I am trying to calculate an integral that can be expressed in terms of infinite hypergeometric series by using transforms and Residue method, the integral is $$ I_{n,m}(\alpha,\sigma,\omega,r)=\int_0^\infty x^{\alpha -1}Li_n (-\sigma x) Li_m(-\omega x^r)dx $$ where n,m are positive integers. The complex parameters are $\alpha,\sigma,\omega$ and real $r\neq 0$ are defined so that the integral exists. This is an integral containing polyLogs so we define the Polylogarithm function for $|z|\leq 1$ and $n\geq 2$ by a power series expression of the form $$ Li_n(z)=\sum_{j=1}^\infty \frac{z^j}{j^n}, \ (|z| \leq 1). $$ For $|z| > 1$ we obtain $$ Li_n(z)=(-1)^{n+1}Li_n(z^{-1})-\frac{1}{n!}\ln^n(-z)-\sum_{j=0}^{n-2}\frac{1}{j!}(1+(-1)^{n-j})(1-2^{1-n+j})\zeta(n-j)\ln^j(-z) $$ where the Riemann Zeta function is given by $\zeta(n-j)$ and the logarithm is defined on the principle sheet. I was thinking of also using an integral representation of the polyLogs which are given by $$ Li_n(-z)=\frac{1}{2\pi i}\int_{\gamma -i\infty}^{\gamma+i\infty} \frac{\Gamma(s)\Gamma(-s)}{(-s)^{n-1}} z^{-s} ds, \ (-1<\gamma <0, |\arg z| < \pi, \ n=0,1,2,...) $$ and inserting into this $I_{n,m}$...Any ideas on what do, I need a complete solution. Thanks.

  • $\begingroup$ Bounty Again Soon? $\endgroup$ – draks ... May 16 '14 at 21:01
  • $\begingroup$ @draks... Yes I am putting it up as a bounty (again) since nobody could help last time I put it up for +500. $\endgroup$ – Jeff Faraci May 17 '14 at 20:02

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