Prove that that two integrals are asymptotically equal I am currently reading a paper where the following claim is used without justification $$\int_0^\infty \exp\left[-\frac{1}{2}x^2\left(1-\frac{3}{n}\right)-\frac{x^4}{n^2}\right]x^{n-1}dx\sim e^{-1}\int_0^\infty \exp\left[-\frac{1}{2}x^2\left(1-\frac{3}{n}\right)\right]x^{n-1}dx$$where $\sim$ denotes asymptotic equality i.e. $f(n)\sim g(n)$ implies $$\lim_{n\to \infty}\frac{f(n)}{g(n)}=1$$ provided $g(n)\neq 0$. I think the author implicitly uses Laplace's method; however, the maximizer of the integrand seems pretty complicated. It'd be great if someone could explicate the steps in this (asymptotic) equality.
 A: Essentially, the idea is that when using Laplace's method, since $n \to \infty$, only the leading-order approximation to the maximum will appear in the non-dominant terms at leading order. The logarithm of the integrand is
$$ (n-1)\log x - \frac{x^2}{2}(1+3n^{-1}) - x^4 n^{-2} , $$
and for large $n$ the dominant terms are, $n\log x - x^2/2$ (the other parts of these factors are less important because they will be smaller than the "main" parts). Differentiating these terms gives
$$ \frac{n}{x} - x = 0 \implies x = \sqrt{n} . $$
Plugging back in, we can verify this is consistent by finding the error term: including all of the terms in the derivative,
$$ \frac{n}{x} - x + \frac{1}{x} - \frac{3x^2}{2n} - \frac{x^4}{n^2}: $$
put $x=\sqrt{n}+u$, and then
$$ 0 = \frac{n}{\sqrt{n}+u} - \sqrt{n}-u + \frac{1}{\sqrt{n}+u} - \frac{3(n+\sqrt{n}u+u^2)}{2n} - \frac{(\sqrt{n}+u)^4}{n^2} , $$
and expanding in $n$ gives
$$ -2u -(4-u^2) \frac{1}{\sqrt{n}} + O(1/n), $$
and then if $u = -2/\sqrt{n}+O(1/n)$, the error is $o(1/\sqrt{n})$ as we might hope. (This rather suspicious-looking method of guess-and-correct is rigourisable, but one does not bother with this in practice.)
The upshot is that we get the correct answer for the leading-order term if we do this approximation in the non-dominant terms in the integrand. In the paper you cite they confine themselves to doing this for the $\exp(-x^4/n^2) \sim e^{-1}$ for some reason, rather than also including the $-3x^2/(2n)$ term, which is the same size. One can then resume applying Laplace's method to get the rest of the leading-order term.
