{c}

For a simple <saddle point>, a nearby <pole> coalesces with its contributing region when their separation is of order the inverse square root of the large parameter. A local quadratic phase coordinate changes the singular part to a <Gaussian pole integral>. Adding any residue crossed by the <contour deformation> gives a <Faddeeva function> rather than two separately singular approximations. The regular part of the transformed amplitude remains smaller by the ordinary saddle width.

Gaussian saddle-pole transition

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> is
$$
\boxed{z_d=z_s=i,\qquad z_0-i=O(k^{-1/2}).}
$$
Use the exact quadratic phase coordinate $s=2i\sqrt{k}(u-1)$ and define
$$
\boxed{s_d=2i\sqrt{k}\bigl(e^{-i\pi/4}\sqrt{z_0}-1\bigr)
=\sqrt{k}(z_0-i)+O\bigl(\sqrt{k}(z_0-i)^2\bigr).}
$$
On the descending contour, $s$ runs along the real axis from $-\infty$ to $+\infty$. The transformed differential has the useful <partial fraction decomposition>
$$
\frac{dz}{z-z_0}=\left(\frac1{u-u_0}+\frac1{u+u_0}\right)du,
\qquad u_0=e^{-i\pi/4}\sqrt{z_0}.
$$
The 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 $\operatorname{Im}s_d<0$. The <residue theorem> then gives the <Gaussian saddle-pole transition>
$$
J(z_0)=e^{-k}\left[\int_{-\infty}^{\infty}\frac{e^{-s^2/4}}{s-s_d}\,ds+2\pi i e^{-s_d^2/4}\right]+O(e^{-k}k^{-1/2}).
$$
The exponent is $-s^2/4$, as in the original PDF; the converted TeX's $-s^2/2$ is a transcription error. The exact identity $k\phi(z_0)=-k-s_d^2/4$ 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 is
$$
\boxed{J(z_0)=i\pi e^{-k}w(s_d/2)+O(e^{-k}k^{-1/2}).}
$$
The 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 336 / 1 / c / Solution

{c}
{title2=$G(a)=\int_{-\infty}^{\infty}e^{-s^2/4}/(s-a)\,ds$}

For a pole off the real axis, substitution $s=2t$ into the <Faddeeva function> integral gives $G(a)=i\pi w(a/2)$ when $\operatorname{Im}a>0$, and $G(a)=-i\pi w(-a/2)$ when $\operatorname{Im}a<0$. The difference of boundary values is the <residue theorem> jump $2\pi i e^{-a^2/4}$. In particular, for a lower-half-plane pole, $G(a)+2\pi i e^{-a^2/4}=i\pi w(a/2)$, which packages a crossed residue and a nearby saddle into one smooth expression.

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 Incoming links: Gaussian pole integral