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Analytic continuation of an integral involving the mittag-leffler function

  1. Feb 18, 2012 #1
    greetings . we have the integral :

    [tex] I(s)=\int_{0}^{\infty}\frac{s(E_{s}(x^{s})-1)-x}{x(e^{x}-1)}dx [/tex]

    which is equivalent to

    [tex] =I(s)=\frac{1}{4}\int_{0}^{\infty}\frac{\theta(ix)\left(sE_{s/2} ((\pi x)^{s/2})-s-2x^{1/2}\right)}{x}dx [/tex]

    [itex]E_{\alpha}(z)[/itex] being the mittag-leffler function

    and [itex] \theta(x) [/itex] is the jacobi theta function

    the integral above behaves well for Re(s)>1 . i am trying to extend the domain of [itex]I(s)[/itex] to the whole complex plane except for some points. but i have no idea where to start !!
     
    Last edited: Feb 18, 2012
  2. jcsd
  3. Feb 18, 2012 #2
    the mittag-leffler function admits the beautiful continuation :

    [tex]E_{\alpha}(z)=1-E_{-\alpha}(z^{-1}) [/tex]

    using the fact that
    [tex]I(s)=\frac{1}{4}\int_{0}^{\infty}\frac{\theta(ix) \left(sE_{s/2} ((\pi x)^{s/2})-s-2x^{1/2}\right)}{x}dx [/tex]

    and :

    [tex]\theta(-\frac{1}{t})=(-it)^{1/2}\theta(t) [/tex]

    one can split the integration much like the one concerning the Riemann zeta . but i am not sure this will yield a meromorphic integral . hence, the problem !!
     
  4. Feb 19, 2012 #3
    here is what i've been trying to do

    [tex] I(s)= 1 -\frac{1}{2}\left[\ln(x)\right]_{1}^{\infty}-\int_{1}^{\infty}\omega(x)\left(x^{-1}+\frac{sx^{-1}}{2}+x^{-1/2} \right ) dx +\int_{1}^{\infty}\frac{s\omega(x)}{2}\left[x^{-1}E_{\frac{s}{2}}(x\pi)^{s/2}-x^{-1/2} E_{\frac{-s}{2}}\left(\frac{x}{\pi}\right)^{s/2} \right ]dx +\int_{1}^{\infty}\frac{s}{4}\left[x^{-1} E_{\frac{-s}{2}}\left(\frac{x}{\pi} \right )^{s/2}+x^{-1/2}E_{\frac{-s}{2}}\left(\frac{x}{\pi} \right )^{s/2} \right ]dx[/tex]

    where[tex] \omega(x)=\frac{\theta(ix)-1}{2}[/tex]


    i would like some help here !!!
     
    Last edited: Feb 19, 2012
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