Engineering Mathematics: May 27, 2011

Friday, May 27, 2011

Raabe's integral

This post aims to show the derivation of Raabe's integral

$\int _{0}^{1} \log \Gamma (t)\, dt = \frac{1}{2}\log 2\pi$

Duplication formula of gamma function is :

$\Gamma (t)\Gamma (t+\frac{1}{2})= 2^{1-2t} \pi ^{\frac{1}{2}}\Gamma (2t)$

replacing t=2t

$\Gamma (\frac{t}{2})\Gamma (\frac{t+1}{2})= 2^{1-t} \pi ^{\frac{1}{2}}\Gamma (t)$

(1)

$2^{t-1}\Gamma (\frac{t}{2})\Gamma (\frac{t+1}{2})= \pi ^{\frac{1}{2}}\Gamma (t)$

(2)

$\log 2^{t-1} + \log\Gamma (\frac{t}{2})+ \log\Gamma (\frac{t+1}{2})= \log \pi ^{\frac{1}{2}}+ \log\Gamma (t)$

(3)

$(t-1)\log 2 + \log\Gamma (\frac{t}{2})+ \log\Gamma (\frac{t+1}{2})= \frac{1}{2}\log \pi + \log\Gamma (t)$

(4)

$\log 2\int _{0}^{1}(t-1) + \int _{0}^{1}\log\Gamma (\frac{t}{2})+ \int _{0}^{1}\log\Gamma (\frac{t+1}{2})= \frac{1}{2}\int _{0}^{1}\log \pi + \int _{0}^{1}\log\Gamma (t)$

(5)

$C= \int _{0}^{1} \log \Gamma (t)\, dt$

(6)

from (5) and (6) ,

$\frac{1}{2}\log \pi +C = -\frac{1}{2}\log 2 + 2\int _{0}^{\frac{1}{2}} \Gamma (u)\, du + 2\int _{\frac{1}{2}}^{1} \Gamma (u)\, du$

(7)

$\frac{1}{2}\log \pi +C = -\frac{1}{2}\log 2 + 2\int _{0}^{1} \Gamma (u)\, du$

(8)

from (6) and (8) ,

$\frac{1}{2}\log \pi +C = -\frac{1}{2}\log 2 + 2C$ $C=\frac{1}{2}\log 2\pi$

therefore

$\int _{0}^{1} \log \Gamma (t)\, dt = \frac{1}{2}\log 2\pi$

Beta Function

firstly , I recommend you to read these Beta function to have fundamental understanding of the Gamma function.

G.ID.1: integral representation

$\beta (x,y)=\int _{0}^{1}t^{x-1}(1-t)^{y-1}\, dt$

G.ID.1: Recurrence

$\beta (x,y+1)= \beta (x,y)-\beta (x+1,y)$

$\beta (x,y)= \frac{y}{x}\beta (x+1,y) + \beta (x+1,y) = \frac{y+x}{x} \beta (x+1,y)$

$\beta (x,y)=\frac{y+x}{y}\beta (x,y+1)$

Proof

$\beta (x,y+1)=\int _{0}^{1}t^{x-1}(1-t)^{y}\, dt$

(1)

$\beta (x,y+1)=\int _{0}^{1}t^{x-1}(1-t)^{y-1}(1-t)\, dt$

(2)

$\beta (x,y+1)=\int _{0}^{1}(t^{x-1}-tt^{x-1})(1-t)^{y-1}\, dt$

(3)

$\beta (x,y+1)=\int _{0}^{1}t^{x-1}(1-t)^{y-1} \, dt- \int _{0}^{1} t^{x}(1-t)^{y-1}\, dt$

(4)

$\beta (x,y+1)= \beta (x,y)-\beta (x+1,y)$

(5)

integrating by parts of (1)

$\int _{0}^{1}t^{x-1}(1-t)^{y}\, dt = \left [ \frac{t^{x}}{x}(1-t)^{y} \right ]_{0}^{1} + \frac{y}{x} \int _{0}^{1}t^{x}(1-t)^{y-1}\, dt$

(6)

from (1) and (6)

$\beta (x,y+1)= \frac{y}{x}\beta (x+1,y)$

(7)

from (5) and (7)

$\beta (x,y)= \frac{y}{x}\beta (x+1,y) + \beta (x+1,y) = \frac{y+x}{x} \beta (x+1,y)$

$\frac{x}{y+x}\beta (x,y)=\beta (x+1,y)$

(8)

from (7) and (8),

$\frac{x}{y+x}\beta (x,y)=\frac{x}{y}\beta (x,y+1)$

(9)

$\beta (x,y)=\frac{x+y}{x}\frac{x}{y}\beta (x,y+1)$

(10)

$\beta (x,y)=\frac{y+x}{y}\beta (x,y+1)$

(11)

G.ID.2: integral representation

$\beta (x,y)=\int_{0}^{\infty }\frac{u^{x-1}}{(1+u)^{x+y}} \, du$

given

$\beta (x,y)=\int _{0}^{1}t^{x-1}(1-t)^{y-1}\, dt$

proof

$u=\frac{t}{1-t},\, t=\frac{u}{u+1},\, dt=\frac{1}{(1+u)^{2}} \, du$

(1)

substitute,

$\beta (x,y)=\int_{0}^{\infty }\left (\frac{u}{u+1} \right )^{x-1}\left(1-\frac{u}{u+1} \right )^{y-1}\times \frac{1}{(1+u)^{2}} \, du$

(2)

$\beta (x,y)=\int_{0}^{\infty }\left (\frac{u}{u+1} \right )^{x-1}\left(\frac{1}{u+1} \right )^{y-1}\times \frac{1}{(1+u)^{2}} \, du$

(3)

$\beta (x,y)=\int_{0}^{\infty }\left u^{x-1}\left(\frac{1}{u+1} \right )^{x+y} \, du$

(4)

$\beta (x,y)=\int_{0}^{\infty }\frac{u^{x-1}}{(1+u)^{x+y}} \, du$

(5)