At $\theta=1$, the two saddles have receded to infinity and the ordinary quadratic saddle approximation is nonuniform. One should first use an <asymptotic expansion> of the phase and amplitude for large $z$:
$$
\sqrt{z^2-1}
=z-\frac1{2z}+O(z^{-3})
\quad(z\to+\infty),
$$
so
$$
\Phi(z)=-\frac1{2z}+O(z^{-3}).
$$
The phase varies on the scale $z=O(\lambda)$ rather than in an $O(\lambda^{-1/2})$ neighbourhood of a finite saddle. Rescaling $z=\lambda Z$ in the positive and negative tails, retaining the corresponding large-$z$ amplitude, and matching these tail integrals to the finite part of the deformed contour produces the leading approximation. This is an <endpoint at infinity> problem; a uniform calculation may equivalently begin from the $\theta<1$ saddle representation and take the coalescing-at-infinity limit.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 336 / 1 / c / Solution

An endpoint at infinity contributes when the region controlling an asymptotic integral recedes without bound as the large parameter changes. Expanding the phase and amplitude at infinity reveals the rescaling that converts the distant tail into a finite limiting integral.


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