= Solution
Choose $N$ near the least term, so $\gamma=N+2/3=R+O(1)$ with $R=|z|^3$. On writing $\beta=3(\arg z-\pi/3)$, the parameter is $\lambda=Re^{i\beta}$. For $\beta=O(R^{-1/2})$,
$$
\lambda=R+iR\beta+O(1)=\gamma+i\mu\sqrt\gamma+O(1),\qquad \mu\sim\beta\sqrt R.
$$
The supplied <Gaussian pole transition for a large gamma parameter> therefore gives $I\sim i\pi[1+\operatorname{erf}(\mu/\sqrt2)]$. Substitution in <a/solution> gives the smoothing formula in <2/solution> once the amplitude identity in <c/solution> is used.
The origin of the <error function> can also be seen locally. Set $t=1+s/\sqrt\gamma$ in the Borel integral. Then
$$
(\gamma-1)\log t+\lambda(1-t)=-\frac{s^2}{2}-i\mu s+O(\gamma^{-1/2}),\qquad \frac{dt}{1-t}=-\frac{ds}{s}.
$$
The saddle has become a <Gaussian integral> meeting a simple pole. With the upper-pole contour, the leading integral is $-\int_{\mathcal C_+}e^{-s^2/2-i\mu s}ds/s$. Its value at $\mu=0$ is $i\pi$ from the indentation. Differentiating with respect to $\mu$ gives $i\sqrt{2\pi}e^{-\mu^2/2}$, so integration from zero gives $i\pi[1+\operatorname{erf}(\mu/\sqrt2)]$. This independently fixes the sign and the half-jump on the <Stokes line>.
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