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

1周前 (05-04) 5次浏览

##### 2.1Bearbeiten
{displaystyle {frac {1}{2pi i}}int _{a-iinfty }^{a+iinfty }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}{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 {text{Re}}(alpha )>0,,,left|{text{Re}}(b)right|<{frac {pi }{2}}}
Beweis

In der Formel

{displaystyle {frac {1}{2pi i}}int _{-iinfty }^{iinfty }{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(2cos {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(2cos {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 {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}}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 {2pi }}}int _{-{frac {pi }{2}}}^{frac {pi }{2}}{frac {left(2cos tright)^{mu +nu -2}}{Gamma (mu +nu -1)}},e^{-i(mu -nu +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 tright)^{mu +nu -2}}{Gamma (mu +nu -1)}},cos(mu -nu +2x)t,dt},

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

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