Question

I am trying to say something about the asymptotics of $$\int_{\mathbb{R}} e^{cx - x^{4/3}}dx$$ as $c \to +\infty$, and need a sanity check. As I understand it, Laplace's method is to write $$q(x) = x-c^{-1}x^{4/3}$$ and note that $q(x)$ has a global maximum at $x_{0} = \frac{27c^{3}}{81}$. Then it follows $$\int_{\mathbb{R}}e^{cq(x)}dx \lesssim e^{c q(x_{0}})\int_{\mathbb{R}} e^{-c|q''(x_{0})|(x-x_{0})^{2}/2}dx$$ And because the right hand integral is a Gaussian, in fact $$\int_{\mathbb{R}}e^{cq(x)}dx \lesssim e^{c q(x_{0})}\sqrt{\frac{2\pi}{c|q''(x_{0})|}}$$ So explicitly in my case $$\int_{\mathbb{R}} e^{cx - x^{4/3}}dx \lesssim \frac{9\sqrt{2\pi}}{8}ce^{\frac{27}{256}c^{4}}$$ My concern is about $q''(x) = -\frac{4}{9c}x^{-2/3}$. The proofs of Laplace's method I have seen only APPEAR to require $q''(x)$ be continuous in a neighborhood of $x_{0}$, with $q''(x_{0}) < 0$. This will certainly hold for me when $c > 0$. But am I missing something?

Answer 1

The version of Laplace's Method I know uses that $q(x)$ does not depend on $c$. If you try to extend it, you need some extra uniformity condition to show that the rest of the function does not contribute. Here is a counterexample when you only assume that the maximum is always at $x_0$ and the second derivative is constant there.

Let the interval be $[-1,1]$ and let $q(x,c) = \max (-x^2,-1/c^2).$ This means $q(x,c) = -x^2$ in a neighborhood of $0$ whose width depends on $c$, $[-1/c, 1/c]$ and it is flat outside. Let $f(x,c) = c q(x,c).$

For each $c$, $q(x,c)$ has a global maximum at $x=0$ and the second derivative with respect to x there is -2. However,

$$\int_{-1}^1 e^{f(x,c)} dx ~\large{\nsim} ~ e^{c q(x_0,c)} \sqrt{\frac{2\pi}{c|q''(x_0,c)|}}= \sqrt{\frac{\pi}{c}}.$$

In fact,

$$\lim_{c \to \infty}\int_{-1}^1 e^{f(x,c)} dx = 2.$$

The part of $\exp(f(x,c))$ near $0$ doesn't dominate the integral.