Engineering Mathematics: Dirichlet Beta

DBG.ID.1: Integral representation

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

given

Dirchelet Beta :

$\beta (s)= \sum _{0}^{\infty }\frac{(-1)^{n}}{(2n+1)^{s}}$

(1)

$\beta (s)\Gamma (s)= \sum _{0}^{\infty }(-1)^{n} \int _{0}^{\infty }\frac{t^{s-1}e^{-t}}{(2n+1)^{s}} dt$

(2)

$t=(2n+1)x \, , \, dt= (2n+1)dx$

(3)

replacing (2) with (3)

$\beta (s)\Gamma (s)= \sum _{0}^{\infty }(-1)^{n} \int _{0}^{\infty }\frac{(2n+1)^{s-1}x^{s-1}e^{-(2n+1)x}}{(2n+1)^{s}} (2n+1)dx$

(4)

$\beta (s)\Gamma (s)= \int _{0}^{\infty }x^{s-1}\sum _{0}^{\infty }(-1)^{n}e^{-(2n+1)x})dx$

(5)

using geometric series

$\beta (s)\Gamma (s)= \int _{0}^{\infty }\frac{x^{s-1} e^{-x}}{1+e^{-2x}}dx$

(6)

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

(7)

DBG.ID.2: Integral representation

$\beta (s)\Gamma (s+1)= \int _{0}^{\infty } \frac{e^{x}-e^{-x} }{(e^{x}+e^{-x})^{2}}x^{s}dx$

proof

multiply both sides by s

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

(8)

$\beta (s)\Gamma (s+1)= \int _{0}^{\infty }\left ( \frac{d}{dx} x^{s}\right )\frac{1 }{e^{x}+e^{-x}}dx$

(9)

$\beta (s)\Gamma (s+1)= \int _{0}^{\infty }\left ( \frac{d}{dx} \frac{1 }{e^{x}+e^{-x}}\right )x^{s}dx$

(10)

$\beta (s)\Gamma (s+1)= \int _{0}^{\infty } \frac{e^{x}-e^{-x} }{(e^{x}+e^{-x})^{2}}x^{s}dx$

(11)

DBG.ID.3:Recurrence Relation

$\beta (s+1)-\beta (s) = \frac{2}{\Gamma (s+1)}\int _{0}^{\infty } \frac{e^{-x}}{(e^{x}+e^{-x})^{2}} x^{s} dx$

given

from DBG.ID.1

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

(1)

$\beta (s+1)\Gamma (s+1) = \int _{0}^{\infty } \frac{e^{x}+e^{-x}}{(e^{x}+e^{-x})^{2}} x^{s} dx$

(2)

from DBG.ID.2

$\beta (s)\Gamma (s+1)= \int _{0}^{\infty } \frac{e^{x}-e^{-x} }{(e^{x}+e^{-x})^{2}}x^{s}dx$

(3)

(3) - (2) results in :

$[\beta (s+1)-\beta (s)]\Gamma (s+1) = 2\int _{0}^{\infty } \frac{e^{-x}}{(e^{x}+e^{-x})^{2}} x^{s} dx$

(4)

$\beta (s+1)-\beta (s) = \frac{2}{\Gamma (s+1)}\int _{0}^{\infty } \frac{e^{-x}}{(e^{x}+e^{-x})^{2}} x^{s} dx$

(5)

DBG.ID.5: Integral representation by natural log

$\beta (s)\Gamma (s)=\int _{0}^{1}\frac{In(\frac{1}{t})^{s-1}}{1+t^{2}} dt$

from DBG.ID.1

$\beta (s)\Gamma (s)= \int _{0}^{\infty }\frac{x^{s-1} e^{-x}}{1+e^{-2x}}dx$

(1)

Change of variable

$x= In(\frac{1}{t}) \: ,\: dx = -\frac{1}{t}dt$

Therefore,

$\beta (s)\Gamma (s)=-\int _{1}^{0}\frac{In(\frac{1}{t})^{s-1}e^{-In(\frac{1}{t})}}{1+e^{-2In(\frac{1}{t})}t} dt$

(2)

$\beta (s)\Gamma (s)=\int _{0}^{1}\frac{In(\frac{1}{t})^{s-1}}{1+t^{2}} dt$

(3)

if s=2

$\beta (2)=\int _{0}^{1}\frac{In(\frac{1}{t})}{1+t^{2}} dt$

$\beta (2)=-\int _{0}^{1}\frac{In(t)}{1+t^{2}} dt$

(4)

Engineering Mathematics

Saturday, June 2, 2012

Dirichlet Beta - Gamma

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