Engineering Mathematics: 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

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

Engineering Mathematics

Friday, May 27, 2011

Raabe's integral

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