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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.

Repeated integration by parts gives the Gamma integralThe logarithm of its integrand is $n\log t-t$, whose unique maximum is at $t=n$. Write $t=n+\sqrt{n}s$. For fixed $s$ and large $n$, a Taylor expansion givesThe Big O notation here is for a fixed $s$, or uniformly for $s$ in a fixed bounded interval. It does not claim a uniform approximation over the entire growing integration domain. Consequently the peak, divided by its height, approaches $e^{-s^2/2}$.

The exact substitution givesReplacing the integral by the Gaussian integral ${\int }_{\mathbb{R}}{e}^{-s^2/2}ds=\sqrt{2\pi }$ produces $S_n$. Thus the estimate is peak height $n^n{e}^{-n}$ times width $\sqrt{n}$ times the area of a standard Gaussian peak. This is Laplace's method in a concrete form; controlling the integral's tails justifies the replacement. The next terms, including their products after exponentiation, produce the Stirling correction by Gaussian moments.

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.

A unique nondegenerate interior minimum does not by itself ensure that a integral treated by Laplace's method is localized there. A sufficient condition is a positive gap above the minimum outside every sufficiently small fixed neighbourhood, together with finiteness of the integral for some positive exponent. The exterior contribution is then exponentially negligible. Without localization, $g(y)=y^2(1-y{)}^4$ on $(-1,1)$ has its unique interior minimum at zero with $g^{\prime \prime }(0)=2$, but its boundary contribution is at least $e^{-1}{n}^{-1/4}$ and dominates the nominal $n^{-1/2}$ interior Gaussian contribution.

If a nonnegative smooth phase has an isolated cubic zero at one endpoint and an isolated quadratic zero at the other, their leading Laplace's method contributions scale respectively as ${\lambda }^{-1/3}$ and ${\lambda }^{-1/2}$. The first two global terms therefore come from different endpoints, rather than from the first two terms at the dominant endpoint. When the cubic endpoint has no quartic phase term and constant amplitude, its next local term starts at ${\lambda }^{-1}$; cubic corrections at the quadratic endpoint also start at that order.

An Edgeworth expansion refines a central normal approximation using higher cumulants. Let $Y_i$ be independent and identically distributed, with mean $\mu$, positive variance ${\sigma }^2$, and cumulants ${\kappa }_r$. DefineThe logarithm of its characteristic function has the expansionExpanding the exponential adds the squared cubic term ${\gamma }_3^2(it{)}^6/(72n)$. Fourier inversion turns $(it{)}^r{e}^{-t^2/2}$ into $\varphi (z)H_r(z)$, where the probabilists' Hermite polynomials satisfyThus the density Edgeworth expansion through order $n^{-1}$ isHere $H_3=z^3-3z$, $H_4=z^4-6z^2+3$, and $H_6=z^6-15z^4+45z^2-15$. Since $(\varphi H_{r-1}{)}^{\prime }=-\varphi H_r$, the corresponding cumulative distribution function expansion isThese remainder statements require adequate moments and smooth nonlattice conditions allowing transform inversion and error control, not merely a formal cumulant expansion. The density version needs the corresponding local smoothness. A symmetric distribution has ${\gamma }_3=0$, removing the first correction. Truncation can produce negative densities or nonmonotone distribution functions far from the center; it is not automatically a uniformly reliable relative tail approximation. A lattice distribution needs its lattice correction.

A Laplace approximation instead approximates an integral dominated by a sharp maximum. Suppose $h$ has a unique interior maximizer $\hat{\theta }\in {\mathbb{R}}^d$, with $A=-h^{\prime \prime }(\hat{\theta })$ positive definite, and the integral is localized near that point. For smooth amplitude $g$,Set $\theta =\hat{\theta }+z/\sqrt{n}$. A Taylor expansion givesExpanding $g$ and integrating the resulting Gaussian integral givesUnder sufficient smoothness and localization the relative correction is $O(n^{-1})$: odd terms of order $n^{-1/2}$ integrate to zero over the limiting symmetric Gaussian region. Higher terms use derivatives of $h,g$ and Gaussian moments.

