Hankel function expansion for large arguement

Question

The leading order behaviour of the Hankel function for large arguments is known to be $$ H_{n}^{(1)}(z)\sim\sqrt{\frac{2}{\pi z}}e^{i\left(z-\frac{n\pi}{2}-\frac{\pi}{4}\right)} $$ as $z\to\infty$. I would like to know what the analytical form of the full expansion is i.e. if we write $$ H_{n}^{(1)}(z)=\sqrt{\frac{2}{\pi z}}e^{i\left(z-\frac{n\pi}{2}-\frac{\pi}{4}\right)}f_{n}\left(\frac{1}{z}\right) $$ then I would like to know the power series expansion for $f_{n}(z)$. I can get the first few terms in Mathematica but I don't see how to derive a general form for the coefficients. Thanks in advance for any help.

Answer 1

The asymptotic expansion for the Hankel function can be found in DLMF:

\begin{align} &{H^{(1)}_{\nu}}\left(z\right)\sim\left(\frac{2}{\pi z}\right)^{\frac{1}{2}}e^{i\omega}\sum_{k=0}^{\infty}i^{k}\frac{a_{k}(\nu)}{z^{k}}\\ &\omega=z-\tfrac{1}{2}\nu\pi-\tfrac{1}{4}\pi\\ &a_{k}(\nu)=\frac{(4\nu^{2}-1^{2})(4\nu^{2}-3^{2})\cdots(4\nu^{2}-(2k-1)^{2})}{ k!8^{k}}\text{ for }k\ge1\\ &a_0(\nu)=1 \end{align} for $-\pi<\operatorname{arg}z<2\pi$. It is obtained using a contour integral representation for the function (see here section 7, for example).