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Math Help - analytic continuation of an integral

  1. #1
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    analytic continuation of an integral

    greetings. we have the following integral :


    I(s)=s\int_{0}^{\infty}\frac{E_{s}(x^{s})-1}{xe^{x}(e^{x}-1)}dx

    where

    E_{s}(x) is the mittag-leffler function

    the integral is well defined for \Re(s)>1

    i was wondering if we can apply Riemann's trick, and replace this integral with a contour integral to obtain a meromorphic integral - one that is analytic almost everywhere in the complex plane- !?

    namely, consider the contour integral :

    K(s)=-s\oint _{\gamma}\frac{E_{s}((-x)^{s})-1}{xe^{x}(e^{x}-1)}dx

    where the contour starts and ends at + and circles the origin once. using this contour along with the Mellin-Barnes integral rep. of the mittag-leffler function, can we start working the analytic continuation of the original integral ?
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  2. #2
    Junior Member
    Joined
    Apr 2008
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    Re: analytic continuation of an integral

    following Riemann's trick, here is what i did :

    start with the contour integral :

    I(s)=-s\oint_{c}\frac{E_{s}((-x)^{s})-1}{xe^{x}(e^{x}-1)}dx
    the contour is the usual Hankel contour. consider I(-s) :

    I(-s)=s\oint_{c}\frac{E_{-s}((-x)^{-s})-1}{xe^{x}(e^{x}-1)}dx=-s\oint_{c}\frac{E_{s}((-x)^{s})}{xe^{x}(e^{x}-1)}dx
    - the Mittag-Leffler function admits the beautiful continuation : E_{z}(x^{-1})=1-E_{-z}(x)

    or

    I(s)-I(-s)=s\oint_{c}\frac{dx}{xe^{x}(e^{x}-1)}=s\oint_{c}(-x)^{-1}e^{-x}dx-s\oint_{c}\frac{(-x)^{-1}dx}{e^{x}-1}

    now : \oint_{c}(-x)^{-1}e^{-x}dx=\frac{-2\pi i}{\Gamma(1)}=-2\pi i
    and the second integral could be thought of as:

    \oint_{c}\frac{(-x)^{-1}dx}{e^{x}-1}=\lim_{z\rightarrow 0}\oint_{c}\frac{(-x)^{z-1}dx}{e^{x}-1}=-2i\lim_{z\rightarrow 0}\sin(\pi z)\Gamma(z)\zeta(z)=i\pi

    or :

    I(s)-I(-s)=-3\pi is

    lets go back to the 1st integral, and expand the Mittag-leffler function :

    I(s)=-s\oint_{c}\frac{E_{s}((-x)^{s})-1}{xe^{x}(e^{x}-1)}dx=-s\sum_{n=1}^{\infty}\frac{1}{\Gamma(1+ns)}\oint_{c  }\frac{(-x)^{sk-1}dx}{e^{x}(e^{x}-1)}

    =s\sum_{n=1}^{\infty}\frac{2i \sin(k\pi s)\Gamma(ks)}{\Gamma(1+ns)}\left(\zeta(ks)-1\right)=2i\sum_{n=1}^{\infty}\sin(k\pi s)\frac{\zeta(ks)-1}{k}

    now the problem becomes finding a function of the variable s -lets call it A(s)- such that:

    \sum_{n=1}^{\infty}\sin(k\pi s)\frac{\zeta(ks)-1}{k}=sA(s)\int_{0}^{\infty}\frac{E_{s}(x^{s})-1}{xe^{x}(e^{x}-1)}dx

    define:

    k(s)=s\int_{0}^{\infty}\frac{E_{s}(x^{s})-1}{xe^{x}(e^{x}-1)}dx
    then :

    A(s)k(s)-A(-s)k(-s)=-\frac{3}{2}\pi  s
    and the problem becomes proving the existence of A(s) for all s, and of course, finding it !!

    here is my strategy for finding  A(s) :

    k(s)=s\int_{1}^{\infty}\Omega(x)x^{-s-1}dx

    for some function/distribution \Omega(x) . clearly the relation above defines a Mellin pair. Using Mellin inversion theorem, along with Mellin convolution, one can find a function b(x) , such that:

     b(x)=\Omega(x)\star a(x)

    where :

    a(x)\rightleftharpoons A(s)

    under Mellin transform .

    that being said, i'm stuck on doing the following Mellin type integral :

    \frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}\sin(k\pi s)\frac{\zeta(ks)-1}{sk}x^{s}ds

    any insights are more than welcome
    Last edited by mmzaj; May 6th 2012 at 10:19 AM.
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