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

 

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)\,dx\qquad \left(\Rightarrow \,y(0)=\int _{0}^{\infty }J_{0}(x)\,dx=1\right)}

{\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)\,dx\qquad {\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(x\cos t)\,\sin ^{2\nu }t\,dt\qquad {\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(x\cos t)\,dx\,\sin ^{2\nu }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 ^{2\nu }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 2\int _{0}^{\frac {\pi }{2}}(\sin t)^{2\left(\nu +{\frac {1}{2}}\right)-1}\,(\cos t)^{2\left({\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