Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,sin,cos)

 

1.1Bearbeiten

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

Beweis (Poissonsche Darstellungen der Besselfunktion)

In der Formel

{\displaystyle J_{\nu }(z)={\frac {2}{\Gamma \left({\frac {1}{2}}\right)\Gamma \left(\nu +{\frac {1}{2}}\right)}}\left({\frac {z}{2}}\right)^{\nu }\int _{0}^{1}\cos(zx)\,(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx\qquad {\text{Re}}(\nu )>-{\frac {1}{2}}}

substituiere {\displaystyle x\mapsto \cos x}.

 

 

 

1.2Bearbeiten

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

Beweis

In der Formel

{\displaystyle J_{\nu }(z)={\frac {2}{\Gamma \left({\frac {1}{2}}\right)\Gamma \left(\nu +{\frac {1}{2}}\right)}}\left({\frac {z}{2}}\right)^{\nu }\int _{0}^{1}\cos(zx)\,(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx\qquad {\text{Re}}(\nu )>-{\frac {1}{2}}}

substituiere {\displaystyle x\mapsto \sin x}.

 

 

 

1.3Bearbeiten

{\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {\sin \alpha x}{\sin x}}\,\cos ^{\alpha -1}x\,dx={\frac {\pi }{2}}\qquad {\text{Re}}(\alpha )>0}

Beweis (Liouvillesches Integral)

Aus der Formel {\displaystyle \sin 2\alpha x=\alpha \sum _{n=0}^{\infty }(-1)^{n}\,{\frac {\left(\alpha -{\frac {1}{2}}+n\right)!}{\left(\alpha -{\frac {1}{2}}-n\right)!}}\,{\frac {(2\sin x)^{2n+1}}{(2n+1)!}}} für {\displaystyle {\text{Re}}(\alpha )>0\,} und {\displaystyle 0\leq x\leq {\frac {\pi }{2}}} folgt

{\displaystyle I:=\int _{0}^{\frac {\pi }{2}}{\frac {\sin 2\alpha x}{\sin x}}\,\cos ^{2\alpha -1}x\,dx=\alpha \sum _{n=0}^{\infty }(-1)^{n}\,{\frac {\left(\alpha -{\frac {1}{2}}+n\right)!}{\left(\alpha -{\frac {1}{2}}-n\right)!}}\,{\frac {2^{2n}}{(2n)!}}\,{\frac {2}{2n+1}}\,\int _{0}^{\frac {\pi }{2}}\sin ^{2n}x\,\cos ^{2\alpha -1}x\,dx},

dabei ist {\displaystyle 2\int _{0}^{\frac {\pi }{2}}\sin ^{2n}x\,\cos ^{2\alpha -1}x\,dx=B\left(n+{\frac {1}{2}},\alpha \right)={\frac {\left(n-{\frac {1}{2}}\right)!\,(\alpha -1)!}{\left(n-{\frac {1}{2}}+\alpha \right)!}}}.

Also ist {\displaystyle I=\alpha !\,\sum _{n=0}^{\infty }(-1)^{n}\,{\frac {2^{2n}\,\left(n-{\frac {1}{2}}\right)!}{\left(\alpha -{\frac {1}{2}}-n\right)!\,(2n)!}}\,{\frac {1}{2n+1}}}.

Nach der Legendreschen Verdopplungsformel lässt sich {\displaystyle {\frac {2^{2n}\,\left(n-{\frac {1}{2}}\right)!}{(2n)!}}} durch {\displaystyle {\frac {\sqrt {\pi }}{n!}}} ersetzen.

{\displaystyle I={\sqrt {\pi }}\,\alpha !\,\sum _{n=0}^{\infty }{\frac {1}{\left(\alpha -{\frac {1}{2}}-n\right)!\,n!}}\,{\frac {1}{2n+1}}={\frac {\sqrt {\pi }}{2}}\,{\frac {\alpha !}{\left(\alpha -{\frac {1}{2}}\right)!}}\,\sum _{n=0}^{\infty }(-1)^{n}\,{\alpha -{\frac {1}{2}} \choose n}\,{\frac {1}{n+{\frac {1}{2}}}}}.

