# Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,log,Gamma)

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##### 0.1
${\displaystyle \int _{0}^{1}\log \Gamma (x)\,dx=\log {\sqrt {2\pi }}}$
1. Beweis

${\displaystyle 2\int _{0}^{1}\log \Gamma (x)\,dx=\int _{0}^{1}\log \Gamma (x)\,dx+\int _{0}^{1}\log \Gamma (1-x)\,dx=\int _{0}^{1}\log {\Big (}\Gamma (x)\,\Gamma (1-x){\Big )}\,dx}$

${\displaystyle =\int _{0}^{1}\log \left({\frac {\pi }{\sin \pi x}}\right)dx=\log \pi -\int _{0}^{1}\log \sin \pi x\,dx=\log \pi +\log 2\,\Rightarrow \,\int _{0}^{1}\log \Gamma (x)\,dx={\frac {1}{2}}\log(2\pi )}$

2. Beweis

Die Riemannsche Approximationssumme ${\displaystyle \sum _{k=1}^{n-1}\log \Gamma \!\left({\frac {k}{n}}\right)\cdot {\frac {1}{n}}}$ vereinfacht sich zu

${\displaystyle \log \left(\prod _{k=1}^{n-1}\Gamma \!\left({\frac {k}{n}}\right)\right)\cdot {\frac {1}{n}}=\log \left({\frac {{\sqrt {2\pi }}^{\,n-1}}{\sqrt {n}}}\right)\cdot {\frac {1}{n}}={\frac {(n-1)\log {\sqrt {2\pi }}-\log {\sqrt {n}}}{n}}}$,

und konvergiert daher gegen ${\displaystyle \log {\sqrt {2\pi }}}$ für ${\displaystyle n\to \infty \,}$.

##### 0.2
${\displaystyle \int _{1/4}^{3/4}\log \Gamma (x)\,dx={\frac {1}{2}}\left(\log {\sqrt {2\pi }}-{\frac {G}{\pi }}\right)}$
1. Beweis

${\displaystyle I:=\int _{1/4}^{3/4}\log \Gamma (x)\,dx=\int _{1/4}^{3/4}\log \Gamma (1-x)\,dx}$

${\displaystyle \Rightarrow \,2I=\int _{1/4}^{3/4}\log {\Big (}\Gamma (x)\Gamma (1-x){\Big )}\,dx=\int _{1/4}^{3/4}\log \left({\frac {\pi }{\sin \pi x}}\right)dx={\frac {1}{2}}\log \pi -\int _{1/4}^{3/4}\log(\sin \pi x)\,dx}$,

wobei ${\displaystyle \int _{1/4}^{3/4}\log(\sin \pi x)\,dx=\int _{-1/4}^{1/4}\log(\cos \pi x)\,dx=2\int _{0}^{1/4}\log(\cos \pi x)\,dx}$ ${\displaystyle ={\frac {1}{\pi }}\int _{0}^{\frac {\pi }{2}}\log \left(\cos {\frac {x}{2}}\right)dx={\frac {G}{\pi }}-{\frac {1}{2}}\log 2}$ ist.

Also ist ${\displaystyle 2I={\frac {1}{2}}\log(2\pi )-{\frac {G}{\pi }}}$.

2. Beweis
##### 1.1
${\displaystyle \int _{u}^{u+1}\log \Gamma (x)\,dx=u\,{\Big (}\log(u)-1{\Big )}+\log {\sqrt {2\pi }}\qquad u>0}$
Beweis (Raabesche Formel)

${\displaystyle \int _{u}^{u+1}\log \Gamma (x)\,dx=\int _{0}^{u+1}\log \Gamma (x)\,dx-\int _{0}^{u}\log \Gamma (x)\,dx}$

${\displaystyle =\int _{0}^{1}\log \Gamma (x)\,dx+\int _{1}^{u+1}\log \Gamma (x)\,dx-\int _{0}^{u}\log \Gamma (x)\,dx}$

${\displaystyle =\log {\sqrt {2\pi }}+\int _{0}^{u}\log \Gamma (x+1)\,dx-\int _{0}^{u}\log \Gamma (x)\,dx}$.

Wegen ${\displaystyle \log \Gamma (x+1)-\log \Gamma (x)=\log x\,}$ ist

${\displaystyle \int _{0}^{u}\log \Gamma (x+1)dx-\int _{0}^{u}\log \Gamma (x)dx=u\,{\Big (}\log(u)-1{\Big )}}$.