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

Zurück zu Bestimmte Integrale

1.1

$\int _{0}^{1}\log \Gamma (x)\,\sin(2n\pi x)\,dx={\frac {\gamma +\log(2n\pi )}{2n\pi }}\qquad n\in \mathbb {Z} ^{>0}$

Beweis

Multipliziere die Formel $\log \Gamma (x)=\int _{0}^{\infty }\left({\frac {x-1}{t\,e^{t}}}-{\frac {1-e^{-t(x-1)}}{t\,(e^{t}-1)}}\right)dt$

mit $\sin(2n\pi x)\,$ durch und integriere nach $x\,$ von 0 bis 1 :

$I_{n}:=\int _{0}^{1}\log \Gamma (x)\,\sin(2n\pi x)\,dx$

$=\int _{0}^{1}\int _{0}^{\infty }\left({\frac {x-1}{t\,e^{t}}}\,\sin(2n\pi x)-{\frac {1-e^{-t(x-1)}}{t\,(e^{t}-1)}}\,\sin(2n\pi x)\right)dt\,dx$

Vertausche anschließend die Integrationsreihenfolge.

Wegen $\int _{0}^{1}\sin(2n\pi x)\,dx=0\,\,,\,\,\int _{0}^{1}x\,\sin(2n\pi x)\,dx=-{\frac {1}{2n\pi }}$

und $\int _{0}^{1}e^{-t(x-1)}\,\sin(2n\pi x)\,dx={\frac {2n\pi }{(2n\pi )^{2}+t^{2}}}\,(e^{t}-1)$

ist daher $I_{n}=\int _{0}^{\infty }\left(-{\frac {1}{2n\pi }}\,{\frac {1}{t\,e^{t}}}+{\frac {1}{t}}\,{\frac {2n\pi }{(2n\pi )^{2}+t^{2}}}\right)dt$.

$2n\pi \,I_{n}=\int _{0}^{\infty }\left({\frac {(2n\pi )^{2}}{(2n\pi )^{2}+t^{2}}}-{\frac {1}{e^{t}}}\right)\,{\frac {dt}{t}}=\int _{0}^{\infty }\left({\frac {1}{1+t^{2}}}-{\frac {1}{e^{2n\pi t}}}\right)\,{\frac {dt}{t}}$

Und das ist $\gamma +\log(2n\pi )\,$ wegen $\int _{0}^{\infty }\left({\frac {t^{z-1}}{1+t^{2}}}-{\frac {t^{z-1}}{e^{2n\pi t}}}\right)\,dt={\frac {\pi }{2}}\,{\frac {1}{\sin {\frac {\pi z}{2}}}}-{\frac {\Gamma (z)}{(2n\pi )^{z}}}$

$=\left({\frac {\pi }{2}}\,{\frac {1}{\sin {\frac {\pi z}{2}}}}-{\frac {1}{z}}\right)+{\frac {1}{(2n\pi )^{z}}}\,\left(\underbrace {{\frac {1}{z}}\,(2n\pi )^{z}} _{{\frac {1}{z}}+\log(2n\pi )+O(z)}-\Gamma (z)\right)\to \gamma +\log(2n\pi )$ für $z\to 0^{+}\,$.