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

3.1Bearbeiten

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

ohne Beweis

 

4.1Bearbeiten

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

Beweis (Additionstheorem für die Gammafunktion)

Für {displaystyle k=1,2\,} sei {displaystyle u_{k}(t)={frac {Gamma (alpha _{k}+it)}{eta _{k}^{alpha _{k}+it}}}} und {displaystyle f_{k}(z)={frac {Gamma (z)}{(eta _{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)}{eta _{k}^{alpha _{k}+it}}}\,e^{-iomega t}\,dt}

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

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

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

Also ist

{displaystyle int _{-infty }^{infty }{frac {Gamma (alpha _{1}-it)}{eta _{1}^{alpha _{1}-it}}}\,{frac {Gamma (alpha _{2}+it)}{eta _{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}{2pi }}int _{-infty }^{infty }{mathcal {F}}[u_{1}](omega )cdot {mathcal {F}}[u_{2}](omega )\,domega =2pi int _{-infty }^{infty }{frac {e^{omega alpha _{1}}}{e^{eta _{1}e^{omega }}}}\,{frac {e^{omega alpha _{2}}}{e^{eta _{2}e^{omega }}}}\,domega }.

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

 

4.2Bearbeiten

{displaystyle {frac {1}{2pi i}}int _{-iinfty }^{iinfty }{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)_{ngeq 0}} und {displaystyle (-b-n)_{ngeq 0}} von den Polstellen {displaystyle (c-n)_{ngeq 0}} und {displaystyle (b-n)_{ngeq 0}} trennt, so gilt

{displaystyle {frac {1}{2pi 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 { ext{Re}}(a+b+c+d)>3}

ohne Beweis (Ramanujans Beta Integral)