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Riemann zeta function

  1. Aug 25, 2008 #1
    Can someone show me the steps to evaluating [tex]\zeta(c + xi)[/tex], where [tex] 0 \leq c<1[/tex]?
  2. jcsd
  3. Aug 26, 2008 #2

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


    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.
  4. Aug 26, 2008 #3
    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 [tex]re(\zeta(s)) + im(\zeta(s))=0[/tex], and are said to be equal to [tex]\zeta(1/2+xi)[/tex], right?

    Hopefully you can see through my confusion
  5. Aug 27, 2008 #4


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    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, [itex]\zeta(s)=0[/itex] at [itex]s=1/2+xi[/itex] for some real values of x.
  6. Sep 3, 2008 #5
    I'm not quite sure how to evaluate the function where s equals say, 1/2 + i, mainly the i part.
  7. Sep 4, 2008 #6
  8. Sep 5, 2008 #7
    What is the definition of the zeta function anyway? I only found this:[tex]\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}[/tex]
    on wikipedia, but then I read that zeros occur at even negative integers, but I cant see how that fits with this definition....
  9. Sep 5, 2008 #8
    That is the definition for Re(s)>1 only. You have to use the analytic continuation to evaluate for the negative even integers.
  10. Sep 5, 2008 #9
    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
  11. Sep 5, 2008 #10
    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...
  12. Sep 5, 2008 #11


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    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.
  13. Sep 5, 2008 #12
    Which means??? :blushing:
  14. Sep 6, 2008 #13
    Let's recap here,

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

    [tex]\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}[/tex]

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

    [tex]\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}[/tex]

    Which satisfies the relationship:


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

    [tex]\zeta(s)=2^s \pi^{s-1} \sin\left(\frac{\pi s}{2}\right) \Gamma(1-s) \zeta(1-s)[/tex]
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