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

 

2.1Bearbeiten

{\displaystyle {\frac {1}{2\pi i}}\int _{a-i\infty }^{a+i\infty }\Gamma (s)\,t^{-s}\,ds=e^{-t}\qquad a>0\,,\,{\text{Re}}(t)>0}

Beweis (Cahen-Mellin Integral)

Diese Formel ergibt sich aus der Mellin-Rücktransformation.

Aus {\displaystyle \Gamma (s)={\mathcal {M}}{\big [}e^{-t}{\big ]}(s)=\int _{0}^{\infty }e^{-t}\,t^{s-1}\,dt}

folgt {\displaystyle e^{-t}={\mathcal {M}}^{-1}[\Gamma (s)](t)={\frac {1}{2\pi i}}\int _{a-i\infty }^{a+i\infty }\Gamma (s)\,t^{-s}\,ds}.

 

2.2Bearbeiten

{\displaystyle {\frac {1}{2\pi i}}\int _{-i\infty }^{i\infty }e^{2bix}\,|\Gamma (\alpha +x)|^{2}\,dx={\frac {\Gamma (2\alpha )}{(2\cos b)^{2\alpha }}}\qquad {\text{Re}}(\alpha )>0\,,\,\left|{\text{Re}}(b)\right|<{\frac {\pi }{2}}}

Beweis

In der Formel

{\displaystyle {\frac {1}{2\pi i}}\int _{-i\infty }^{i\infty }{\frac {\Gamma (\alpha _{1}+x)}{\beta _{1}^{\alpha _{1}+x}}}\,{\frac {\Gamma (\alpha _{2}-x)}{\beta _{2}^{\alpha _{2}-x}}}\,dx={\frac {\Gamma (\alpha _{1}+\alpha _{2})}{(\beta _{1}+\beta _{2})^{\alpha _{1}+\alpha _{2}}}}\qquad {\text{Re}}(\alpha _{1}),{\text{Re}}(\alpha _{2}),{\text{Re}}(\beta _{1}),{\text{Re}}(\beta _{2})>0}

setze {\displaystyle \alpha _{1}=\alpha _{2}=\alpha \,,\,\beta _{1}=e^{ib}\,} und {\displaystyle \beta _{2}=e^{-ib}\,}.

 

3.1Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itx}}{\Gamma (\mu +x)\Gamma (\nu -x)}}\,dx=\left\{{\begin{matrix}{\frac {\left(2\cos {\frac {t}{2}}\right)^{\mu +\nu -2}}{\Gamma (\mu +\nu -1)}}\,e^{-i(\mu -\nu )t/2}&-\pi <t<\pi \\0&{\text{sonst}}\end{matrix}}\right.\qquad {\text{Re}}(\mu +\nu )>1}

Beweis

Setze {\displaystyle f(t)=\left\{{\begin{matrix}{\frac {\left(2\cos {\frac {t}{2}}\right)^{\mu +\nu -2}}{\Gamma (\mu +\nu -1)}}\,e^{-i(\mu -\nu )t/2}&-\pi <t<\pi \\0&{\text{sonst}}\end{matrix}}\right.}

und berechne davon die Fouriertransformierte {\displaystyle {\hat {f}}(x)={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }f(t)e^{-ixt}\,dt}.

Das ist {\displaystyle {\frac {1}{\sqrt {2\pi }}}\int _{-\pi }^{\pi }{\frac {\left(2\cos {\frac {t}{2}}\right)^{\mu +\nu -2}}{\Gamma (\mu +\nu -1)}}\,e^{-i(\mu -\nu )t/2}\,e^{-ixt}\,dt}, da {\displaystyle f(t)\,} für {\displaystyle |t|\geq \pi } verschwindet.

Und das ist {\displaystyle {\frac {2}{\sqrt {2\pi }}}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}{\frac {\left(2\cos t\right)^{\mu +\nu -2}}{\Gamma (\mu +\nu -1)}}\,e^{-i(\mu -\nu +2x)t}\,dt} nach der Substitution {\displaystyle t\mapsto 2t}.

Der ungerade Anteil hebt sich auf; somit ist {\displaystyle {\hat {f}}(x)={\frac {2}{\sqrt {2\pi }}}\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}{\frac {\left(2\cos t\right)^{\mu +\nu -2}}{\Gamma (\mu +\nu -1)}}\,\cos(\mu -\nu +2x)t\,dt},

was sich aufgrund der Symmetrie auch als {\displaystyle {\frac {1}{\sqrt {2\pi }}}\,{\frac {2^{\mu +\nu }}{\Gamma (\mu +\nu -1)}}\,\int _{0}^{\frac {\pi }{2}}\left(\cos t\right)^{\mu +\nu -2}\,\cos(\mu -\nu +2x)t\,dt} schreiben lässt.

Nach der Cauchyschen Cosinus-Integralformel {\displaystyle \int _{0}^{\frac {\pi }{2}}(\cos t)^{\alpha -1}\,\cos \beta t\,dt={\frac {\pi }{2^{\alpha }}}\,{\frac {\Gamma (\alpha )}{\Gamma \left({\frac {\alpha +\beta -1}{2}}\right)\,\Gamma \left({\frac {\alpha +\beta -1}{2}}\right)}}} ist nun

{\displaystyle {\hat {f}}(x)={\frac {1}{\sqrt {2\pi }}}\,{\frac {2^{\mu +\nu }}{\Gamma (\mu +\nu -1)}}\,{\frac {\pi }{2^{\mu +\nu -1}}}\,{\frac {\Gamma (\mu +\nu -1)}{\Gamma \left({\frac {(\mu +\nu -1)+(\mu -\nu +2x)+1}{2}}\right)\,\Gamma \left({\frac {(\mu +\nu -1)-(\mu -\nu +2x)+1}{2}}\right)}}={\frac {\sqrt {2\pi }}{\Gamma (\mu +x)\,\Gamma (\nu -x)}}}.

Die behauptete Gleichung ist dann die Rücktransformation {\displaystyle {\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }{\hat {f}}(x)\,e^{itx}\,dx=f(t)}.