2 $n\gg 1$の時の $J_{2n}(2n\xi )$の振る舞い

$n\gg 1$の時以下の式が成り立つ：

$$J_{2n}(2n\xi) \sim\frac{1}{n^{1/3}\sqrt{\pi}} \Phi\left( n^{2/3} \left[ 1-\xi^2 \right] \right) .$$ (56)

1 証明

$n\gg 1,, \xi \sim 1$

$$J_{2n}(2n\xi) = \frac{1}{\pi} \int_0^{\infty} \cos\left[ 2n \left( \frac{1-\xi^... ...ght) \zeta \right] d\zeta \, ; \quad n\zeta^3=\eta^3, \, d\zeta = n^{-1/3}d\eta$$

$$= \frac{n^{-1/3}}{\pi} \int_0^\infty \cos\left[ \frac{\eta^3}{3} ... ...= \frac{1}{n^{1/3}\sqrt{\pi}} \Phi\left( n^{2/3} \left[ 1-\xi^2 \right] \right)$$

Eq.(56)の両辺を$2n\xi$で微分すると以下を得る：

$$\dI{(2n\xi)} J_{2n}(2n\xi) =J_{2n}'(2n\xi) = \frac{1}{n^{1/3}\sqrt{\pi}} \Phi'\left(n^{2/3}\... ...\frac{1}{n^{2/3}\sqrt{\pi}} \xi \Phi'\left(n^{2/3}\left[1-\xi^2\right]\right) .$$ (57)

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp