Here's a fun challenge!
Evaluate Raabe's integral:
$\displaystyle \int_{0}^{1} \log\Gamma(x)dx $
(This used to be my signature. )
it's quite easy, first note that $\displaystyle \int_{0}^{1}{\ln \left( \Gamma (x) \right)\,dx}=\frac{1}{2}\left( \ln \pi -\int_{0}^{1}{\ln (\sin \pi x)\,dx} \right),$ for the last integral put $\displaystyle t=\pi x$ so the last challenge is to compute $\displaystyle \int_{0}^{\pi }{\ln (\sin x)\,dx},$ which comes from the fact that $\displaystyle \int_{0}^{\frac{\pi }{2}}{\ln (\sin x)\,dx}=-\frac{\pi }{2}\ln 2.$
eventually, your integral equals $\displaystyle \frac12\ln(2\pi).$
He used the Gamma function's reflection formula.
Gamma function - Wikipedia, the free encyclopedia
Perform $\displaystyle u $ substitution by letting $\displaystyle u=1-x $ to get
$\displaystyle \int_{0}^{1} \log\Gamma(1-x)dx = \int_{0}^{1} \log\Gamma(u)du $.
You could also look at it this way: $\displaystyle \{\Gamma(x)|x\in[0,1]\} = \{\Gamma(1-x)|x\in[0,1]\} $, so when we integrate we can swap the two functions.
Then shouldn't it be the following:
$\displaystyle \int^{1/2}_{0} \ln \Gamma(x) \ dx + \int^{1/2}_{0} \ln \Gamma(1-x) \ dx = \int_{0}^{1/2} (\ln \pi - \ln \sin \pi x) \ dx $
$\displaystyle \int^{1/2}_{0} \ln \Gamma(x) \ dx + \int^{1}_{1/2} \ln \Gamma(t) \ dt = \int_{0}^{1/2} (\ln \pi - \ln \sin \pi x) \ dx $
EDIT: $\displaystyle \int^{1}_{0} \ln \Gamma(x) \ dx = \frac{1}{2} \ln \pi - \int^{1/2}_{0} \ln \sin \pi x \ dx $
Actually, I think that might be equivalent.
Take a look here , the integral can be generalized to that has a new lower and upper limit :
http://www.mathhelpforum.com/math-he...egral-bee.html
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