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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.

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 discrete Laplace method estimates a sum whose positive terms concentrate near the maximum of a smooth exponent. When the Gaussian concentration width contains asymptotically many lattice points, the sum is a Riemann-sum version of Laplace's method with the corresponding lattice-spacing factor.

If a Laplace integral has finitely many separated nondegenerate global maxima with the same phase value, its leading term is the sum of their local Gaussian contributions. An interior maximum contributes a full Gaussian integral, while a stationary endpoint maximum contributes half of one.

For fixed $x>0$,as $\nu \to +\infty$. Applying Laplace's method to the integral representation of the modified Bessel function of the second kind proves the formula.

The stationary points of the exponent are $t=1/3$ and $t=1$, because ${\varphi }^{\prime }(t)=(3t-1)(t-1)$. The first is a local maximum with $\varphi (1/3)=4/27$ and ${\varphi }^{\prime \prime }(1/3)=-2$; the second is a local minimum. Alsoso the endpoint overtakes the interior maximum only when $b>4/3$.

For fixed $0<b<1/3$, the upper endpoint is the unique maximum. Linearizing the exponent there gives the Laplace endpoint estimate $e^{\lambda \varphi (b)}/(\lambda {\varphi }^{\prime }(b))$. At $b=1/3$, only half of the local Gaussian integral is included. For $1/3<b\le 4/3$, the whole Gaussian near $1/3$ contributes. At $b=4/3$ an endpoint has the same exponential height, but its prefactor is ${\lambda }^{-1}$, smaller than the interior ${\lambda }^{-1/2}$ prefactor. For $b>4/3$, the endpoint has strictly greater height and dominates.

Consequently all fixed-$b$ regimes areThe estimate follows from Laplace's method; the tail toward negative infinity is exponentially harmless since the cubic tends to negative infinity.

Use the real integral representation of the modified Bessel function of the second kindThe unique stationary point is the strict maximumThe Laplace approximation is consequentlyFor fixed $x>0$, as $\nu \to +\infty$,Substitution gives the required large-order asymptotic of the modified Bessel function of the second kind:

The intended integral representation of the modified Bessel function isDifferentiation under the integral sign givesSinceintegration by parts and the vanishing endpoint term yield $x{I}_0^{\prime \prime }+I_0^{\prime }-x{I}_0=0$.

For large positive $x$, Laplace's method localizes the integral near the endpoint $\theta =0$. Put $\theta =u/\sqrt{x}$ and useExtending the exponentially small tail to infinity and evaluating the resulting Gaussian integrals givesand therefore

Now write $y=x^{-1/2}w$. Direct differentiation givesso substitution into the ordinary differential equation givesFor the Liouville–Green approximation, put $Q(x)=1-1/(4x^2)$. The two approximate solutions areso at leading order $y_{\pm }(x)\sim x^{-1/2}{e}^{\pm x}$. The growing solution agrees with the preceding asymptotic expansion, including its leading normalization $I_0(x)\sim e^x/\sqrt{2\pi x}$.

For $0<a<1$, the phase has an interior minimum atThe inverse hyperbolic cosine selects the positive stationary point. At this point,Expanding the phase to second order and evaluating the resulting Gaussian integral by Laplace's method yieldsThe exponential grows because the minimum of $S$ is negative; the original integral remains convergent because $a\sinh t$ dominates $t$ at infinity.

Let $S(t)=a\sinh t-t$. For $a>1$, $S^{\prime }(t)=a\cosh t-1>0$ on the integration interval, so its minimum is the endpoint $t=0$. The Taylor series isOn the contributing scale $t=O({\nu }^{-1})$, the higher terms are negligible. The endpoint form of Laplace's method, or the Watson lemma, therefore gives

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.

The phase function $\cos t$ attains its global maximum $1$ on $[0,3\pi ]$ at the stationary endpoint $t=0$ and at the interior point $t=2\pi$. At either point,By Laplace's method at a nondegenerate stationary endpoint, the endpoint contributesBy Laplace's method, the interior maximum contributesThe remaining parts of the interval are exponentially smaller. Summing the two contributions according to Laplace's method with multiple global maxima gives

Write $\sinh (x\Phi )=(e^{x\Phi }-e^{-x\Phi })/2$. The global maximum of $\Phi$ is the nonstationary endpoint $t=1$, whereEndpoint Laplace's method therefore givesThe second exponential has its maximum at $t=1/2$ and is only of order $x^{-1/2}{e}^{x/8}$, exponentially smaller. Hence