Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 30D (Asymptotic Methods)

With phase $\varphi (z)=(1+i)z-z^2/2$, the original ray is a contour integral. At the endpoint $z=-1$, the steepest direction solves $\mathrm{Im}(\varphi (z)-\varphi (-1))=0$ with decreasing real part; locally it is the direction for which ${\varphi }^{\prime }(-1)(z+1)$ is negative real.

The saddle is $z_s=1+i$. Completing the square gives $\varphi (z)=i-(z-1-i{)}^2/2$, so its steepest-descent line is the horizontal line $z=1+i+s$, $s\in \mathbb{R}$.

Deform the real ray upward to the saddle line, joining it near the endpoint and at a large right cutoff. The integrand is entire. The far bridge vanishes by Gaussian decay; the endpoint bridge is $O(x^{-1}{e}^{-3x/2})$ because $\Re \varphi (-1)=-3/2$.

The saddle contributes the full GaussianThus $\alpha =\sqrt{2\pi }$. Since the phase is exactly quadratic, there are no further algebraic saddle terms; the next contribution is the endpoint term $O(x^{-1}{e}^{-3x/2})$.