Nach der Formel {\displaystyle \sum _{n=0}^{\infty }(-1)^{n}\,{x-1 \choose n}\,{\frac {1}{n+y}}=B(x,y)} ist dies {\displaystyle {\frac {\sqrt {\pi }}{2}}\,{\frac {\alpha !}{\left(\alpha -{\frac {1}{2}}\right)!}}\,B\left(\alpha +{\frac {1}{2}},{\frac {1}{2}}\right)={\frac {\sqrt {\pi }}{2}}\,\Gamma \left({\frac {1}{2}}\right)={\frac {\pi }{2}}}.

 

 

 

1.4Bearbeiten

{\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {\sin 2\alpha x}{\sin x}}\,\cos x\,dx={\frac {\pi }{2}}+{\frac {\sin \alpha \pi }{2}}\left(\psi \left({\frac {\alpha }{2}}\right)-\psi \left({\frac {\alpha }{2}}+{\frac {1}{2}}\right)+{\frac {1}{\alpha }}\right)}

ohne Beweis    

 

 

1.5Bearbeiten

{\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {k\,\cos \,x}{\sqrt {1-k^{2}\sin ^{2}x}}}\,dx=\arcsin k}

ohne Beweis    

 

 

1.6Bearbeiten

{\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {k\,\sin x\,\cos ^{2}x}{\sqrt {1-k^{2}\sin ^{2}x}}}\,dx={\frac {1}{2k}}+{\frac {k^{2}-1}{2k^{2}}}\,{\text{artanh}}\,k}

ohne Beweis    

 

 

2.1Bearbeiten

{\displaystyle \int _{0}^{\pi }\sin nx\,\cos mx\,dx=\left\{{\begin{matrix}0&,&m\equiv n\mod 2\\\\{\frac {1}{n+m}}+{\frac {1}{n-m}}&,&\mathrm {sonst} \end{matrix}}\right.}

ohne Beweis    

 

 

2.2Bearbeiten

{\displaystyle \int _{-\pi }^{\pi }\sin ^{n}x\,\cos ^{m}x\,dx={\frac {{\frac {n!}{2^{n}\left({\frac {n}{2}}\right)!}}\,{\frac {m!}{2^{m}\left({\frac {m}{2}}\right)!}}}{\left({\frac {n+m}{2}}\right)!}}} wenn {\displaystyle n,m\in \mathbb {N} } beide gerade sind, andernfalls ist das Integral 0.

ohne Beweis    

 

 

2.3Bearbeiten

{\displaystyle J_{n}(z)={\frac {1}{\pi }}\int _{0}^{\pi }\cos(z\sin x-nx)\,dx\qquad n\in \mathbb {Z} \;,\;z\in \mathbb {C} }

1. Beweis (Bessel Integral)

Multipliziere die Jacobi-Anger Entwicklung {\displaystyle e^{iz\sin x}=\sum _{n\in \mathbb {Z} }J_{n}(z)\,e^{inx}}

mit {\displaystyle e^{-imx}\,} durch und integriere anschließend beide Seiten nach {\displaystyle x\,} von {\displaystyle -\pi \,} bis {\displaystyle \pi \,}:

{\displaystyle \int _{-\pi }^{\pi }e^{iz\sin x}\,e^{-imx}\,dx=\sum _{n\in \mathbb {Z} }J_{n}(z)\int _{-\pi }^{\pi }e^{i(n-m)x}\,dx=\sum _{n\in \mathbb {Z} }J_{n}(z)\,\delta _{nm}\cdot 2\pi =J_{m}(z)\cdot 2\pi }

Somit ist {\displaystyle \int _{-\pi }^{\pi }\cos(z\sin x-mx)\,dx+i\int _{-\pi }^{\pi }\sin(z\sin x-mx)\,dx=J_{m}(z)\cdot 2\pi }.

Der erste Integrand ist gerade und der zweite ungerade.

