3.1Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }(\alpha -ix)^{n}\,\Gamma (\beta +ix)\,dx={\frac {2\pi }{e}}\sum _{k=0}^{n}{n \choose k}\,(\alpha +\beta )^{k}\,\phi _{n-k}(-1)}

ohne Beweis

 

 

4.1Bearbeiten

{\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}

Beweis (Additionstheorem für die Gammafunktion)

Für {\displaystyle k=1,2\,} sei {\displaystyle u_{k}(t)={\frac {\Gamma (\alpha _{k}+it)}{\beta _{k}^{\alpha _{k}+it}}}} und {\displaystyle f_{k}(z)={\frac {\Gamma (z)}{(\beta _{k}\,e^{\omega })^{z}}}} mit {\displaystyle \omega \in \mathbb {R} }.

Berechne die Fouriertransformierte {\displaystyle {\mathcal {F}}[u_{k}](\omega )=\int _{-\infty }^{\infty }{\frac {\Gamma (\alpha _{k}+it)}{\beta _{k}^{\alpha _{k}+it}}}\,e^{-i\omega t}\,dt}

{\displaystyle =-i\,e^{\omega \alpha _{k}}\int _{-\infty }^{\infty }{\frac {\Gamma (\alpha _{k}+it)}{(\beta _{k}\,e^{\omega })^{\alpha _{k}+it}}}\,i\,dt=-i\,e^{\omega \alpha _{k}}\int _{\alpha _{k}+i\mathbb {R} }f_{k}\,dz}

{\displaystyle =-i\,e^{\omega \alpha _{k}}\,\lim _{N\to \infty }\oint _{\gamma _{N}}f_{k}\,dz=2\pi \,e^{\omega \alpha _{k}}\,\sum _{n=0}^{\infty }{\text{res}}(f_{k},-n)}

{\displaystyle =2\pi \,e^{\omega \alpha _{k}}\,\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\,\left(\beta _{k}\,e^{\omega }\right)^{n}=2\pi \,{\frac {e^{\omega \alpha _{k}}}{e^{\beta _{k}\,e^{\omega }}}}}.

Also ist

{\displaystyle \int _{-\infty }^{\infty }{\frac {\Gamma (\alpha _{1}-it)}{\beta _{1}^{\alpha _{1}-it}}}\,{\frac {\Gamma (\alpha _{2}+it)}{\beta _{2}^{\alpha _{2}+it}}}\,dt=\int _{-\infty }^{\infty }u_{1}(0-t)\,u_{2}(t)\,dt=(u_{1}*u_{2})(0)}

nach der Faltungsformel {\displaystyle {\frac {1}{2\pi }}\int _{-\infty }^{\infty }{\mathcal {F}}[u_{1}](\omega )\cdot {\mathcal {F}}[u_{2}](\omega )\,d\omega =2\pi \int _{-\infty }^{\infty }{\frac {e^{\omega \alpha _{1}}}{e^{\beta _{1}e^{\omega }}}}\,{\frac {e^{\omega \alpha _{2}}}{e^{\beta _{2}e^{\omega }}}}\,d\omega }.

Und das ist nach Substitution {\displaystyle e^{\omega }=t\,} gleich {\displaystyle 2\pi \int _{0}^{\infty }{\frac {t^{\alpha _{1}}}{e^{\beta _{1}t}}}\,{\frac {t^{\alpha _{2}}}{e^{\beta _{2}t}}}\,{\frac {dt}{t}}=2\pi \int _{0}^{\infty }t^{\alpha _{1}+\alpha _{2}-1}\,e^{-(\beta _{1}+\beta _{2})t}\,dt=2\pi {\frac {\Gamma (\alpha _{1}+\alpha _{2})}{(\beta _{1}+\beta _{2})^{\alpha _{1}+\alpha _{2}}}}}.

 

4.2Bearbeiten

{\displaystyle {\frac {1}{2\pi i}}\int _{-i\infty }^{i\infty }{\frac {\Gamma (a+x)\,\Gamma (b-x)}{\Gamma (c+x)\,\Gamma (d-x)}}\,dx={\frac {\Gamma (a+b)\,\Gamma (c+d-a-b-1)}{\Gamma (c+d-1)\,\Gamma (c-a)\,\Gamma (d-b)}}}

ohne Beweis

 

 

4.3Bearbeiten

Sind {\displaystyle a,b,c,d\,} komplexe Zahlen und ist {\displaystyle \gamma \,} eine Kurve, welche die Polstellen

 

{\displaystyle (-a-n)_{n\geq 0}} und {\displaystyle (-b-n)_{n\geq 0}} von den Polstellen {\displaystyle (c-n)_{n\geq 0}} und {\displaystyle (b-n)_{n\geq 0}} trennt, so gilt

 

{\displaystyle {\frac {1}{2\pi i}}\int _{\gamma }\Gamma (a+z)\,\Gamma (b+z)\,\Gamma (c-z)\,\Gamma (d-z)\,dz={\frac {\Gamma (a+c)\,\Gamma (a+d)\,\Gamma (b+c)\,\Gamma (b+d)}{\Gamma (a+b+c+d)}}}

ohne Beweis (Lemma von Barnes)

 

 

4.4Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{\Gamma (a+x)\,\Gamma (b+x)\,\Gamma (c-x)\,\Gamma (d-x)}}={\frac {\Gamma (a+b+c+d-3)}{\Gamma (a+c-1)\,\Gamma (a+d-1)\,\Gamma (b+c-1)\,\Gamma (b+d-1)}}\qquad {\text{Re}}(a+b+c+d)>3}

ohne Beweis (Ramanujans Beta Integral)