Laplace's method

If a smooth real phase $\varphi$ has a unique nondegenerate interior maximum at $t_0$, thenFor a maximum at an endpoint $b$ with ${\varphi }^{\prime }(b)>0$, the leading factor is $a(b)e^{x\varphi (b)}/(x{\varphi }^{\prime }(b))$.

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.

At a unique interior maximum with positive-definite negative Hessian matrix, a quadratic expansion and rescaling by $\sqrt{n}$ give the displayed Gaussian integral, provided the tails are negligible. In a regular Bayesian inference it approximates the integrated likelihood by the likelihood at its dominant point times the prior density and inverse square root of local curvature. Boundary modes, singular Hessians, and multiple equally dominant modes need modified formulas.

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.

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.

If an odd amplitude multiplies an even Gaussian phase on an interval asymmetric about its interior minimum, the symmetric central integral cancels exactly. The surviving leading term can be exponentially small and comes from the nearer unmatched endpoint. For example,An interior Laplace method expansion alone yields zero to every algebraic order and does not capture this term.

If a real smooth phase has its strict maximum at the upper endpoint $b$ and $\varphi \prime (b)>0$, then ${\int }_a^{b}a(t)e^{\lambda \varphi (t)}dt\sim a(b)e^{\lambda \varphi (b)}/(\lambda \varphi \prime (b))$. The endpoint coordinate $b-t$ has scale $1/\lambda$. At a nondegenerate stationary endpoint the scale is instead ${\lambda }^{-1/2}$ and the leading contribution is half a Gaussian integral.

A Laplace integral contains a large parameter in an exponential weight, typically in the form $\int a(t)e^{x\varphi (t)}dt$ or ${\int }_0^b{e}^{-xt}f(t)dt$. Its asymptotic behavior is controlled by the points where the real part of the phase is largest.

For a contour integral with factor $e^{{\zeta }^4/4+z\zeta }$, the quartic term dominates the linear term on rays at large radius. Writing $\zeta =r{e}^{i\theta }$ gives real part $r^4\cos (4\theta )/4$ for the leading exponent. There are four decay sectors, centered at $\theta =\pi /4+j\pi /2$ and each of angular width $\pi /4$. On a closed subsector the integrand and all its $z$ derivatives decay faster than any exponential in $r$, locally uniformly in $z$. Contours joining two such sectors therefore allow differentiation under the integral sign and have vanishing infinite endpoint terms. Distinct geometric contours still need an independence proof; their initial moments often give a Vandermonde matrix calculation.

A linear differential-equation solution may be sought as ${\int }_C{e}^{xt}f(t)dt$. Integration by parts transfers the factor $x$ to derivatives of $f$, giving an equation for the amplitude and boundary terms. Admissible contours must cancel those terms and make the integral converge.

A null weight contributes zero to a Laplace integral on its domain of convergence. It can be added to a weight without changing the represented function, so an integration-by-parts differential equation for the weight is not an unconditional uniqueness theorem.

For example, let $\Gamma$ enter along angle $\pi /3$ from infinity to zero and leave along angle $2\pi /3$ to infinity. The entire weight $h(\zeta )=e^{{\zeta }^2}$ is null for every complex $z$. Throughout the wedge between those rays, $\mathrm{Re}({\zeta }^2)\le -r^2/2$; the factor $e^{z\zeta }$ is dominated by this quadratic decay. Closing the contour by a large circular arc and applying the Cauchy integral theorem therefore proves the zero integral. All derivatives with respect to $z$ converge locally uniformly as well.

On the same contour, the quartic weight $e^{{\zeta }^4/4}$ gives a nontrivial entire solution. Its rays can be deformed within the quartic decay sectors for a contour integral to angles $\pi /4$ and $3\pi /4$, where the value at $z=0$ is $-\sqrt{2}{4}^{-3/4}\Gamma (1/4)\ne 0$. Adding the null weight changes neither this integral nor its vanishing endpoint terms, but the sum does not satisfy $g^{\prime }={\zeta }^{3}g$. Thus the usual Laplace method selects a convenient weight by cancelling the interior integrand separately; a necessity claim needs additional injectivity assumptions.

If the maximum $t_0$ of the smooth real phase $\varphi$ is an endpoint, ${\varphi }^{\prime }(t_0)=0$, and ${\varphi }^{\prime \prime }(t_0)<0$, the local Gaussian is integrated over only one side of its peak. Consequentlywith the analogous formula for a right endpoint.

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.

Suppose a real phase has its unique minimum at the left endpoint and $\varphi (t)-\varphi (0)\sim c{t}^m$, with $c>0$ and integer $m\ge 1$. For a continuous amplitude $a$ with $a(0)\ne 0$, localization followed by $s=(\nu c{)}^{1/m}t$ givesThis uses the Gamma integral. It assumes sufficient decay away from the minimum to make the remaining integral negligible. A cubic endpoint has width ${\nu }^{-1/3}$; an ordinary quadratic endpoint has width ${\nu }^{-1/2}$.

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 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.

After $t=zs$, the phase in the integral for ${\Gamma }^{\prime }(z)$ becomes $\log s-s$, with maximum $-1$ and second derivative $-1$ at $s=1$. Hence