# analytic continuation of an integral

• May 1st 2012, 11:48 AM
mmzaj
analytic continuation of an integral
greetings. we have the following integral :

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

where

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

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

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 :

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

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 ?
• May 6th 2012, 10:05 AM
mmzaj
Re: analytic continuation of an integral
following Riemann's trick, here is what i did :

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

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

or

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

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

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

or :

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

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

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

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

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

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

define:

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

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

here is my strategy for finding $\displaystyle A(s)$ :

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

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

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

where :

$\displaystyle a(x)\rightleftharpoons A(s)$

under Mellin transform .

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

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

any insights are more than welcome