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

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1.1
${\displaystyle \int _{0}^{\infty }{\frac {x}{\sqrt {\alpha ^{2}+x^{2}}}}\,J_{0}(x)\,dx=e^{-\alpha }\qquad {\text{Re}}(\alpha )\geq 0}$
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 ${\displaystyle y(\alpha )=e^{-\alpha }}$.

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

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

Verwende die Poissonsche Darstellung

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

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

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

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

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

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

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