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Analytic continuation of an integral involving the mittag-leffler function

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mmzaj
#1
Feb18-12, 06:05 AM
P: 99
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 !!
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mmzaj
#2
Feb18-12, 01:43 PM
P: 99
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 !!
mmzaj
#3
Feb19-12, 07:18 PM
P: 99
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 !!!


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