Notes on Whittaker & Watson, Chapter XVII, part 2

Whittaker & Watson, A Course of Modern Analysis. Chapter XVII, Bessel Functions.

17.3 Hankel’s contour integral for $J_n(z)$​.

The integral on a figure of eight contour around $t=\pm1$​, with the starting point $A$​ on the right of $t=1$​ and $\arg(t-1)=\arg(t+1)=0$​ at $t=A$​​ $$ \begin{align*}& z^n\int_A^{1+,-1-}(t^2-1)^{n-\frac12}\cos(zt),dt \\ = {} & \sum_{r=0}^\infty\frac{(-1)^rz^{n+2r}}{(2r)!}\int_A^{1+,-1-}t^{2r}(t^2-1)^{n-\frac12}dt \\ = {} & \sum_{r=0}^\infty\frac{(-1)^rz^{n+2r}}{(2r)!}(-4i)\sin{\left(n-\frac12\right)\pi}\int_0^1t^{2r}(t^2-1)^{n-\frac12}dt\\ ={}&\sum_{r=0}^\infty\frac{(-1)^rz^{n+2r}}{(2r)!}\frac{2i\sin{\left(n-\frac12\right)\pi}\Gamma(r+\frac12)\Gamma(n+\frac12)}{\Gamma(n+r+1)}\\ ={}&2^{n+1}i\sin{\left(n-\frac12\right)\pi}\Gamma\left(r+\frac12\right)\Gamma\left(n+\frac12\right)J_n(z) \end{align*} $$ gives us the contour integral of the Bessel function $J_n(z)$ $$ J_n(z)=\frac{\Gamma(\frac12-n)(\frac12z)^n}{2\pi i\Gamma(\frac12)}\int_A^{1+,-1-}(t^2-1)^{n-\frac12}\cos(zt),dt. $$ 17.4 Connexion between Bessel coefficients and Legendre functions.

In the limit as $n\to\infty$, $$ \lim_{n\to\infty}n^{-m}P_n^m\left(1-\frac{z^2}{2n^2}\right)=J_m(z) $$ which is the result of obtaining the associated Legendre functions $P_n^m(z)$​​​​ by differentiating the Murphy’s expression of $P_n(z)$​​​​ (§15.22).

17.5 Asymptotic series for $J_n(z)$ when $|z|$ is large.

Instead of expressing $J_n(z)$​​ in terms of $M_{0,n}(2iz)$​​, $J_n(z)$​​ can be expressed in $W_{0,n}$​​​​ $$ J_n(z)=\frac1{(2\pi z)^\frac12}\left(e^{\frac12(n+\frac12)\pi i}W_{0,n}(2iz)+e^{-\frac12(n+\frac12)\pi i}W_{0,n}(-2iz)\right). $$ Using the asymptotic expasion of $W_{0,n}$ we have $$ J_n(z)\sim\left(\frac{2}{\pi z}\right)^\frac12\left(\cos(z-\frac12n\pi-\frac14\pi)U_n(z)-\sin(z-\frac12n\pi-\frac14\pi)V_n(z)\right) $$ where $$ \begin{gather*} U_n(z)=1+\sum_{r=1}^\infty\frac{(-1)^r(4n^2-1)(4n^2-3^2)\cdots(4n^2-(4r-1)^2)}{(2r)!2^{6r}z^{2r}},\\ V_n(z) = \sum_{r=1}^\infty\frac{(-1)^{r-1}(4n^2-1)(4n^2-3^2)\cdots(4n^2-(4r-3)^2)}{(2r-1)!2^{6r-3}z^{2r-1}}. \end{gather*} $$