Laplace's Method
Table of Contents
For "sharply peaked" integrands, we can approximate the integral since the dominant contribution comes from the neighborhood of a single point (the peak). Laplace-type integrals are of the form
\begin{equation} I(x) = \int^{b}_{a} f(t)\E^{x\phi(t)}\,\D t \end{equation}as \(x\to\infty\) and are prime candidates for using Laplace's method.
1. Examples
Consider, as \(x\to\infty\),
\begin{equation} I(x) = \int^{10}_{-10}\E^{-xt^{2}}\,\D t. \end{equation}This is a Laplace-type integral, with \(f(t)=1\) and \(\phi(t)=-t^{2}\). But we can't apply integration-by-parts since \(\phi'(0)=0\).
A crude estimate: draw the curve then determine its width when
\begin{equation} \E^{-xt^{2}}\sim 1 \end{equation}which happens when \(t=1/\sqrt{x}\). If we pretend the integrand is a rectangle, we get \(I(x)\sim\mathcal{O}(1/\sqrt{x})\). And it's correct!
Let's examine the tail. The tail region where \(|t|\gg 1/\sqrt{x}\) contribute exponentially small terms. We don't need to examine the tails: they're transcendentally small terms, they do not affect the asymptotic expansion. They're "subdominant".
Now, we take
\begin{equation} \begin{split} I(x) &= \int^{10}_{-10}\E^{-xt^{2}}\,\D t=\left(\int^{\infty}_{-\infty}\E^{-xt^{2}}\,\D t\right) + \mbox{(TST)}\\ &= \sqrt{\frac{\pi}{x}} + \mbox{(TST)} \end{split} \end{equation}where "TST" stand for "transcendentally small terms".
Question: How do we know the remainder terms really are "transcendentally small terms"?
Solution. We observe the remainder term is
\begin{equation} R(x) = -2\int^{\infty}_{10}\E^{-xt^{2}}\,\D t. \end{equation}We need to come up with a bound to reflect the remainder term really is a transcendentally small term. For \(t\gt 10\) we see \(t^{2}\gt 10t\), hence
\begin{equation} \begin{split} |R(x)| &\leq 2\int^{\infty}_{10}\E^{-x10t}\,\D t = \left.\frac{\E^{-10xt}}{-10x}\right|^{t=\infty}_{t=10}\\ &\leq\frac{\E^{-100x}}{100x}. \end{split} \end{equation}Hence we have our hard bound
\begin{equation} R(x)\lt \frac{\E^{-100x}}{100x} \end{equation}which is clearly as transcendentally small term.
Now consider
\begin{equation} I(x) = \int^{\infty}_{-\infty}\E^{-x\cosh(t)}\,\D t. \end{equation}To leading order,
\begin{equation} \cosh(t)\approx 1 + \frac{t^{2}}{2} + \mathcal{O}(t^{4}) \end{equation}as \(t\to 0\). When \(t\) is large, we're out in the tails, and it does not matter that this approximation is no good! Then we have
\begin{equation} \begin{split} I(x) &\sim\int^{\infty}_{-\infty}\exp\left[-x\left(1 + \frac{t^{2}}{2}\right)\right]\D t\\ &\sim\E^{-x}\int^{\infty}_{-\infty}\E^{-xt^{2}/2}\,\D t=\sqrt{\frac{2\pi}{x}}\E^{-x}. \end{split} \end{equation}Again, the remainder is a transcendentally small term.