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 { ext{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=1 ight)}

{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) ight]_{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}}}} ight)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) ight]_{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 { ext{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) ight]_{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_{ u }(x)\,x^{s-1}\,dx={frac {2^{s-1}\,Gamma left({frac { u +s}{2}} ight)}{Gamma left(1+{frac { u -s}{2}} ight)}}qquad -{ ext{Re}}( u )<{ ext{Re}}(s)<{frac {3}{2}}}

Beweis

Verwende die Poissonsche Darstellung

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


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

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

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

{displaystyle ={frac {Gamma ( u +s)\,{frac {pi }{Gamma left({frac {1}{2}}+{frac { u +s}{2}} ight)\,Gamma left({frac {1}{2}}-{frac { u +s}{2}} ight)}}}{2^{ u }\,{sqrt {pi }}\,Gamma left( u +{frac {1}{2}} ight)}}cdot {frac {Gamma left( u +{frac {1}{2}} ight)\,Gamma left({frac {1}{2}}-{frac { u +s}{2}} ight)}{Gamma left(1+{frac { u -s}{2}} ight)}}}

{displaystyle ={frac {sqrt {pi }}{2^{ u }}}\,{frac {Gamma ( u +s)}{Gamma left({frac {1}{2}}+{frac { u +s}{2}} ight)}}cdot {frac {1}{Gamma left(1+{frac { u -s}{2}} ight)}}={frac {2^{s-1}\,Gamma left({frac { u +s}{2}} ight)}{Gamma left(1+{frac { u -s}{2}} ight)}}}

 

3.1Bearbeiten

{displaystyle int _{0}^{infty }J_{mu }(x)\,J_{ u }(x)\,x^{s-1}\,dx=2^{s-1}\,{frac {Gamma (1-s)cdot Gamma left({frac {mu + u +s}{2}} ight)}{Gamma left(1+{frac {mu - u -s}{2}} ight)cdot Gamma left(1+{frac { u -mu -s}{2}} ight)cdot Gamma left(1+{frac {mu + u -s}{2}} ight)}}}

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