Also ist {\displaystyle 2\int _{0}^{\pi }\cos(z\sin x-mx)\,dx=J_{m}(z)\cdot 2\pi }.

  2. Beweis

Multipliziere die Jacobi-Reihe {\displaystyle \cos(z\sin x)=J_{0}(z)+2\sum _{n=1}^{\infty }J_{2n}(z)\cos 2nx}

mit {\displaystyle \cos mx\,} durch und integriere anschließend beide Seiten nach {\displaystyle x\,} von {\displaystyle 0\,} bis {\displaystyle \pi \,}:

{\displaystyle \int _{0}^{\pi }\cos(z\sin x)\,\cos mx\,dx=J_{0}(z)\int _{0}^{\pi }\cos mx\,dx+2\sum _{n=1}^{\infty }J_{2n}(z)\int _{0}^{\pi }\cos 2nx\,\cos mx\,dx}

{\displaystyle =J_{0}(z)\,\delta _{m,0}\cdot \pi +\sum _{n=1}^{\infty }J_{2n}(z)\,\delta _{m,2n}\cdot \pi =\left\{{\begin{matrix}J_{m}(z)\cdot \pi &m\;{\text{gerade}}\\0&m\;{\text{ungerade}}\end{matrix}}\right.}.


Multipliziere die Jacobi-Reihe {\displaystyle \sin(z\sin x)=2\sum _{n=0}^{\infty }J_{2n+1}(z)\sin(2n+1)x}

mit {\displaystyle \sin mx\,} durch und integriere anschließend beide Seiten nach {\displaystyle x\,} von {\displaystyle 0\,} bis {\displaystyle \pi \,}:

{\displaystyle \int _{0}^{\pi }\sin(z\sin x)\,\sin mx\,dx=2\sum _{n=0}^{\infty }J_{2n+1}(z)\int _{0}^{\pi }\sin(2n+1)x\,\sin mx\,dx}

{\displaystyle =\sum _{n=0}^{\infty }J_{2n+1}(z)\,\delta _{m,2n+1}\cdot \pi =\left\{{\begin{matrix}0&m\;{\text{gerade}}\\J_{m}(z)\cdot \pi &m\;{\text{ungerade}}\end{matrix}}\right.}.


Also ist {\displaystyle J_{m}(z)\cdot \pi =\int _{0}^{\pi }\cos(z\sin x)\,\cos mx\,dx+\int _{0}^{\pi }\sin(z\sin x)\,\sin mx\,dx=\int _{0}^{\pi }\cos(z\sin x-mx)\,dx}.

 

 

 

2.4Bearbeiten

{\displaystyle 2\int _{0}^{\frac {\pi }{2}}\sin ^{2\alpha -1}x\,\cos ^{2\beta -1}x\,dx=B(\alpha ,\beta )\qquad {\text{Re}}(\alpha ),{\text{Re}}(\beta )>0}

Beweis

In der Formel {\displaystyle B(\alpha ,\beta )=\int _{0}^{1}x^{\alpha -1}(1-x)^{\beta -1}\,dx} substituiere {\displaystyle x\,} durch {\displaystyle \sin ^{2}x\,}:

{\displaystyle B(\alpha ,\beta )=\int _{0}^{1}(\sin ^{2}x)^{\alpha -1}\,(\cos ^{2}x)^{\beta -1}\cdot 2\,\sin x\,\cos x\,dx=2\int _{0}^{\frac {\pi }{2}}\sin ^{2\alpha -1}x\,\cos ^{2\beta -1}x\,dx}

 

 

 

3.1Bearbeiten

{\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {dx}{(a^{2}\,\cos ^{2}x+b^{2}\,\sin ^{2}x)^{n+1}}}={\frac {\pi }{2ab}}\,\sum _{k=0}^{n}{\frac {2k \choose k}{(2a)^{2k}}}\,{\frac {2(n-k) \choose n-k}{(2b)^{2(n-k)}}}}

ohne Beweis

标签：Integrale,cos,right,frac,Form,sin,alpha,pi,displaystyle
来源： https://www.cnblogs.com/Eufisky/p/14730824.html