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

2天前 3次浏览

##### 1.1Bearbeiten
{displaystyle int _{0}^{infty }{frac {x}{sqrt {alpha ^{2}+x^{2}}}},J_{0}(x),dx=e^{-alpha }qquad {text{Re}}(alpha )geq 0}
Beweis

{displaystyle y(alpha ):=int _{0}^{infty }{frac {x}{sqrt {alpha ^{2}+x^{2}}}},J_{0}(x),dxqquad left(Rightarrow ,y(0)=int _{0}^{infty }J_{0}(x),dx=1right)}

{displaystyle y'(alpha )=int _{0}^{infty }{frac {-alpha x}{{sqrt {alpha ^{2}+x^{2}}}^{,3}}},J_{0}(x),dx=underbrace {left[{frac {alpha }{sqrt {alpha ^{2}+x^{2}}}},J_{0}(x)right]_{0}^{infty }} _{=-1}-int _{0}^{infty }{frac {alpha }{sqrt {alpha ^{2}+x^{2}}}},J_{0}'(x),dxqquad {Big (}Rightarrow ,y'(0)=-1{Big )}}

{displaystyle y”(alpha )=int _{0}^{infty }left({frac {alpha ^{2}}{{sqrt {alpha ^{2}+x^{2}}}^{,3}}}-{frac {1}{sqrt {alpha ^{2}+x^{2}}}}right)J_{0}'(x),dx=int _{0}^{infty }{frac {alpha ^{2}}{{sqrt {alpha ^{2}+x^{2}}}^{,3}}},J_{0}'(x),dx-int _{0}^{infty }{frac {1}{sqrt {alpha ^{2}+x^{2}}}},J_{0}'(x),dx}

{displaystyle =underbrace {left[{frac {x}{sqrt {alpha ^{2}+x^{2}}}},J_{0}'(x)right]_{0}^{infty }} _{=0}-int _{0}^{infty }{frac {x}{sqrt {alpha ^{2}+x^{2}}}},J_{0}”(x),dx-int _{0}^{infty }{frac {1}{sqrt {alpha ^{2}+x^{2}}}},J_{0}'(x),dx}

Nachdem {displaystyle J_{0}(x)} die Differenzialgleichung {displaystyle x^{2},J_{0}”(x)+xJ_{0}'(x)+x^{2},J_{0}(x)=0} löst, ist {displaystyle xJ_{0}”(x)+J_{0}'(x)=-x,J_{0}(x)}.

Und daher ist {displaystyle y”(alpha )=int _{0}^{infty }{frac {x}{sqrt {alpha ^{2}+x^{2}}}},J_{0}(x),dx=y(alpha ),,Rightarrow ,y(alpha )=C_{1},e^{alpha }+C_{2},e^{-alpha }}.

Wegen {displaystyle y(0)=1} und {displaystyle y'(0)=-1} ist {displaystyle C_{1}=0} und {displaystyle C_{2}=1}; also {displaystyle y(alpha )=e^{-alpha }}.

##### 1.2Bearbeiten
{displaystyle int _{0}^{infty }{frac {alpha }{sqrt {alpha ^{2}+x^{2}}}},J_{1}(x),dx=1-e^{-alpha }qquad {text{Re}}(alpha )geq 0}
Beweis

Betrachte folgende Formel:

{displaystyle int _{0}^{infty }{frac {x}{sqrt {alpha ^{2}+x^{2}}}},J_{0}(x),dx=e^{-alpha }}

Differenziere nach {displaystyle alpha ,}:

{displaystyle int _{0}^{infty }{frac {-alpha x}{{sqrt {alpha ^{2}+x^{2}}}^{,3}}},J_{0}(x),dx=-e^{-alpha }}

Das Integral ist nach partieller Integration

{displaystyle left[{frac {alpha }{sqrt {alpha ^{2}+x^{2}}}},J_{0}(x)right]_{0}^{infty }-int _{0}^{infty }{frac {alpha }{sqrt {alpha ^{2}+x^{2}}}},J_{0}'(x),dx}, wobei {displaystyle J_{0}'(x)=-J_{1}(x)} ist.

