Laplace's method
/ lah-PLAHSS /
Laplace's method is the real-variable ancestor of steepest descent, and the simplest member of the whole asymptotic family. Imagine an integral I(s) = (integral of e^(s h(t)) dt) over a real interval, with s large and h a real, smooth function. The factor e^(s h(t)) is overwhelmingly largest wherever h(t) reaches its maximum; raise s and that peak becomes a needle-thin spike. So to leading order the integral only cares about a small neighborhood of the maximizing point t_0.
Suppose h has a single interior maximum at t_0, so h'(t_0) = 0 and h''(t_0) < 0. Near t_0 we have h(t) approximately h(t_0) + (1/2) h''(t_0)(t - t_0)^2, a downward parabola, so e^(s h(t)) is approximately e^(s h(t_0)) times a Gaussian bump of width about 1 over the square root of (s |h''(t_0)|). Integrating that Gaussian gives the classic result: I(s) approximately e^(s h(t_0)) times the square root of (2 pi / (s |h''(t_0)|)) as s -> infinity. In one sentence: the value is the height of the peak times the width of the peak, and the width shrinks like 1 over the square root of s.
Laplace's method is everywhere statisticians and physicists look. It is the basis of the Laplace approximation to posterior distributions in Bayesian statistics, of the large-deviation and partition-function estimates in statistical mechanics, and of the derivation of Stirling's formula. The difference from steepest descent is just the setting: Laplace's method works on the real line with a real maximum; steepest descent lifts the same idea into the complex plane so it can handle oscillatory or complex exponents by deforming the contour through a saddle. The same honesty applies — it yields an asymptotic estimate, sharpened to all orders by Watson's lemma, not an exact value.
For Stirling's formula write n! = (integral from 0 to infinity of t^n e^(-t) dt) = (integral of e^(n log t - t) dt). With t = n s the exponent becomes n(log s - s), whose maximum over s is at s = 1 with second derivative -1. Laplace's method gives n! approximately e^(-n) n^(n+1) times the square root of (2 pi / n), i.e. n! approximately (square root of 2 pi n)(n/e)^n.
A sharp real peak at t_0 = n controls the factorial integral.
If the maximum sits at an endpoint of the interval rather than inside it, the peak is only half a Gaussian, so the formula picks up an extra factor of 1/2; and a degenerate maximum where h''(t_0) = 0 needs a different power of s.