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

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

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

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 ${\displaystyle {\frac {\sqrt {\pi }}{2}}\,{\frac {\alpha !}{\left(\alpha -{\frac {1}{2}}\right)!}}\,B\left(\alpha +{\frac {1}{2}},{\frac {1}{2}}\right)={\frac {\sqrt {\pi }}{2}}\,\Gamma \left({\frac {1}{2}}\right)={\frac {\pi }{2}}}$.

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

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

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

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