= Solution
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.
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