Engineering Mathematics: Sep 9, 2012

Sunday, September 9, 2012

HZ.ID.1: Hurwitz Zeta- Gamma relation

$\zeta (s,a)\Gamma (s)= \int_{0}^{\infty } \frac{x^{s-1}e^{-ax}}{1-e^{-x}} dx$

Proof

$\zeta (s,a)= \sum _{n=0}^{\infty } \frac{1}{(n+a)^{s}}$

(1)

From (), the multiplication of (1) with gamma integral definition

$\zeta (s,a)\Gamma (s)= \sum _{n=0}^{\infty }\int_{0}^{\infty } \frac{t^{s-1}e^{-t}}{(n+a)^{s}} dt$

(2)

change of variable

$t=(n+a)x , dt = (n+a)dx$

(3)

substituting (3) into (2)

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

(4)

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

(5)

$\zeta (s,a)\Gamma (s)= \sum _{n=0}^{\infty }e^{-nx}\int_{0}^{\infty } x^{s-1} e^{-ax} dx$

(6)

using geometric series

$\zeta (s,a)\Gamma (s)= \int_{0}^{\infty } \frac{x^{s-1}e^{-ax}}{1-e^{-x}} dx$

(7)