Asymptotic behavior of the Bessel function $J_\nu$ as $\nu \to \infty$ is given by $$ J_\nu(z) = \frac 1 {\sqrt {2 \pi}} \left( \frac {e z} {2 \nu} \right)^\nu \left( \nu^{-1/2} - \frac {3 z^2 + 1} {12} \nu^{-3/2} + O(\nu^{-5/2}) \right). $$ This post very nicely shows how to derive this from the integral representation of the Bessel function on the real axes $$ J_\nu(z) =\frac {z^\nu} {2^\nu \sqrt \pi \, \Gamma {\left( \nu + \frac 1 2 \right)}} \int_{-1}^1 (1 - t^2)^{\nu - 1/2} \cos z t \, dt. $$ Is there a way to obtain the same expansion from the integral representation in the complex plane $$ J_\nu (z) = \frac{1}{2\pi i} \int_{\mathcal C} e^{z \sinh t - \nu t} dt~? $$ Here the contour $\mathcal C$ goes from $u_x \to \infty$ and $u_y = -\pi$, continues to $u_x \to -\infty$ and $u_y = 0$ and finally returns to $u_x \to \infty$ and $u_y = \pi$.
How would one approach constructing this expansion here without returning to the real axes integral representation? Can one use something like a saddle point method or Watson's lemma or something like that? This is usually done for obtaining the $z\to \infty$ expansion.
If this is for some reason a particularly bad approach, or not possible, I'd be interested in knowing the reasons why?