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

2.1Bearbeiten

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

Beweis (Cahen-Mellin Integral)

Diese Formel ergibt sich aus der Mellin-Rücktransformation.

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

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

 

2.2Bearbeiten

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

Beweis

In der Formel

{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}

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

 

3.1Bearbeiten

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

Beweis

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

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

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

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

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

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

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

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

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