Engineering Mathematics: Gamma-Polylogarithm

Friday, June 1, 2012

GP.ID.1:

$\int _{0}^{}\infty \frac{x^{s-1}}{e^{x-a}+1} dx = -\Gamma (s) Li_{s}(-e^{a})$

Proof

Given

$\frac{1}{e^{t}+1} = \sum _{n=1}^{\infty } (-1)^{n-1} e^{-nt}$

(1)

therefore ,

$\int _{0}^{}\infty \frac{x^{s-1}}{e^{x-a}+1} dx = \int _{0}^{}\infty x^{s-1} \sum _{n=1}^{\infty } (-1)^{n-1} e^{-n(x-a)} dx$

(2)

$\int _{0}^{}\infty \frac{x^{s-1}}{e^{x-a}+1} dx = \sum _{n=1}^{\infty } (-1)^{n-1} e^{na} \int _{0}^{}\infty x^{s-1} e^{-nx} dx$

(3)

$\int _{0}^{}\infty \frac{x^{s-1}}{e^{x-a}+1} dx = \sum _{n=1}^{\infty } (-1)^{n-1} e^{na} \frac{\Gamma (s)}{n^{s}}$

(4)

$\int _{0}^{}\infty \frac{x^{s-1}}{e^{x-a}+1} dx = \Gamma (s)\sum _{n=1}^{\infty } \frac{(-1)^{n-1} e^{na}}{n^{s}}$

(5)

$\int _{0}^{}\infty \frac{x^{s-1}}{e^{x-a}+1} dx = -\Gamma (s)\sum _{n=1}^{\infty } \frac{ (-e^{a})^{n}}{n^{s}}$

(6)

Based on the identity:

$Li_{s}(z)= \sum _{n=1}^{\infty } \frac{ z^{n}}{n^{s}}$

(7)

From (5) and (6) ,

$\int _{0}^{}\infty \frac{x^{s-1}}{e^{x-a}+1} dx = -\Gamma (s) Li_{s}(-e^{a})$

Engineering Mathematics