1. ## Raabe's Integral

Here's a fun challenge!

Evaluate Raabe's integral:

$\displaystyle \int_{0}^{1} \log\Gamma(x)dx$

(This used to be my signature. )

2. 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 $\displaystyle \frac12\ln(2\pi).$

3. Originally Posted by Krizalid
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 $\displaystyle \frac12\ln(2\pi).$
Correct! The hard part of the problem though is showing $\displaystyle \int_{0}^{\frac{\pi }{2}}{\ln (\sin x)\,dx}=-\frac{\pi }{2}\ln 2$.

4. $\displaystyle \int_{0}^{1}{\ln \left( \Gamma (x) \right)\,dx}=\frac{1}{2}\left( \ln \pi -\int_{0}^{1}{\ln (\sin \pi x)\,dx} \right)$

What exactly was done here? Integration by parts?

5. Originally Posted by Random Variable
$\displaystyle \int_{0}^{1}{\ln \left( \Gamma (x) \right)\,dx}=\frac{1}{2}\left( \ln \pi -\int_{0}^{1}{\ln (\sin \pi x)\,dx} \right)$

What exactly was done here? Integration by parts?
He used the Gamma function's reflection formula.

Gamma function - Wikipedia, the free encyclopedia

6. Originally Posted by chiph588@
Correct! The hard part of the problem though is showing $\displaystyle \int_{0}^{\frac{\pi }{2}}{\ln (\sin x)\,dx}=-\frac{\pi }{2}\ln 2$.
i've solved that one before, it should be somewhere.

no time now to search for, but i'd take a look tomorrow.

Originally Posted by chiph588@
He used the Gamma function's reflection formula.

Gamma function - Wikipedia, the free encyclopedia
yeah, that's why i said it's an easy problem, because it turns to be to another known problem.

7. Originally Posted by chiph588@
He used the Gamma function's reflection formula.

Gamma function - Wikipedia, the free encyclopedia
$\displaystyle \Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin \pi x}$

$\displaystyle \ln \Gamma(x) + \ln \Gamma(1-x) = \ln \pi - \ln \sin \pi x$

What do you do with the second term on the left?

8. Originally Posted by Random Variable
$\displaystyle \Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin \pi x}$

$\displaystyle \ln \Gamma(x) + \ln \Gamma(1-x) = \ln \pi - \ln \sin \pi x$

What do you do with the second term on the left?
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.

9. 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.

10. Originally Posted by Random Variable
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.
Your method works too, thus showing $\displaystyle \int^{\frac{1}{2}}_{0} \log (\sin (\pi x)) dx = \frac{1}{2}\int^{1}_{0} \log (\sin (\pi x)) dx$.

11. 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