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greetings. we have the following integral :

$$I(s)=s\int_{0}^{\infty} \frac{dx}{2x}\left(E_{s/2}((\pi x)^{s/2})-1\right)\omega(x)-\left(E_{s/2}(( x)^{s/2})-1\right)e^{-x}$$

where : $E_{\alpha}(z)$ is the mittag-leffler fuction

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

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

and $\zeta(s)$ is the Riemann zeta function

the integral behaves well for $Re(s)>1$ . i am trying to extend the domain of the integral to the whole complex plane except for some points. i have tried the identities:

$\omega(x^{-1})=-\frac{1}{2}+\frac{1}{2}x^{1/2}+x^{1/2}\omega(x)$

and

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

to split the integration at 1 , and replace $x$ by $x^{-1}$ in the interval [0,1], but it seems i am lost !!! hence, the post .

edit: another representation of the integral above is - for a minute assume i'm correct about this equivalence - : $$I(s)=s\int_{0}^{\infty}\frac{E_{s}(x^{s})-1}{xe^{x}(e^{x}-1)}dx$$

i was wondering if we can apply Riemann's trick, and replace this integral with a contour integral to obtain a meromorphic integral !?


following Riemann's trick, here is what i did :

start with contour integral :

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

the contour is the usual Hankel contour. consider $I(-s)$ :

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

or

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

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

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

or :

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

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

$$I(s)=-s\oint_{c}\frac{E_{s}((-x)^{s})-1}{xe^{x}(e^{x}-1)}dx=-s\sum_{n=1}^{\infty}\frac{1}{\Gamma(1+ns)}\oint_{c}\frac{(-x)^{sk-1}dx}{e^{x}(e^{x}-1)}$$ $$=s\sum_{n=1}^{\infty}\frac{2i \sin(k\pi s)\Gamma(ks)}{\Gamma(1+ns)}\left(\zeta(ks)-1\right)=2i\sum_{n=1}^{\infty}\sin(k\pi s)\frac{\zeta(ks)-1}{k}$$

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

$$\sum_{n=1}^{\infty}\sin(k\pi s)\frac{\zeta(ks)-1}{k}=A(s)\sum_{n=1}^{\infty}\frac{\zeta(ks)-1}{k}$$

if we define :

$$k(s)=\sum_{n=1}^{\infty}\frac{\zeta(ks)-1}{k}$$

then :

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

share|improve this question
    
one thing that might help is the Mellin-Barnes integral rep. of the mittag-Leffler function :$$E_{\alpha}(z)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-\alpha s)}(-z)^{-s}ds$$ –  mohammad-83 Apr 29 '12 at 11:54
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