 Solution

At a simple saddle point, ${\varphi }^{\prime }(z_0)=0$ and ${\varphi }^{\prime \prime }(z_0)\ne 0$. On a method of steepest descent ray leaving the saddle, choose the oriented local coordinatewith the square-root branch determined by that ray. Then $\varphi (z)=\varphi (z_0)-s^2/2+O(s^3)$ and $dz=ids/\sqrt{{\varphi }^{\prime \prime }(z_0)}$ to leading order. For large positive $x$, the effective range is $s=O(x^{-1/2})$. The half-Gaussian integral givesAs usual this saddle contribution presumes the remaining contour is lower in real phase and integrable. Reversing the contour orientation reverses the sign; the square-root convention encodes the chosen outgoing ray.

For the second integral set $t=\sqrt{n}z$. Using the specified logarithm branch,The saddles are $z=\pm i$, and only $z_0=i$ is in the upper half-plane. HereThe contour can be deformed to the horizontal line through $i$, oriented from left to right, without crossing the pole at zero. Along $z=s+i$,which has a unique strict maximum at $s=0$ and decays at infinity. This explicitly verifies saddle dominance. Locally $\varphi (i+s)=1-i\pi -2s^2+O(s^3)$. Both sides of the saddle contribute a full Gaussian, soFor integer $n$, $e^{-i\pi n}=(-1{)}^n$, and henceThere is no extra factor of two from another saddle: the full Gaussian already accounts for the incoming and outgoing parts at the upper saddle.