For a likelihood function with ${\ell }_n=nh$ and a smooth positive prior density, this gives the integrated likelihood approximationwhere $j=-{\ell }_n^{\prime \prime }$ is the observed information. Ratios of such integrals give posterior expectation by Laplace approximation; the same method is useful when integrating nuisance variables. Boundary maxima, singular Hessian matrices, nonlocalized tails and several dominant modes require altered formulas, with contributions summed when appropriate.

Both methods start with a quadratic Gaussian approximation and then retain higher-order terms. The Edgeworth series corrects a sampling distribution through its cumulants and Fourier inversion; Laplace's method corrects an integral through local derivatives at its dominant point. Neither should be confused with simply replacing a likelihood by a normalized posterior density.

The phase $f(x)=x(1-\cos x)$ is nonnegative on the interval and vanishes only at its two endpoints. An interval away from both endpoints contributes exponentially little, so both endpoint contributions must be expanded before their orders are ranked.

At the left endpoint,The Laplace's method scale is $x={\sigma }^{-1/3}z$. Hence its leading contribution isThe substitution $v=z^3/2$ evaluates the integral by the Gamma function. Since there is no quartic term in the phase, the next left-end correction is relatively $O({\sigma }^{-2/3})$, and thus absolutely $O({\sigma }^{-1})$.

At the right endpoint put $r=2\pi -x$. Thereso $r={\sigma }^{-1/2}z$ givesThe leading right-end integral is a Gaussian integral. Its first correction is also absolutely $O({\sigma }^{-1})$. The mixed cubic and quadratic Laplace endpoints therefore produce the first two global termsKeeping only the dominant cubic endpoint would miss the second term, which is larger than that endpoint's own first correction.

Let $c=g^{\prime \prime }(\tilde{y})>0$. Near the interior minimum, Taylor expansion givesOn the scale $y=\tilde{y}+z/\sqrt{nc}$, the leading exponent is $ng(\tilde{y})+z^2/2$ and $dy=dz/\sqrt{nc}$. If the integral is localized at this minimum, the rescaled integration endpoints tend to infinity in both directions, so the Gaussian integral gives Laplace's method:Formally expanding the cubic term gives an order-$n^{-1/2}$ correction odd in $z$, whose Gaussian integral is zero. Quartic terms and the square of the cubic term produce the order-$n^{-1}$ correction. Sufficient smoothness near the minimum and control of the exterior integral justify this formal calculation.

The printed local assumptions alone do not guarantee localization. A useful localization condition for Laplace's method is that $g(y)\ge g(\tilde{y})+{\delta }_{\epsilon }$ outside each sufficiently small fixed neighbourhood of $\tilde{y}$, for some ${\delta }_{\epsilon }>0$, and that ${\int }_a^b{e}^{-n_{\ast }g(y)}dy<\infty$ for some $n_{\ast }>0$. For $n>n_{\ast }$, the exterior contribution is at mostwhich is exponentially smaller than the displayed local term. To see why something of this kind is needed, take $g(y)=y^2(1-y{)}^4$ on $(-1,1)$. Its unique minimum is at zero and $g^{\prime \prime }(0)=2$, yet on $1-n^{-1/4}<y<1$ we have $ng(y)\le 1$. This boundary interval alone contributes at least $e^{-1}{n}^{-1/4}$, dominating the proposed $n^{-1/2}$ local approximation. The Laplace formula therefore has its usual implicit tail assumption, rather than following from uniqueness of the interior minimum alone.

For the gamma function, set $y=nu$ to obtainHere $g(u)=u-\log u$ has its unique minimum at $u=1$, with $g(1)=1$ and $g^{\prime \prime }(1)=1$. It tends to infinity at both endpoints, and its integral is finite for positive exponents, so localization holds. Laplace's method yields Stirling's formula:

