Zurück zu Bestimmte Integrale
I := ∫ 0 1 log Γ ( x ) cos ( 2 n π x ) d x = ∫ 0 1 log Γ ( 1 − x ) cos ( 2 n π x ) d x {\displaystyle I:=\int _{0}^{1}\log \Gamma (x)\,\cos(2n\pi x)\,dx=\int _{0}^{1}\log \Gamma (1-x)\,\cos(2n\pi x)\,dx} ⇒ 2 I = ∫ 0 1 log ( Γ ( x ) Γ ( 1 − x ) ) cos ( 2 n π x ) d x = ∫ 0 1 log ( π sin π x ) cos ( 2 n π x ) d x {\displaystyle \Rightarrow \,2I=\int _{0}^{1}\log {\Big (}\Gamma (x)\Gamma (1-x){\Big )}\,\cos(2n\pi x)\,dx=\int _{0}^{1}\log \left({\frac {\pi }{\sin \pi x}}\right)\cos(2n\pi x)\,dx} log ( 2 π ) ∫ 0 1 cos ( 2 n π x ) d x ⏟ = 0 + ∫ 0 1 − log ( 2 sin π x ) cos ( 2 n π x ) d x {\displaystyle \log(2\pi )\underbrace {\int _{0}^{1}\cos(2n\pi x)\,dx} _{=0}+\int _{0}^{1}-\log {\big (}2\sin \pi x{\big )}\,\cos(2n\pi x)\,dx} , wobei − log ( 2 sin π x ) = ∑ k = 1 ∞ cos 2 k π x k {\displaystyle -\log {\big (}2\sin \pi x{\big )}=\sum _{k=1}^{\infty }{\frac {\cos 2k\pi x}{k}}} ist. ⇒ 2 I = ∑ k = 1 ∞ 1 k ∫ 0 1 cos ( 2 k π x ) cos ( 2 n π x ) d x ⏟ = 1 2 δ n k = 1 2 n {\displaystyle \Rightarrow 2I=\sum _{k=1}^{\infty }{\frac {1}{k}}\underbrace {\int _{0}^{1}\cos(2k\pi x)\,\cos(2n\pi x)\,dx} _{={\frac {1}{2}}\,\delta _{nk}}={\frac {1}{2n}}} .
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