I'm trying to answer this question, but more questions arise. Let's consider the integral in order to estimate it with the method of steepest descent
\begin{equation}
I(p) = \int\limits_C \frac{e^{ip(v-\frac{1}{2}v^2+\log(v))}}{v}dv,
\end{equation}
where contour $C$ goes around the cut $[0, e^{\frac{3\pi}{4}i}\infty]$ and $p$ is large. Saddle points:
\begin{equation}
f'(v)=(v-\frac{1}{2}v^2+\log(v))'=0 \Rightarrow v_{\pm}=\frac{1}{2} \pm \frac{\sqrt{5}}{2}
\end{equation}
As I understand, now we should deform the path to the lines of constant phase of the exponent in the intergand: $Re(f(v))=Re(f(v_{\pm})$. I've tried to draw them in Wolfram Mathematica:
Saddle points are inersection points of these lines, orange line corresponds to $v_-$, blue line corresponds to $v_+$ and green line is the cut. Here I have some troubles. First, orange line crosses branch cut and goes trough the saddle point twice. Second, in the regions $\arg(v)\in(-\pi, -\frac{\pi}{2}) \cup (0,\frac{\pi}{2})$ there is a singularity at infinity due to the term $-ip\frac{1}{2}v^2$ in the exponent, so there should be some problems if we deform the contour in this region. So it seems there is no way to deform $C$ to any of these lines.
Does it all mean that we can't use the method in this situation? Can we treat these difficulties some way? Any advice would be helpful very much.