 Solution

Let $P(z)=1/(2\sqrt{\pi }{z}^{1/4})$, choose a consistent branch, and truncate the corrected series after $r=n-1$, with $n=|\sigma |+O(1)$. The leading factorial tail in part (iii) is represented by the same indented integral as in part (ii):Here the divergent sum means its lateral Borel summation, not ordinary summation. To see why the pole approximation controls this remainder, write $a_r={\sigma }^r{Y}_r$ and insert $\Gamma (r)={\int }_0^{\infty }{e}^{-v}{v}^{r-1}dv$ in the leading late terms. Summing the geometric tail creates $1/(1-v/\sigma )$; $v=\sigma t$ produces $I(\sigma ,n)$. The relative $O(r^{-1})$ corrections to the late coefficients give lower-order contributions on its saddle scale. More concretely, the Borel transform of the coefficient series isThe simple pole supplies the leading Stokes phenomenon; its weaker logarithmic singularity supplies the smaller corrections. This is the integral justification for retaining the factorial tail.

For $\arg \sigma =\varphi /\sqrt{|\sigma |}$, expand $\sigma =n+i\varphi \sqrt{n}+O(1)$. Part (i), with $\mu =\varphi$, now givesThis formula uses the upper-pole prescription and the upper Airy function switching ray: near $\arg z=2\pi /3$, take $\arg \sigma =3\arg z/2-\pi$ near zero. The coefficient of the subdominant exponential changes smoothly from zero to $i$ across this ray and has value $i/2$ on it. The angular transition width is $O(|\sigma {|}^{-1/2})=O(|z{|}^{-3/4})$. The conjugate lower ray has the conjugate contour sign, rather than the same $+i$ on both rays. This is error-function smoothing of a Stokes multiplier, not a discontinuity of the analytic Airy function. With the literal printed $r=1$ series, the missing $P(z)e^{\sigma /2}$ would invalidate this remainder formula.