# Continuation of zeta function

## Main Question or Discussion Point

Can anybody of you people recommend me the best, most pedagocical, clearest, easiest, but detailed enough explanation of how to analytical continue the zeta function to the whole complex plane (except 1, of course!)?

In a book, notes on the net, whatever!

thank you

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You mean analytically continue the Euler sum, which converges only for Re(s)>1, to the left-half-plane Re(s)<=1. That analytic continuation is called the "zeta" function. I think the most elegant explanation is the integral derivation described by Riemann and explained in Edwards book, "The Zeta Function". However, that is really tough to follow unless you're up on Complex Analysis. In general, start with an integral that converges for Re(s)>1 and reduces to the Euler sum when Re(s)>1, then show the integral converges for all s not equal to 1. Then by the principle of Analytic Continuation, conclude the integral and the Euler sum both represent the same (analytic) function where ever they are both analytic. That is, the integral expression is the analytic continuation of the Euler sum to $\mathbb{C}\backslash \{1\}$.

If you want, get the book and try to work through (muscle through) the first few pages of the first chapter. Also see "The Distribution of Prime Numbers" by Ingham. That one's tough too but a little easier than Edwards.

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As jackmell pointed out, you need to know complex analysis to follow any explanation of how to analytically continue the zeta function. What is your background in this area?

Hi guys. I'd like to go over the analysis just for fun ok. I'm no expert and perhaps those more familiar with it can make some further comments and/or corrections. I think the derivation is quite beautiful:

I'll start it by beginning with the Euler sum definition which is valid for $Re(s)>1$:

Let:

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

Now consider the Gamma function and let $y=nx$:

$$\Gamma(s)=\int_0^{\infty}y^{s-1}e^{-y}dy=n^s\int_0^{\infty}x^{s-1}e^{nx}dx,\quad Re(s)>1, n>0$$

Dividing by $n^s$ and summing from one to infinity:

$$\Gamma(s)E(s)=\sum_{n=1}^{\infty} x^{s-1}e^{-nx}dx=\int_0^{\infty}\frac{x^{s-1}}{e^{x}-1}dx$$

which is valid for $Re(s)>1$. Now we turn to Complex Analysis and consider the Hankel Integral over the mirror-image Hankel Contour $C^{-}$ (since that is more natural and avoids the $(-z)^s$ in the numerator originally used by Riemann):

$$I(s)=\frac{1}{2\pi i}\int_{C^-} \frac{z^{s-1}}{e^{-z}-1}dz$$

That is just a branch-cut integral over the principal branch cut of $\log$ along the negative real axis. We can show the integral function $I(s)$ converges for all $s$ and so is an analytic function of $s$. Now using the standard substitutions $z=re^{\pi i}$ over the top trace, $z=re^{-\pi i}$ over the bottom trace and $z=\rho e^{it}$ around the origin, we obtains after some further analysis:

$$\pi I(s)=\sin(\pi s)\int_0^{\infty}\frac{r^{s-1}}{e^{r}-1}dr=\sin(\pi s)\Gamma(s)E(s)[/itex] and therefore: [tex]E(s)=\frac{\pi I(s)}{\sin(\pi s)\Gamma(s)}=\Gamma(1-s)I(s)$$

Now here is where we use the Principle of Analytic Continuation:

We have just shown this analysis is valid for $Re(s)>1$. However, the right side of the last expression is analytic for all $s$ (because gamma and I are) except possibly at the (simple) poles of $\Gamma(1-s)$ or $s=1,2,3,\cdots$

And it can be further shown that $I(s)$ has simple zeros at $s=2,3,\cdots$.
(may need to confirm this part)

Therefore the function given by:

$$\Gamma(1-s)I(s)$$

is a meromorphic function with a single simple pole at $s=1$ and reduces to the Euler Sum $E(s)$ for $Re(s)>1$. This function was named the zeta function by Riemann:

$$\zeta(s)=\Gamma(1-s)I(s)$$

and represents the analytic continuation of the Euler sum.

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Thanks a lot, Jackmell. I saw a similiar explanation in the book 'Gamma' by Havil. I think the part where you go from the branch-cut integral to the expression with the sinus is the hardest. Havil gives some more details, as you mentioned before, someone has to be up on Complex analysis to follow it all. But the outline of the derivation is clear for me now, I think.

But there are other ways to continue the Euler sum. One involves the Bernoullie numbers and is supposed to be quite easy. Then one which involves a functional equation. So there are really many ways of doing this?

I think the part where you go from the branch-cut integral to the expression with the sinus is the hardest.
Absolutely. That part is tough for me to follow. Also the part where the zeros of I(s) cancel the poles of Gamma is also tough for me.

But there are other ways to continue the Euler sum. One involves the Bernoullie numbers and is supposed to be quite easy. Then one which involves a functional equation. So there are really many ways of doing this?
Yes.

What I outlined is the original derivation used by Riemann which I think is beautiful, except he did not use the mirror-image contour and so had to deal with the $(-z)^s$ in the numerator which I think is quite awkward to deal with.

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mathwonk