analytic continuation of an integral involving the mittagleffler functionby mmzaj Tags: analytic, continuation, function, integral, involving, mittagleffler 

#1
Feb1812, 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})s2x^{1/2}\right)}{x}dx [/tex] [itex]E_{\alpha}(z)[/itex] being the mittagleffler 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 !! 



#2
Feb1812, 01:43 PM

P: 99

the mittagleffler function admits the beautiful continuation :
[tex]E_{\alpha}(z)=1E_{\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})s2x^{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 !! 



#3
Feb1912, 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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