For a prior density $p$ and independent observations, Bayes' theorem writes the desired posterior expectation as a ratio of integrals:Suppose the log-posterior kernel has a unique concentrating interior mode $\tilde{\theta }$, with curvature $J_n=-H_n^{\prime \prime }(\tilde{\theta })>0$ of order $n$, and the smoothness and tail conditions hold for both integrals. Expanding $H_n$ quadratically gives a common Gaussian factor $e^{H_n(\tilde{\theta })}\sqrt{2\pi /J_n}$. Replacing the smooth amplitude $g(\theta )$ by $g(\tilde{\theta })$ givesThis approximation does not require $g$ to be positive: it treats $g$ as an amplitude, without taking its logarithm. For more accuracy, expand the amplitude and the cubic term in the log-posterior kernel. With $s=\theta -\tilde{\theta }$, the Gaussian moments are $\mathbb{E}{s}^2=J_n^{-1}$ and $\mathbb{E}{s}^4=3J_n^{-2}$. Terms multiplying $g(\tilde{\theta })$ cancel between numerator and denominator; the two remaining first corrections areunder the stronger smoothness, derivative-scaling and integrability assumptions needed for this expansion. The first term accounts for the curvature of the amplitude and the second for posterior skewness. This is posterior expectation by Laplace approximation. If there are several modes, boundary modes, or nonconcentrating tails, the appropriate separate contributions must be included; a single interior Gaussian approximation cannot then be assumed.

Multiply the equation by $x^2$. Its left side is an exact derivative, yielding the first integralFor a solution continuous at the singular endpoint, the limiting value is $C=-2{ϵ}^3\gamma$. This can also be established from the exact integrating-factor formula below, without assuming uniform bounds on derivatives near zero.

Away from zero, the outer expansion has leading term $y\sim C/x^2$. The endpoint value at one requires $C\sim {ϵ}^3$ if an exponentially amplified homogeneous term is to be avoided. It follows already thatTo exhibit the matching and explain the required boundedness, first set $x=ϵX$, $y=ϵY(X)$ for a representative without an order-one homogeneous contribution. The leading inner expansion obeysIntegration giveswhere $\mathrm{Ei}$ is the exponential integral. For $X\to \infty$, $Y\sim X^{-2}$, matching the outer term ${ϵ}^3/x^2$. For $X\to 0$, $Y\sim -1/X$, so the intermediate solution behaves as $-{ϵ}^2/x$ and reaches order one on a second scale $x={ϵ}^2\xi$.

On this second scale, the first integral becomes, at leading order, ${\xi }^2{Y}_0^{\prime }-2Y_0=1$. Equivalently its differentiated equation is $Y_0^{\prime \prime }+(2/\xi -2/{\xi }^2)Y_0^{\prime }=0$, as in the hint. The matching solution that tends to zero as $\xi \to \infty$ isIt has $Y_0\sim -1/\xi$ at large $\xi$, matching the intermediate region, and $Y_0(0)=-1/2$. Thus the nested layers at an essential singular endpoint connect an order-${ϵ}^3$ outer solution to an order-one endpoint value without an unbounded overshoot.

The issue of uniqueness is clarified by exact integration. PutFor every fixed $ϵ>0$ and every prescribed exact value of $\gamma$, the solution satisfying the right endpoint isAs $x\to 0$, the integral is asymptotic to $\mu (x)/(2{ϵ}^2)$, proving $y(0)=\gamma$. Since $h(1)>0$, the two exact endpoint values fix the two constants: for a prescribed exact $\gamma$, this finite-$ϵ$ boundary-value problem is unique.

However, specifying only the leading value $\gamma \sim -1/2$ and uniform order-one size does not select a unique asymptotic family. To see this constructively, choose $x_m=\sqrt{2}{ϵ}^{3/2}$ and an arbitrary bounded real constant $A$. WriteThe right endpoint fixesEndpoint Laplace's method gives $I=ϵ\mu (1)[1+2ϵ+O({ϵ}^2)]$. Therefore every bounded choice of $A$ haswith changes in $A$ affecting $\gamma$ only by $O({ϵ}^{-3}{e}^{-1/ϵ})$. These solutions are uniformly order one: near zero their particular part has the finite limit $\gamma$, near one it follows the small outer solution, and the freely chosen value at $x_m$ is $Ah(x_m)=O(1)$. The inner-layer homogeneous contribution is also bounded. Indeed $h$ vanishes at zero, is order one near $x_m$, and is exponentially small at one.