Also ist {displaystyle -1+int _{0}^{infty }{frac {alpha }{sqrt {alpha ^{2}+x^{2}}}},J_{1}(x),dx=-e^{-alpha }}

##### 2.1Bearbeiten
{displaystyle int _{0}^{infty }J_{nu }(x),x^{s-1},dx={frac {2^{s-1},Gamma left({frac {nu +s}{2}}right)}{Gamma left(1+{frac {nu -s}{2}}right)}}qquad -{text{Re}}(nu )<{text{Re}}(s)<{frac {3}{2}}}
Beweis

Verwende die Poissonsche Darstellung

{displaystyle J_{nu }(x)={frac {2}{Gamma left({frac {1}{2}}right)Gamma left(nu +{frac {1}{2}}right)}}left({frac {x}{2}}right)^{nu }int _{0}^{frac {pi }{2}}cos(xcos t),sin ^{2nu }t,dtqquad {text{Re}}(nu )>-{frac {1}{2}}}

{displaystyle int _{0}^{infty }J_{nu }(x),x^{s-1},dx={frac {2}{2^{nu },Gamma left({frac {1}{2}}right)Gamma left(nu +{frac {1}{2}}right)}}int _{0}^{frac {pi }{2}}int _{0}^{infty }x^{nu +s-1},cos(xcos t),dx,sin ^{2nu }t,dt}

{displaystyle ={frac {2}{2^{nu },Gamma left({frac {1}{2}}right)Gamma left(nu +{frac {1}{2}}right)}}int _{0}^{frac {pi }{2}}{frac {Gamma (nu +s)}{(cos t)^{nu +s}}},cos {frac {(nu +s)pi }{2}},sin ^{2nu }t,dt}

{displaystyle ={frac {Gamma (nu +s),cos {frac {(nu +s)pi }{2}}}{2^{nu },{sqrt {pi }},Gamma left(nu +{frac {1}{2}}right)}}cdot 2int _{0}^{frac {pi }{2}}(sin t)^{2left(nu +{frac {1}{2}}right)-1},(cos t)^{2left({frac {1}{2}}-{frac {nu +s}{2}}right)-1},dt}

{displaystyle ={frac {Gamma (nu +s),{frac {pi }{Gamma left({frac {1}{2}}+{frac {nu +s}{2}}right),Gamma left({frac {1}{2}}-{frac {nu +s}{2}}right)}}}{2^{nu },{sqrt {pi }},Gamma left(nu +{frac {1}{2}}right)}}cdot {frac {Gamma left(nu +{frac {1}{2}}right),Gamma left({frac {1}{2}}-{frac {nu +s}{2}}right)}{Gamma left(1+{frac {nu -s}{2}}right)}}}

{displaystyle ={frac {sqrt {pi }}{2^{nu }}},{frac {Gamma (nu +s)}{Gamma left({frac {1}{2}}+{frac {nu +s}{2}}right)}}cdot {frac {1}{Gamma left(1+{frac {nu -s}{2}}right)}}={frac {2^{s-1},Gamma left({frac {nu +s}{2}}right)}{Gamma left(1+{frac {nu -s}{2}}right)}}}

##### 3.1Bearbeiten
{displaystyle int _{0}^{infty }J_{mu }(x),J_{nu }(x),x^{s-1},dx=2^{s-1},{frac {Gamma (1-s)cdot Gamma left({frac {mu +nu +s}{2}}right)}{Gamma left(1+{frac {mu -nu -s}{2}}right)cdot Gamma left(1+{frac {nu -mu -s}{2}}right)cdot Gamma left(1+{frac {mu +nu -s}{2}}right)}}}
ohne Beweis