# Analytic continuation of an integral involving the mittag-leffler function

1. Feb 18, 2012

### mmzaj

greetings . we have the integral :

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

which is equivalent to

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

$E_{\alpha}(z)$ being the mittag-leffler function

and $\theta(x)$ is the jacobi theta function

the integral above behaves well for Re(s)>1 . i am trying to extend the domain of $I(s)$ to the whole complex plane except for some points. but i have no idea where to start !!

Last edited: Feb 18, 2012
2. Feb 18, 2012

### mmzaj

the mittag-leffler function admits the beautiful continuation :

$$E_{\alpha}(z)=1-E_{-\alpha}(z^{-1})$$

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

and :

$$\theta(-\frac{1}{t})=(-it)^{1/2}\theta(t)$$

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. Feb 19, 2012

### mmzaj

here is what i've been trying to do

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

where$$\omega(x)=\frac{\theta(ix)-1}{2}$$

i would like some help here !!!

Last edited: Feb 19, 2012