Consequently the order-one selected value is $-1/2$, but algebraic asymptotic endpoint data leave an order-one interior freedom. This is beyond-all-orders boundary-value sensitivity. If “constant” were interpreted as fixing $\gamma =-1/2$ exactly and independently of $ϵ$, the omitted algebraic corrections would excite an exponentially large interior homogeneous term; a genuinely uniformly bounded exact family requires the higher-order tuning of $\gamma$. Constancy in $x$ does not preclude that parameter dependence.

Under local integrability and suitable bounds away from zero, Watson's lemma givesEach term follows by $u=\lambda t$ and the Gamma function integral; a finite local remainder is bounded in the same way, and the region away from zero is exponentially small.

For Laplace's method, assume an isolated nondegenerate global minimum $t_{\ast }$ of $p$, with $p^{\prime \prime }(t_{\ast })>0$ and suitable tail bounds. Expand $p,q$ near $t_{\ast }$ and set $t-t_{\ast }=v/\sqrt{z}$. The leading term is $e^{-zp(t_{\ast })}q(t_{\ast })\sqrt{2\pi /(z{p}^{\prime \prime }(t_{\ast }))}$; successive terms come from Gaussian moments of the Taylor expansions. Several equally low minima contribute additively, while degenerate minima need a different scale.

Here the integrand is entire in $t$, so deform the contour to the line at imaginary part $-\pi$, the vertical segment from $-i\pi$ to $i\pi$, and the line at imaginary part $\pi$. The two horizontal tails sum to $2{\int }_0^{\infty }{e}^{-z\cosh x}dx=O(e^{-z}{z}^{-1/2})$. The vertical segment contributes $i{\int }_{-\pi }^{\pi }{e}^{z\cos \theta }d\theta$. Its maximum is at zero; putting $\theta =v/\sqrt{z}$ and using $\cos \theta =1-{\theta }^2/2+{\theta }^4/24+\cdots$ givesThe fourth Gaussian moment is three times the zeroth, soSince $\Gamma (1/2)=\sqrt{\pi }$, this has the required normalization.

Watson's lemma gives, under integrability away from zero and the stated endpoint asymptotic expansion,Indeed each power integrates over the positive half-line to $\Gamma (n\beta +1){\lambda }^{-(n\beta +1)}$; replacing a fixed upper endpoint by infinity has exponentially small error. A remainder $o(t^{N\beta })$ near zero gives $o({\lambda }^{-N\beta -1})$ after splitting off a small endpoint neighbourhood and rescaling $u=\lambda t$.

For Laplace's method, assume the phase and amplitude are smooth near the nondegenerate maximum, as required for a full expansion. Localize to a small neighbourhood of $c$: outside it the unique maximum and continuity make the exponential smaller by a factor $e^{-\lambda \epsilon }$. SetNear $c$ this is a smooth increasing coordinate, with ${\zeta }^{\prime }(c)=\sqrt{|{\varphi }^{\prime \prime }(c)|}$. It need not be monotonic on the entire integration interval. In the localized integral put $A(\zeta )=F(t(\zeta ))dt/d\zeta$. Extending its Gaussian integration to both infinite endpoints incurs exponentially small error. Taylor expansion of $A$ and integration of even Gaussian moments givesHere $A(0)=F(c)/\sqrt{|{\varphi }^{\prime \prime }(c)|}$, giving the leading term $\boxed{e^{\lambda \varphi (c)}F(c)\sqrt{2\pi /(\lambda |{\varphi }^{\prime \prime }(c)|)}}$. The asymptotic-equivalence notation for that term presumes $F(c)\ne 0$; if it vanishes the series determines the first nonzero term. The maximum condition alone, without amplitude regularity, would not imply a full expansion.

For the gamma function, set $t=xs$:The phase has unique maximum $-1$ at $s=1$, with second derivative $-1$. To obtain the correction, put $s=1+u/\sqrt{x}$. Expansion givesand hence a Gaussian multiplierThe odd term integrates to zero. The normalized Gaussian fourth and sixth moments are 3 and 15, so the correction is $-3/4+15/18=1/12$. Localization near one, with the exponentially suppressed endpoint tails, justifies termwise expansion. Thus Stirling's formula is

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.