Zurück zu Bestimmte Integrale
2 ∫ 0 1 log Γ ( x ) d x = ∫ 0 1 log Γ ( x ) d x + ∫ 0 1 log Γ ( 1 − x ) d x = ∫ 0 1 log ( Γ ( x ) Γ ( 1 − x ) ) d x {\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} = ∫ 0 1 log ( π sin π x ) d x = log π − ∫ 0 1 log sin π x d x = log π + log 2 ⇒ ∫ 0 1 log Γ ( x ) d x = 1 2 log ( 2 π ) {\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 )}
Die Riemannsche Approximationssumme ∑ k = 1 n − 1 log Γ ( k n ) ⋅ 1 n {\displaystyle \sum _{k=1}^{n-1}\log \Gamma \!\left({\frac {k}{n}}\right)\cdot {\frac {1}{n}}} vereinfacht sich zu log ( ∏ k = 1 n − 1 Γ ( k n ) ) ⋅ 1 n = log ( 2 π n − 1 n ) ⋅ 1 n = ( n − 1 ) log 2 π − log n n {\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 log 2 π {\displaystyle \log {\sqrt {2\pi }}} für n → ∞ {\displaystyle n\to \infty \,} .
I := ∫ 1 / 4 3 / 4 log Γ ( x ) d x = ∫ 1 / 4 3 / 4 log Γ ( 1 − x ) d x {\displaystyle I:=\int _{1/4}^{3/4}\log \Gamma (x)\,dx=\int _{1/4}^{3/4}\log \Gamma (1-x)\,dx} ⇒ 2 I = ∫ 1 / 4 3 / 4 log ( Γ ( x ) Γ ( 1 − x ) ) d x = ∫ 1 / 4 3 / 4 log ( π sin π x ) d x = 1 2 log π − ∫ 1 / 4 3 / 4 log ( sin π x ) d x {\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 ∫ 1 / 4 3 / 4 log ( sin π x ) d x = ∫ − 1 / 4 1 / 4 log ( cos π x ) d x = 2 ∫ 0 1 / 4 log ( cos π x ) d x {\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} = 1 π ∫ 0 π 2 log ( cos x 2 ) d x = G π − 1 2 log 2 {\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 2 I = 1 2 log ( 2 π ) − G π {\displaystyle 2I={\frac {1}{2}}\log(2\pi )-{\frac {G}{\pi }}} .
Für 0 ≤ x ≤ 1 {\displaystyle 0\leq x\leq 1} betrachte folgende Rechteck-Impuls-Funktion: f ( x ) = 4 π ∑ k = 0 ∞ ( − 1 ) k 2 k + 1 cos ( ( 2 k + 1 ) 2 π x ) = { + 1 0 ≤ x < 1 4 ∨ 3 4 < x ≤ 1 0 x = 1 4 ∨ x = 3 4 − 1 1 4 < x < 3 4 {\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.} ∫ 0 1 log Γ ( x ) f ( x ) d x = 4 π ∑ k = 0 ∞ ( − 1 ) k 2 k + 1 ∫ 0 1 log Γ ( x ) cos ( ( 2 k + 1 ) 2 π x ) d x = 4 π ∑ k = 0 ∞ ( − 1 ) k 2 k + 1 1 4 ( 2 k + 1 ) = G π {\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 I. ∫ 0 1 / 4 log Γ ( x ) d x + ∫ 1 / 4 3 / 4 log Γ ( x ) d x + ∫ 3 / 4 1 log Γ ( x ) d x = log 2 π {\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 }}} II. ∫ 0 1 / 4 log Γ ( x ) d x − ∫ 1 / 4 3 / 4 log Γ ( x ) d x + ∫ 3 / 4 1 log Γ ( x ) d x = G π {\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 2 ∫ 1 / 4 3 / 4 log Γ ( x ) d x = log 2 π − G π {\displaystyle 2\int _{1/4}^{3/4}\log \Gamma (x)\,dx=\log {\sqrt {2\pi }}-{\frac {G}{\pi }}} .
∫ u u + 1 log Γ ( x ) d x = ∫ 0 u + 1 log Γ ( x ) d x − ∫ 0 u log Γ ( x ) d x {\displaystyle \int _{u}^{u+1}\log \Gamma (x)\,dx=\int _{0}^{u+1}\log \Gamma (x)\,dx-\int _{0}^{u}\log \Gamma (x)\,dx} = ∫ 0 1 log Γ ( x ) d x + ∫ 1 u + 1 log Γ ( x ) d x − ∫ 0 u log Γ ( x ) d x {\displaystyle =\int _{0}^{1}\log \Gamma (x)\,dx+\int _{1}^{u+1}\log \Gamma (x)\,dx-\int _{0}^{u}\log \Gamma (x)\,dx} = log 2 π + ∫ 0 u log Γ ( x + 1 ) d x − ∫ 0 u log Γ ( x ) d x {\displaystyle =\log {\sqrt {2\pi }}+\int _{0}^{u}\log \Gamma (x+1)\,dx-\int _{0}^{u}\log \Gamma (x)\,dx} . Wegen log Γ ( x + 1 ) − log Γ ( x ) = log x {\displaystyle \log \Gamma (x+1)-\log \Gamma (x)=\log x\,} ist ∫ 0 u log Γ ( x + 1 ) d x − ∫ 0 u log Γ ( x ) d x = u ( log ( u ) − 1 ) {\displaystyle \int _{0}^{u}\log \Gamma (x+1)dx-\int _{0}^{u}\log \Gamma (x)dx=u\,{\Big (}\log(u)-1{\Big )}} .