Subsections

6-3

以下の式を示せ。

$\displaystyle J_{{n}}(z)$ $\displaystyle =\frac{1}{2\pi} \int_0^{2\pi} e^{iz\sin\varphi -in\varphi }d\varphi$ (46)
  $\displaystyle = \frac{i^{-n}}{2\pi} \int_0^{2\pi} e^{iz\cos\varphi +in\varphi } d\varphi$ (47)

6-3解答

$\displaystyle J_{{n}}(z)$ $\displaystyle = \frac{1}{\pi} \int^\pi \cos\left(n\varphi -z\sin\varphi \right)...
...n\varphi } d\varphi + \frac{1}{2\pi} \int_{-\pi}^{0} e^{+iz\sin\xi -in\xi} d\xi$    
  $\displaystyle = \frac{1}{2\pi} \int_0^\pi e^{iz\sin\varphi -in\varphi } d\varph...
...c\vert ccc} \xi & -\pi & \to & 0 \\ \hline \zeta & \pi & \to & 2\pi \end{array}$    
  $\displaystyle = \frac{1}{2\pi} \int_0^\pi e^{iz\sin\varphi -in\varphi } d\varphi + \frac{1}{2\pi} \int_{\pi}^{2\pi} e^{+iz\sin\zeta -in\zeta} d\zeta$    
  $\displaystyle =\frac{1}{2\pi} \int_0^{2\pi} e^{iz\sin\varphi -in\varphi }d\varphi$    

$\displaystyle J_{{n}}(z)$ $\displaystyle = \frac{1}{2\pi} \int_0^{2\pi} e^{iz\sin\varphi -in\varphi }d\var...
...ccc} \varphi & 0 & \to & 2\pi \\ \hline \xi & \pi/2 & \to & -3/2\pi \end{array}$    
  $\displaystyle = \frac{i^{-n}}{2\pi} \int\{-3/2\pi\}^{\pi/2} e^{iz\cos\xi+in\xi}...
...{2\pi} +\int_{2\pi}^{\pi/2} +\int_{-3/2\pi}^{0} \right) e^{iz\cos\xi+in\xi}d\xi$    
  $\displaystyle \qquad \int_{-3/2\pi}^{0} e^{iz\cos\xi+in\xi} d\xi = \int_{\pi/2}...
...ccc} \xi & -3/2\pi & \to & 0 \\ \hline \varphi & \pi/2 & \to & 2\pi \end{array}$    
  $\displaystyle = \frac{i^{-n}}{2\pi} \left( \int_{0}^{2\pi} +\int_{2\pi}^{\pi/2} -\int_{2\pi}^{\pi/2} \right) e^{iz\cos\xi+in\xi}d\xi$    
  $\displaystyle = \frac{i^{-n}}{2\pi} \int_0^{2\pi} e^{iz\cos\varphi +in\varphi } d\varphi$    

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp