크레이지J의 탐구생활

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

We know this.
$\Gamma(x) = \int_{0}^{\infty}t^{x-1}e^{-t} dt \\ \Gamma(x) \Gamma(1-x) = \frac {\pi}{sin(\pi x)}$
take log.
$ln(\Gamma(x) \Gamma(1-x)) = ln { \frac {\pi}{sin(\pi x)} }\\ ln(\Gamma(x))+ln(\Gamma(1-x)) = ln \pi - ln (sin(\pi x))$
take integral.
$\int ln(\Gamma(x))+ \int ln(\Gamma(1-x)) = \int ln \pi - \int ln (sin(\pi x))\\ \int_0^1 ln(\Gamma(x)) dx+\int_0^1 ln(\Gamma(1-x)) dx = \int_0^1 ln \pi dx- \int_0^1 ln (sin(\pi x)) dx \\ \int_0^1 ln(\Gamma(x))dx+\int_0^1 ln(\Gamma(u)) du=\int_0^1 ln \pi - \int_0^1 ln (sin(\pi x))\\ 2\int_0^1 ln(\Gamma(x))dx=\int_0^1 ln \pi dx- \int_0^1 ln (sin(\pi x))dx\\ 2\int_0^1 ln(\Gamma(x))dx=ln\pi - \int_0^1 ln (sin(\pi x))dx$
And watch the last part.
$\int_0^1 ln (sin(\pi x))dx\\ (u=\pi x , du=\pi dx) \\ =\int_0^{\pi} ln(sin(u)) \frac{1}{\pi} du\\ =\frac{1}{\pi} \int_0^{\pi} ln(sin(u)) du$
We know
$\int_0^{\frac{\pi}{2}}ln(sin(x))dx = -\frac{\pi}{2}ln2$
So,
$\frac{1}{\pi} \int_0^{\pi} ln(sin(u)) du=\frac{1}{\pi} 2 \int_0^{\frac{\pi}{2}}ln(sin(x))dx \\ =\frac{1}{\pi} 2 (-\frac{\pi}{2}ln2) \\ = - ln 2\\ \int_0^1 ln (sin(\pi x))dx=-ln2$
So,
$2\int_0^1 ln(\Gamma(x))dx=ln\pi - \int_0^1 ln (sin(\pi x))dx\\ =ln\pi - (-ln2) = ln \pi + ln 2$
$\therefore \int_0^1 ln(\Gamma(x))dx = \frac {ln \pi + ln 2} {2} = \frac {ln(2\pi)}{2}$

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