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