6-3

以下の式を示せ。

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

$$= \frac{i^{-n}}{2\pi} \int_0^{2\pi} e^{iz\cos\varphi +in\varphi } d\varphi$$ (47)

6-3解答

$$J_{{n}}(z) = \frac{1}{\pi} \int^\pi \cos\left(n\varphi -z\sin\varphi \right)... ...n\varphi } d\varphi + \frac{1}{2\pi} \int_{-\pi}^{0} e^{+iz\sin\xi -in\xi} d\xi$$

$$= \frac{1}{2\pi} \int_0^\pi e^{iz\sin\varphi -in\varphi } d\varph... ...c\vert ccc} \xi & -\pi & \to & 0 \\ \hline \zeta & \pi & \to & 2\pi \end{array}$$

$$= \frac{1}{2\pi} \int_0^\pi e^{iz\sin\varphi -in\varphi } d\varphi + \frac{1}{2\pi} \int_{\pi}^{2\pi} e^{+iz\sin\zeta -in\zeta} d\zeta$$

$$=\frac{1}{2\pi} \int_0^{2\pi} e^{iz\sin\varphi -in\varphi }d\varphi$$

$$J_{{n}}(z) = \frac{1}{2\pi} \int_0^{2\pi} e^{iz\sin\varphi -in\varphi }d\var... ...ccc} \varphi & 0 & \to & 2\pi \\ \hline \xi & \pi/2 & \to & -3/2\pi \end{array}$$

$$= \frac{i^{-n}}{2\pi} \int\{-3/2\pi\}^{\pi/2} e^{iz\cos\xi+in\xi}... ...{2\pi} +\int_{2\pi}^{\pi/2} +\int_{-3/2\pi}^{0} \right) e^{iz\cos\xi+in\xi}d\xi$$

$$\qquad \int_{-3/2\pi}^{0} e^{iz\cos\xi+in\xi} d\xi = \int_{\pi/2}... ...ccc} \xi & -3/2\pi & \to & 0 \\ \hline \varphi & \pi/2 & \to & 2\pi \end{array}$$

$$= \frac{i^{-n}}{2\pi} \left( \int_{0}^{2\pi} +\int_{2\pi}^{\pi/2} -\int_{2\pi}^{\pi/2} \right) e^{iz\cos\xi+in\xi}d\xi$$

$$= \frac{i^{-n}}{2\pi} \int_0^{2\pi} e^{iz\cos\varphi +in\varphi } d\varphi$$

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp