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

Zurück zu Bestimmte Integrale

0.1

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

1. Beweis

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

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

2. Beweis

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

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

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

0.2

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

1. Beweis

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

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

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

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

2. Beweis

Für $0\leq x\leq 1$ betrachte folgende Rechteck-Impuls-Funktion:

$f(x)={\frac {4}{\pi }}\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}\,\cos {\Big (}(2k+1)\,2\pi x{\Big )}=\left\{{\begin{matrix}+1&&0\leq x<{\frac {1}{4}}\,\vee \,{\frac {3}{4}}<x\leq 1\\\\0&&x={\frac {1}{4}}\,\vee \,x={\frac {3}{4}}\\\\-1&&{\frac {1}{4}}<x<{\frac {3}{4}}\end{matrix}}\right.$

$\int _{0}^{1}\log \Gamma (x)\,f(x)\,dx={\frac {4}{\pi }}\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}\int _{0}^{1}\log \Gamma (x)\,\cos {\Big (}(2k+1)\,2\pi x{\Big )}dx={\frac {4}{\pi }}\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}\,{\frac {1}{4\,(2k+1)}}={\frac {G}{\pi }}$

Aus den Gleichungen

${\text{I.}}\,\quad \int _{0}^{1/4}\log \Gamma (x)\,dx+\int _{1/4}^{3/4}\log \Gamma (x)\,dx+\int _{3/4}^{1}\log \Gamma (x)\,dx=\log {\sqrt {2\pi }}$

${\text{II.}}\quad \int _{0}^{1/4}\log \Gamma (x)\,dx-\int _{1/4}^{3/4}\log \Gamma (x)\,dx+\int _{3/4}^{1}\log \Gamma (x)\,dx={\frac {G}{\pi }}$

folgt     $2\int _{1/4}^{3/4}\log \Gamma (x)\,dx=\log {\sqrt {2\pi }}-{\frac {G}{\pi }}$.

1.1

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

Beweis (Raabesche Formel)

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

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

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

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

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