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Math Help - An integral - Riemann's derivation of number of primes

  1. #1
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    An integral - Riemann's derivation of number of primes

    OK so I am not sure if this problem is analysis, number theory or calculus.

    I am reading an English translation of Riemann's 1859 paper on the number of primes and cannot follow this bit.

    He says consider this integral:

    \int^{+ \inf}_{+ \inf}\frac{(-x)^{s-1}dx}{e^x-1}

    From infinity in a positive sense around a domain that includes 0 but does not include any other point of discontinuity. This is then equal to:

    (e^{-\pi s \i}-e^{\pi s \i)\int^{+ \inf }_0 \frac{x^{s-1}dx}{e^x-1}

    I would like to fill in the details of this step but have been unsuccessful. I have tried:

    Finding the residue at z=0 by solving:
    Res=\lim_{x \rightarrow 0} \frac{x(-x)^{s-1}}{e^x-1}=(-1)^{s-1} lim_{x \rightarrow 0}\frac{s\Pi(s-1)}{e^x}=(-1)^{s-1}s\int^{\inf}_0 \frac{x^{(s-1)}}{e^x}

    I also tried generalising x to z and integrating around the path from inf, \epsilon to -\epsilon,\epsilon to -\epsilon,-\epsilon to \inf,-\epsilon to \inf,\epsilon

    where
    0 < \epsilon < 2\pi

    This gave interesting results when \epsilon either; tended to 0 equals pi or equals pi/2.

    But non-the less I have not been able to get the required result. Is one of my approaches right? Is there a better approach?
    Last edited by Kiwi_Dave; November 13th 2010 at 10:26 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Kiwi_Dave View Post
    OK so I am not sure if this problem is analysis, number theory or calculus.

    I am reading an English translation of Riemann's 1859 paper on the number of primes and cannot follow this bit.

    He says consider this integral:

    \int^{+ \inf}_{+ \inf}\frac{(-x)^{s-1}dx}{e^x-1}

    From infinity in a positive sense around a domain that includes 0 but does not include any other point of discontinuity. This is then equal to:

    (e^{-\pi s \i}-e^{\pi s \i)\int^{+ \inf }_0 \frac{x^{s-1}dx}{e^x-1}

    I would like to fill in the details of this step but have been unsuccessful. I have tried:

    Finding the residue at z=0 by solving:
    Res=\lim_{x \rightarrow 0} \frac{x(-x)^{s-1}}{e^x-1}=(-1)^{s-1} lim_{x \rightarrow 0}\frac{s\Pi(s-1)}{e^x}=(-1)^{s-1}s\int^{\inf}_0 \frac{x^{(s-1)}}{e^x}

    I also tried generalising x to z and integrating around the path from inf, \epsilon to -\epsilon,\epsilon to -\epsilon,-\epsilon to \inf,-\epsilon to \inf,\epsilon

    where
    0 < \epsilon < 2\pi

    This gave interesting results when \epsilon either; tended to 0 equals pi or equals pi/2.

    But non-the less I have not been able to get the required result. Is one of my approaches right? Is there a better approach?
    I'm not sure if I'll be able to help, but could you explain the contour in a little more detail?
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    It's hard to see what is going on here without knowing more of the context, but the expression \int^{+ \inf}_{+ \inf}\frac{(-x)^{s-1}dx}{e^x-1} suggests to me that it means the integral of \frac{(-z)^{s-1}}{e^z-1} around a contour that goes from +\infty to \varepsilon along the real axis, then goes clockwise all the way round a circle of radius \varepsilon, and finally back along the real axis from \varepsilon to +\infty.

    Notice that the two integrals going in opposite directions along the real axis will not cancel each other out, because \frac{(-z)^{s-1}}{e^z-1} = \frac{e^{(s-1)\log(-z)}}{e^z-1}, and \log(-z) will move onto another branch when z goes round the origin.
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    The contour is allowed to be any closed contour starting at x=inf that does not enclose any singularity except for the one at x=0. I only described a contour using \epsilon to demonstrate that I have had a fair crack at solving this myself!

    The exact wording from the paper is:


    If one now considers the integral

    \int\frac{(-x)^{s-1}dx}{e^x-1}

    from +inf to +inf taken in a positive sense around a domain which includes the value 0 but no other point of discontinuity of the integrand in its interior, then ths is easily seen to be equal to

    (e^{-\pi s \i}-e^{\pi s \i})\int^{+ \inf }_0 \frac{x^{s-1}dx}{e^x-1}

    provided that, in the many-valued function (-x)^{s-1}=e^{(s-1)log(-x)}, the logarithm is determined so as to be real when x is negative.
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    I have this sorted now. If we place our branch cut on the positive x axis then as we approach the positive x axis from the positive y side then:

    (-x)^{(s-1)}=e^{(s-1)(ln(x)+i\pi)}=x^{(s-1)}e^{i\pi(s-1)}=x^{(s-1)}e^{i\pi(s-1)}(1)^{(s-1)}

    =x^{(s-1)}e^{i\pi(s-1)}(e^{-2\pi i})^{(s-1)}=x^{(s-1)}e^{-i\pi(s-1)}

    Similarly as we approach the positive x axis from the negative y side then

    (-x)^{(s-1)}=e^{(s-1)(ln(x)+3i\pi)}=x^{(s-1)}e^{3i\pi(s-1)}=x^{(s-1)}e^{3i\pi(s-1)}(1)^{(s-1)}

    =x^{(s-1)}e^{i\pi(s-1)}(e^{-2\pi i})^{(s-1)}=x^{(s-1)}e^{i\pi(s-1)}

    Hence the integral becomes:

    \int ^{+inf} _0 \frac {x^{(s-1)}e^{-i\pi(s-1)}dx}{e^x-1}+\int ^0 _{+inf}  \frac {x^{(s-1)}e^{i\pi(s-1)}dx}{e^x-1}=(e^{-i\pi(s-1)}-e^{i\pi(s-1)}) \int ^{inf} _0 \frac {x^{(s-1)}dx}{e^x-1}
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    How do you define e^{i \pi (s-1)} and e^{-i \pi (s-1)} ? From your first paragraph, these two are equal since
    x^{(s-1)}e^{i \pi (s-1)} = x^{(s-1)}e^{-i \pi (s-1)}, but similarly to your computation we can conclude
    e^{i \pi (s-1)} e^{-i \pi (s-1)} = 1. Then they must equal to 1 or -1, obviously ridiculous.
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  7. #7
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    Quote Originally Posted by Kiwi_Dave View Post
    Is there a better approach?
    Absolutely. Replace the hankel contour with it's mirror image and by doing so, elimiate the awkward (-x)^{s-1} in the numerator and simultaneously convert the integration to the more "normal" principal branch-cut integral along the negative real axis. When you do this, you obtain the integral:

    \displaystyle\mathop\int\limits_{\multimap}\frac{z  ^{s-1}}{e^{-z}-1}dz

    Now make the usual substitutions for a branch-cut integral over the principle branch-cut of log(z) along the negative real axis: starting along the bottom trace at -\infty, let z=re^{-\pi i}, around the origin, let z=\rho e^{it} and along the top trace, let z=re^{\pi i}. Also, don't forget, you need to analyze this under the assumption that \text{Re}(s)>1, then show that as \rho\to 0 the integral around the origin goes to zero.
    Last edited by shawsend; November 15th 2010 at 04:48 AM. Reason: delete analytic continuation part not relevant to question
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