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}
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}={frac {1}{2pi i}}int _{a-iinfty }^{a+iinfty }Gamma (s)\,t^{-s}\,ds}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45c1918cc77f0d58991987fa7152ef0959d820fb)
![{displaystyle Gamma (s)={mathcal {M}}{ig [}e^{-t}{ig ]}(s)=int _{0}^{infty }e^{-t}\,t^{s-1}\,dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a063ef39a165adfc010999fb6303367b2b2db263)
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}}}
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}\,}
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}
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}
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}
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}
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)}}}
{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)}
