# Riemann zeta function

1. Aug 25, 2008

### epkid08

Can someone show me the steps to evaluating $$\zeta(c + xi)$$, where $$0 \leq c<1$$?

2. Aug 26, 2008

### Count Iblis

There are many ways. In this case, as long as x is nonzero, you can simply use the definition of the zeta function

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

If s has a nonzero imaginary part, the summation will converge. You can then analytically continue this to the positive real axis by adding and subtracting the pole at s = 1.

3. Aug 26, 2008

### epkid08

I have some questions.

The definition of the zeta function you posted, doesn't that only work for where re(s)>1?

I don't really understand the correct usage of s. Isn't s defined as n + xi? So then how can s just equal 1? Or does s refer only to the real part of it?

The non-trivial zeros of the zeta function happen when $$re(\zeta(s)) + im(\zeta(s))=0$$, and are said to be equal to $$\zeta(1/2+xi)$$, right?

Hopefully you can see through my confusion

4. Aug 27, 2008

### CRGreathouse

Re(s) is the real part of s, which is complex. I don't know what you mean by "how can s just equal 1", since no one's claimed that to be the case. Were you talking about the pole at s = 1? That's the thing that stops you from evaluating zeta(1) -- which is obvious since you're trying to sum the harmonic sequence there.

Yes, $\zeta(s)=0$ at $s=1/2+xi$ for some real values of x.

5. Sep 3, 2008

### epkid08

I'm not quite sure how to evaluate the function where s equals say, 1/2 + i, mainly the i part.

6. Sep 4, 2008

7. Sep 5, 2008

### Kurret

What is the definition of the zeta function anyway? I only found this:$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$
on wikipedia, but then I read that zeros occur at even negative integers, but I cant see how that fits with this definition....

8. Sep 5, 2008

### Santa1

That is the definition for Re(s)>1 only. You have to use the analytic continuation to evaluate for the negative even integers.

9. Sep 5, 2008

### Count Iblis

That summation defines an analytical function for Re(s)>1. You can then analytically continue that function to the whole complex plane. You then get a (unique) meromorphic function on the complex plane that is identical to the given summation for Re(s) > 1.

Last edited: Sep 5, 2008
10. Sep 5, 2008

### Count Iblis

If Im(s) is different from zero, then doesn't the summation converge for 0<Re(s)<1? You'll have an oscillatory summand with decreasing terms...

11. Sep 5, 2008

### CRGreathouse

It doesn't converge nicely, that's for sure... it's wildly oscillating up to 10^7, at the least, in the OP's example.

12. Sep 5, 2008

### Kurret

Which means???

13. Sep 6, 2008

### Santa1

Let's recap here,

If you want to evaluate the zeta function With Re(s) > 1 you can use the definition.

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

If you want to evaluate for Re(s) > 0 you can use the Dirichlet eta function for an analytic continuation.

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

Which satisfies the relationship:

$$\frac{\eta(s)}{1-2^{1-s}}=\zeta(s)$$

To evaluate for Re(s) < 0 you can use another analytic continuation.

$$\zeta(s)=2^s \pi^{s-1} \sin\left(\frac{\pi s}{2}\right) \Gamma(1-s) \zeta(1-s)$$