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The integral $I(x)={\int }_0^{\infty }{e}^{xt-t^3/3}dt$ is a rescaled Scorer Hi function. It has the three useful limitsFor the negative-argument limit, scale $t=s/M$ and use the dominated convergence theorem. At zero, substitute $s=t^3/3$ and use the Gamma integral. For the positive-argument limit, the exponent has its maximum at $t=\sqrt{M}$, with second derivative $-2\sqrt{M}$; Laplace's method gives the displayed factor. These estimates describe a cubic endpoint-to-saddle transition.

The three fixed-$a$ estimates below cease to be uniform as the minimum of the phase approaches the endpoint. For the cubic endpoint-to-saddle transition, writeThe Taylor series of the hyperbolic sine gives, for bounded $s$ and fixed $l$,The cubic term controls the tail, so localization of the Laplace integral gives the uniform leading formulaHere $I$ is the cubic Laplace transition integral, equal to $\pi \mathrm{Hi}$ in terms of the Scorer Hi function.

To recover the endpoint regime, let $l\to +\infty$. Scale $s=v/(2^{1/3}l)$ in $I$; the cubic term becomes negligible and $I(-2^{1/3}l)\sim (2^{1/3}l{)}^{-1}$. Thereforewhich agrees with part (i) in the overlap $1\ll l\ll {\nu }^{2/3}$. At $l=0$, the Gamma integral gives $I(0)=3^{-2/3}\Gamma (1/3)$, recovering part (iii).

For $l\to -\infty$, set $M=-2^{1/3}l>0$. The exponent $Ms-s^3/3$ has its maximum at $s=\sqrt{M}$, with second derivative $-2\sqrt{M}$. Thus Laplace's method givesand the transition formula becomesFor $a=1-\eta$ with $\eta \downarrow 0$, part (ii) hasPutting $\eta =(-l){\nu }^{-2/3}$ reproduces both the exponential and its prefactor. For relative agreement of these leading exponentials, one may use the overlap $1\ll -l\ll {\nu }^{4/15}$, which makes $\nu {\eta }^{5/2}\to 0$. Thus the same transition integral connects all three regimes.