标签：Integrale frac log Form int dx pi Gamma

 

0.1Bearbeiten

{\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 \,}.

 

0.2Bearbeiten

{\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

Für {\displaystyle 0\leq x\leq 1} betrachte folgende Rechteck-Impuls-Funktion:

{\displaystyle 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.}

{\displaystyle \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

{\displaystyle {\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 }}}

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

 

1.1Bearbeiten

{\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 )}}.

标签：Integrale,frac,log,Form,int,dx,pi,Gamma
来源： https://www.cnblogs.com/Eufisky/p/14730823.html

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