Solution
= Solution
The saddle contributes the full Gaussian
$$
e^{ix}\int_{-\infty}^{\infty}e^{-xs^2/2}ds=\sqrt{2\pi},x^{-1/2}e^{ix}.
$$
\b[Thus $\alpha=\sqrt{2\pi}$. Since the phase is exactly quadratic, there are no further algebraic saddle terms; the next contribution is the endpoint term $O(x^{-1}e^{-3x/2})$.]