Zurück zu Bestimmte Integrale
y ( α ) := ∫ 0 ∞ x α 2 + x 2 J 0 ( x ) d x ( ⇒ y ( 0 ) = ∫ 0 ∞ J 0 ( x ) d x = 1 ) {\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)} y ′ ( α ) = ∫ 0 ∞ − α x α 2 + x 2 3 J 0 ( x ) d x = [ α α 2 + x 2 J 0 ( x ) ] 0 ∞ ⏟ = − 1 − ∫ 0 ∞ α α 2 + x 2 J 0 ′ ( x ) d x ( ⇒ y ′ ( 0 ) = − 1 ) {\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 )}} y ″ ( α ) = ∫ 0 ∞ ( α 2 α 2 + x 2 3 − 1 α 2 + x 2 ) J 0 ′ ( x ) d x = ∫ 0 ∞ α 2 α 2 + x 2 3 J 0 ′ ( x ) d x − ∫ 0 ∞ 1 α 2 + x 2 J 0 ′ ( x ) d x {\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} = [ x α 2 + x 2 J 0 ′ ( x ) ] 0 ∞ ⏟ = 0 − ∫ 0 ∞ x α 2 + x 2 J 0 ″ ( x ) d x − ∫ 0 ∞ 1 α 2 + x 2 J 0 ′ ( x ) d x {\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 J 0 ( x ) {\displaystyle J_{0}(x)} die Differenzialgleichung x 2 J 0 ″ ( x ) + x J 0 ′ ( x ) + x 2 J 0 ( x ) = 0 {\displaystyle x^{2}\,J_{0}''(x)+xJ_{0}'(x)+x^{2}\,J_{0}(x)=0} löst, ist x J 0 ″ ( x ) + J 0 ′ ( x ) = − x J 0 ( x ) {\displaystyle xJ_{0}''(x)+J_{0}'(x)=-x\,J_{0}(x)} . Und daher ist y ″ ( α ) = ∫ 0 ∞ x α 2 + x 2 J 0 ( x ) d x = y ( α ) ⇒ y ( α ) = C 1 e α + C 2 e − α {\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 y ( 0 ) = 1 {\displaystyle y(0)=1} und y ′ ( 0 ) = − 1 {\displaystyle y'(0)=-1} ist C 1 = 0 {\displaystyle C_{1}=0} und C 2 = 1 {\displaystyle C_{2}=1} ; also y ( α ) = e − α {\displaystyle y(\alpha )=e^{-\alpha }} .
Betrachte folgende Formel: ∫ 0 ∞ x α 2 + x 2 J 0 ( x ) d x = e − α {\displaystyle \int _{0}^{\infty }{\frac {x}{\sqrt {\alpha ^{2}+x^{2}}}}\,J_{0}(x)\,dx=e^{-\alpha }} Differenziere nach α {\displaystyle \alpha \,} : ∫ 0 ∞ − α x α 2 + x 2 3 J 0 ( x ) d x = − e − α {\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 [ α α 2 + x 2 J 0 ( x ) ] 0 ∞ − ∫ 0 ∞ α α 2 + x 2 J 0 ′ ( x ) d x {\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 J 0 ′ ( x ) = − J 1 ( x ) {\displaystyle J_{0}'(x)=-J_{1}(x)} ist. Also ist − 1 + ∫ 0 ∞ α α 2 + x 2 J 1 ( x ) d x = − e − α {\displaystyle -1+\int _{0}^{\infty }{\frac {\alpha }{\sqrt {\alpha ^{2}+x^{2}}}}\,J_{1}(x)\,dx=-e^{-\alpha }}
Verwende die Poissonsche Darstellung J ν ( x ) = 2 Γ ( 1 2 ) Γ ( ν + 1 2 ) ( x 2 ) ν ∫ 0 π 2 cos ( x cos t ) sin 2 ν t d t Re ( ν ) > − 1 2 {\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}}} ∫ 0 ∞ J ν ( x ) x s − 1 d x = 2 2 ν Γ ( 1 2 ) Γ ( ν + 1 2 ) ∫ 0 π 2 ∫ 0 ∞ x ν + s − 1 cos ( x cos t ) d x sin 2 ν t d t {\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} = 2 2 ν Γ ( 1 2 ) Γ ( ν + 1 2 ) ∫ 0 π 2 Γ ( ν + s ) ( cos t ) ν + s cos ( ν + s ) π 2 sin 2 ν t d t {\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} = Γ ( ν + s ) cos ( ν + s ) π 2 2 ν π Γ ( ν + 1 2 ) ⋅ 2 ∫ 0 π 2 ( sin t ) 2 ( ν + 1 2 ) − 1 ( cos t ) 2 ( 1 2 − ν + s 2 ) − 1 d t {\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} = Γ ( ν + s ) π Γ ( 1 2 + ν + s 2 ) Γ ( 1 2 − ν + s 2 ) 2 ν π Γ ( ν + 1 2 ) ⋅ Γ ( ν + 1 2 ) Γ ( 1 2 − ν + s 2 ) Γ ( 1 + ν − s 2 ) {\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)}}} = π 2 ν Γ ( ν + s ) Γ ( 1 2 + ν + s 2 ) ⋅ 1 Γ ( 1 + ν − s 2 ) = 2 s − 1 Γ ( ν + s 2 ) Γ ( 1 + ν − s 2 ) {\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)}}}
![{\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 )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52340a29c6fd6eb21006f58cbb0de6fa50d91df2)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d8470206316f5fc9c51780a595640b3716f97c1)
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![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a350c42604ba5fd3c4232a427eab0908e1918cf6)
