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If a Borel summation contour meets a simple pole, an upper or lower indentation fixes the exponentially small contribution. For the transform $(1-t^2{)}^{-1}$ on the positive ray, the relevant pole is $t=1$. A lower indentation gives the Cauchy principal value minus $i\pi e^{-z}/2$. Finite geometric expansion gives an exact remainder integral, which enables optimal truncation and error-function smoothing of a Stokes multiplier.

Near $\arg z=\pi /3$, set $\lambda =-z^3$ on the branch with argument near zero and truncate the endpoint expansion at $N$ terms. Its remaining factorial series, indexed by $r=N+p$, is ${\lambda }^{-1/3}/[3\Gamma (2/3)]$ times the Borel summation of ${\sum }_{p\ge 0}\Gamma (N+2/3+p)e^{\lambda }/{\lambda }^{N+2/3+p}$. The upper-pole contour fixes the stated remainder and its continuation from the sector below the switching line. With $N+2/3\sim |z{|}^3$, the Gaussian pole transition for a large gamma parameter smooths the algebraic coefficient from $C$ to $C{e}^{2\pi i/3}$.

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.

The factorial tail provides the connection with the supplied Borel summation formula. Near the first switching line choose $\lambda =-z^3$ with $\arg \lambda =3(\arg z-\pi /3)$, so $\lambda$ is near the positive real axis and ${\lambda }^{-1/3}=e^{i\pi /3}/z$. The endpoint contribution through $N-1$ terms isWriting $r=N+p$ in its formal remainder, with $\gamma =N+2/3$, givesThus the Borel remainder for a cubic exponential integral, using the contour just above the pole specified in the paper, isThe continuation is chosen from the sector below the switching line. The term $C/z$ fixes that continuation; an alternative lateral pole contour would shift it by the corresponding residue. This is why the contour prescription must accompany the divergent factorial series.