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 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.

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.

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