Given integral to investigate asymptotic behaviour on:
$$A=Re\int_{0}^{\infty} J_0(xs) e^{-iw{\sqrt {gx}}}xdx$$ for large $s$ and $w$
$\sqrt{2 / \pi x} cos(x-(2n+1)\pi/4)=J_n(x)$
Want to investigate assymptotics without Henkel transform. The J function is 0th Bessel function. The $\sqrt{x}$ term in exponential part and complex messes up attempts to approximate.
Any ideas on how to take the transform? Maybe substitution or something
On
Mathematica defines the Hankel transform (see Details and Options here) as
$$\mathcal{H}_{r,v}\left[f(r)\right](s)=\int_0^\infty r\,f(r)\,J_v(r s)\,dr\tag{1}$$
and (using $v=0$ and replacing $r$ with $x$) evaluates the Hankel transform of
$$f(x)=e^{-i\,t\,\sqrt{g\,x}}\tag{2}$$ as
$$\mathcal{H}_{x,0}\left[e^{-i\,t\,\sqrt{g\,x}}\right](s)=\frac{\sqrt{2} G_{1,3}^{3,1}\left(\frac{g^2 t^4}{64 s^2}| \begin{array}{c} 0 \\ \frac{1}{4},\frac{1}{2},\frac{3}{4} \\ \end{array} \right)}{\pi ^{3/2} s^2}\,,\quad s\geq 0\tag{3}$$
where $G$ is the Meijer G-Function.
Since you know what is the form of the $J_0(xs)$, then you can rewrite it into two exponential terms, in which one gives zero contribution. For the other term, you can apply method of stationary phase by finding where the maximum is attained.