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For a Laplace integral with local phase $\varphi (t;\delta )=\delta t+b{t}^3/3+\cdots$, $b>0$, an interior minimum approaches the endpoint as $\delta \uparrow 0$. The distinguished limit is $t=O({\nu }^{-1/3})$, $\delta =O({\nu }^{-2/3})$. With $s=(\nu b{)}^{1/3}t$, its leading local integral isThe cubic Laplace transition integral joins the ordinary endpoint estimate for positive $\delta$, the cubic endpoint estimate at zero, and the interior Laplace's method estimate for negative $\delta$. Separate fixed-parameter formulas fail to be uniform when the minimum is within its own width of the endpoint.

For real $x>0$ and $\nu$, one useful representation isIt turns large-order asymptotics into a Laplace integral with a moving maximum.

Watson lemma expands a Laplace integral by integrating the small-$t$ expansion term by term. More precisely, if $f(t)\sim {\sum }_{j\ge 0}{a}_j{t}^{\lambda +j-1}$ with $\mathrm{Re}\lambda >0$, and the tail satisfies bounds permitting this localization, thenFor ordinary powers $f(t)\sim \sum c_j{t}^j$, the coefficients are $c_{j}j!$. For the Gaussian weight, put $s=t^2$ separately on the positive and negative half-lines. The integral becomes a Laplace integral with amplitude $[f(\sqrt{s})+f(-\sqrt{s})]/(2\sqrt{s})$. Odd terms cancel and Watson lemma givesThese assertions concern asymptotic expansions, not necessarily convergent series.

For the complex integral in the lower half-plane, the identity $(z-t{)}^{-1}=i{\int }_0^{\infty }{e}^{-i(z-t)s}ds$ and the Gaussian Fourier transform giveThe last expression, initially obtained by completing the square, is entire and therefore gives the requested analytic continuation. Expanding the denominator formally, with odd Gaussian moments zero, gives the algebraic expansionHere $(-1)!!=1$. The complementary error function expansion shows $I(z)\sim \mathcal{A}(z)$ on closed subsectors of $-5\pi /4<\arg z<\pi /4$, using an unwrapped argument centered on the lower half-plane.

In the upper half-plane, the directly evaluated real-contour integral is $J(z)=-i\pi e^{-z^2}\mathrm{erfc}(-iz)$. Since $\mathrm{erfc}(w)+\mathrm{erfc}(-w)=2$, continuation from below givesThe upper expansion for $J$ holds on closed subsectors of $-\pi /4<\arg z<5\pi /4$. The added term is the residue acquired by the continued contour passing the pole; it is exponentially dominant in the upper sector $\pi /4<\arg z<3\pi /4$ and exponentially small near either real ray. On the real axis the continued value is the principal-value integral plus $i\pi e^{-x^2}$.

To make the Stokes phenomenon convention explicit, the exponential switches its asymptotic multiplier across the real rays $\arg z=0,\pi$; these are the Stokes switching lines. The diagonal rays $\arg z=\pi /4+j\pi /2$ separate exponential growth from decay and are often called anti-Stokes lines, or critical boundaries when regions are classified by dominance. Thus both sets of critical directions and the exact jump are specified, independently of which terminology is used. This is the Gaussian Cauchy transform and its Stokes jump.

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.

Watson lemma states that ifand the Laplace integral has suitable growth control away from zero, then

The Taylor series of the amplitude isTermwise application of Watson lemma therefore gives the asymptotic expansionIn particular, the leading asymptotic approximation is $I(x)\sim 2x^{-3}$.

For a smooth real phase decreasing away from its endpoint maximum, put $t=\varphi (a)-\varphi (x)$ and write the transformed amplitude as $G(t)=-g(x(t))/{\varphi }^{\prime }(x(t))$. Its first two Taylor coefficients are $G(0)=-g(a)/{\varphi }^{\prime }(a)$ and $G^{\prime }(0)=g^{\prime }(a)/{\varphi }^{\prime }(a{)}^2-g(a){\varphi }^{\prime \prime }(a)/{\varphi }^{\prime }(a{)}^3$. Applying Watson lemma gives the displayed two-term Laplace integral expansion. Localization and a Taylor remainder estimate justify termwise integration.