s=2} = (-1)^{k} \sum_{n=1}^{\infty} \frac{\ln ^{k} n}{n^{2}}$ (2)

... so that we have...

$\zeta(s) = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} (s-2)^{k} \sum_{n=1}^{\infty} \frac{\ln^{k} n}{n^{2}}$ (3)

The series (3) converges inside a circle with center in $s=2$ and radius $1$ [the 'red circle' in the figure...]. Inside the 'red circle' $\zeta(*)$ is analytic so that we can use (3) to obtain the Taylor expansion of $\zeta(*)$ around any internal point... for example $s_{1} = 1+ \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$ , which is $s_{0}$ 'rotated' by $\frac{\pi}{4}$ around $s=1$. In this way we obtain a new Taylor expansion of $\zeta(*)$ and it converges inside the 'green circle', which is the 'red circle' rotated by $\frac{\pi}{4}$ and contains points for which is $\Re (s) < 1$. After four 'rotations' of $\frac{\pi}{4}$ we obtain a Taylor expansion of $\zeta(*)$ that converges inside the 'blue circle', 'centered' in $s=0$ and has radius $1$ and after four more 'rotations' of $\frac{\pi}{4}$ we obtain again the Taylor expansion (3). With the procedure now illustrated we have been able to compute the function $\zeta(*)$ in any point inside a circle 'centered' in $s=1$ and radius $2$ with the only exception of the point $s=1$. Of course if we use a different [possibly greater...] value of $s_{0}$ we arrive to compute $\zeta(*)$ virtually for the whole s plane with the exception of the point $s=1$...

Kind regards

$\chi$ $\sigma$

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