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The ordinary saddle-point approximation with a nearby pole loses uniformity when the pole lies within the width of the saddle's Gaussian function profile. The distinguished limit isUse the exact quadratic phase coordinate $s=2i\sqrt{k}(u-1)$ and defineOn the descending contour, $s$ runs along the real axis from $-\infty$ to $+\infty$. The transformed differential has the useful partial fraction decompositionThe first term contains the nearby pole; the second is regular at the saddle and contributes $O(e^{-k}{k}^{-1/2})$. Approaching from below the descending contour means $\mathrm{Im}{s}_d<0$. The residue theorem then gives the Gaussian saddle-pole transitionThe exponent is $-s^2/4$, as in the original PDF; the converted TeX's $-s^2/2$ is a transcription error. The exact identity $k\varphi (z_0)=-k-s_d^2/4$ recovers the pole exponential in the requested expression.

The Gaussian pole integral expresses the bracket as $i\pi w(s_d/2)$, where $w$ is the Faddeeva function. Thus a convenient uniform leading answer isThe error estimate applies for bounded scaled pole position on the indicated side. In the overlap $1\ll |s_d|\ll \sqrt{k}$ below the real axis, the integral contributes $-2\sqrt{\pi }/s_d$ to leading order while the residue remains $2\pi i{e}^{-s_d^2/4}$. Since $s_d\sim \sqrt{k}(z_0-i)$, this reproduces the near-saddle limit of the separate saddle and residue terms. Boundary values on the descending contour are limits of this combined expression; the isolated real-axis pole integral itself needs a Cauchy principal value prescription.

PutOn the specified branch, $\sqrt{z^2-1}$ is positive for $z>1$ and negative for $z<-1$. The saddle points of the phase are therefore $z=\pm a$, withandApplying the method of steepest descent at the two simple saddles givesThis formula is valid while the pole and positive saddle remain separated by much more than their $O({\lambda }^{-1/2})$ saddle width.

The pole crosses the positive saddle whenBecause the original contour passes above the pole, deformation onto the steepest-descent contour contributeswhen $z_0>a$, or $0<\theta <{\theta }_{\ast }$. It contributes no residue when $z_0<a$, or ${\theta }_{\ast }<\theta <1$. Hence, away from the transition,The residue is exponentially oscillatory and $O(1)$, whereas each ordinary saddle contribution is $O({\lambda }^{-1/2})$.

At $\theta ={\theta }_{\ast }$ the pole and saddle coalesce, so the displayed saddle formula is singular and must not be used. Passing above the coincident point gives one half of the switched residue at leading order:If $|\theta -{\theta }_{\ast }|=O({\lambda }^{-1/2})$, a uniform saddle-point approximation with a nearby pole replaces the discontinuous switch by a complementary-error-function multiplier.

As $\theta \to 1^{-}$, $s\to 0$, $a\to \infty$, and the pole lies in the no-residue regime. Each saddle coefficient is $O(s^{1/2}{\lambda }^{-1/2})$, so the leading approximation tends to