---
tag: 14B9FE49V7
kind: definition
name: Laplace's method
msc: [41A60, 26A42]
defines: 9G7K8DBT9Y
uses: [00SB5EA3VW, 0Y9D2FJRQ4, 04CPG9K1C4]
sources:
  - url: https://en.wikipedia.org/wiki/Laplace%27s_method
---

## Definition

Let $a < b$ be real numbers and let $f \colon [a, b] \to \R$ be twice continuously differentiable.
Assume that $f$ has a unique global [[maximum|04CPG9K1C4]] on $[a, b]$ at a point $x_0$ with $a < x_0 < b$, and that the [[second derivative|0Y9D2FJRQ4]] satisfies $f''(x_0) < 0$.
*Laplace's method* is the approximation of $I(M) = \int_a^b e^{M f(x)}\, dx$ for large $M$ by the integral of $e^{M (f(x_0) + \frac12 f''(x_0)(x - x_0)^2)}$ over $\R$.
Under these assumptions it gives the [[asymptotic equivalence|00SB5EA3VW]]
$$\int_a^b e^{M f(x)}\, dx \sim \sqrt{\frac{2\pi}{M \abs{f''(x_0)}}} \; e^{M f(x_0)} \qquad (M \to \infty).$$

## Remarks

Example: Stirling's formula.
Substituting $t = n s$ gives $n! = \int_0^\infty t^n e^{-t}\, dt = n^{n+1} \int_0^\infty e^{n(\ln s - s)}\, ds$.
The function $f(s) = \ln s - s$ has its maximum at $s = 1$, with $f(1) = -1$ and $f''(1) = -1$, and after a check that the tails are small the method gives $n! \sim \sqrt{2\pi n}\, (n/e)^n$.

Non-example: if the maximum is at an endpoint, for instance $f(x) = -x$ on $[0, 1]$, the hypothesis $a < x_0 < b$ fails.
Here $\int_0^1 e^{-Mx}\, dx = (1 - e^{-M})/M$, which decays like $1/M$ and not like $1/\sqrt{M}$.

The same idea applies to $\int_a^b g(x) e^{M f(x)}\, dx$ with $g$ continuous and $g(x_0) \ne 0$, where the right side gains the factor $g(x_0)$.