Zurück zu Bestimmte Integrale
Multipliziere die Formel logΓ(x)=∫0∞(x−1tet−1−e−t(x−1)t(et−1))dt{\displaystyle \log \Gamma (x)=\int _{0}^{\infty }\left({\frac {x-1}{t\,e^{t}}}-{\frac {1-e^{-t(x-1)}}{t\,(e^{t}-1)}}\right)dt} mit sin(2nπx){\displaystyle \sin(2n\pi x)\,} durch und integriere nach x{\displaystyle x\,} von 0 bis 1 : In:=∫01logΓ(x)sin(2nπx)dx{\displaystyle I_{n}:=\int _{0}^{1}\log \Gamma (x)\,\sin(2n\pi x)\,dx} =∫01∫0∞(x−1tetsin(2nπx)−1−e−t(x−1)t(et−1)sin(2nπx))dtdx{\displaystyle =\int _{0}^{1}\int _{0}^{\infty }\left({\frac {x-1}{t\,e^{t}}}\,\sin(2n\pi x)-{\frac {1-e^{-t(x-1)}}{t\,(e^{t}-1)}}\,\sin(2n\pi x)\right)dt\,dx} Vertausche anschließend die Integrationsreihenfolge. Wegen ∫01sin(2nπx)dx=0,∫01xsin(2nπx)dx=−12nπ{\displaystyle \int _{0}^{1}\sin(2n\pi x)\,dx=0\,\,,\,\,\int _{0}^{1}x\,\sin(2n\pi x)\,dx=-{\frac {1}{2n\pi }}} und ∫01e−t(x−1)sin(2nπx)dx=2nπ(2nπ)2+t2(et−1){\displaystyle \int _{0}^{1}e^{-t(x-1)}\,\sin(2n\pi x)\,dx={\frac {2n\pi }{(2n\pi )^{2}+t^{2}}}\,(e^{t}-1)} ist daher In=∫0∞(−12nπ1tet+1t2nπ(2nπ)2+t2)dt{\displaystyle I_{n}=\int _{0}^{\infty }\left(-{\frac {1}{2n\pi }}\,{\frac {1}{t\,e^{t}}}+{\frac {1}{t}}\,{\frac {2n\pi }{(2n\pi )^{2}+t^{2}}}\right)dt} . 2nπIn=∫0∞((2nπ)2(2nπ)2+t2−1et)dtt=∫0∞(11+t2−1e2nπt)dtt{\displaystyle 2n\pi \,I_{n}=\int _{0}^{\infty }\left({\frac {(2n\pi )^{2}}{(2n\pi )^{2}+t^{2}}}-{\frac {1}{e^{t}}}\right)\,{\frac {dt}{t}}=\int _{0}^{\infty }\left({\frac {1}{1+t^{2}}}-{\frac {1}{e^{2n\pi t}}}\right)\,{\frac {dt}{t}}} Und das ist γ+log(2nπ){\displaystyle \gamma +\log(2n\pi )\,} wegen ∫0∞(tz−11+t2−tz−1e2nπt)dt=π21sinπz2−Γ(z)(2nπ)z{\displaystyle \int _{0}^{\infty }\left({\frac {t^{z-1}}{1+t^{2}}}-{\frac {t^{z-1}}{e^{2n\pi t}}}\right)\,dt={\frac {\pi }{2}}\,{\frac {1}{\sin {\frac {\pi z}{2}}}}-{\frac {\Gamma (z)}{(2n\pi )^{z}}}} =(π21sinπz2−1z)+1(2nπ)z(1z(2nπ)z⏟1z+log(2nπ)+O(z)−Γ(z))→γ+log(2nπ){\displaystyle =\left({\frac {\pi }{2}}\,{\frac {1}{\sin {\frac {\pi z}{2}}}}-{\frac {1}{z}}\right)+{\frac {1}{(2n\pi )^{z}}}\,\left(\underbrace {{\frac {1}{z}}\,(2n\pi )^{z}} _{{\frac {1}{z}}+\log(2n\pi )+O(z)}-\Gamma (z)\right)\to \gamma +\log(2n\pi )} für z→0+{\displaystyle z\to 0^{+}\,} .