Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\varphi \in C^\infty (\mathbb R^n ;\mathbb R)$ such that

1) $\varphi(0)=0$

2) $\varphi(x)>0$ on $\mathbb R^n\setminus 0$

3) $\text{Hess}_{\varphi}(0)>0 $

and let $B_1(0)$ be the unitary ball around $0$. Then the classical Laplace method gives that for $h>0$ small

$\int_{B_1(0)} e^{-\varphi(x)/h} dx \sim h^{n/2}\sum_{k=0}^\infty h^k b_k$

where $(b_k)$ is a sequence in $\mathbb R$.

My question is: if I have also a function $g\in C^\infty (\mathbb R^n ;\mathbb R)$ such that $g(0)=0$, is it still true that there is a sequence $(c_k) \in \mathbb R$ such that

$\int_{B_1(0)} e^{-(\varphi(x) + \sqrt h g(x))/h} dx \sim h^{n/2}\sum_{k=0}^\infty h^k c_k$

or do there appear also half powers of $h$ in the expansion?

share|cite|improve this question
up vote 1 down vote accepted

Why not try the simplest example available: $\varphi(x)=x^2$ and $g(x)=x^2$, in one dimension? We get $$\int_{-1}^1 \exp(-(h^{-1}+h^{-1/2})x^2)\sim \sqrt{\pi}\left(h^{-1}+h^{-1/2}\right)^{-1/2}$$ because the Gaussian tail decays faster than any power of $h$ as $h\to 0$. Here $$h^{1/2}\left(1+h^{1/2}\right)^{-1/2}=h^{1/2}\left(1-\frac{1}{2}h^{1/2}+\frac38 h-\frac{5}{16}h^{3/2}+\dots\right)$$ so the half-integer powers of $h$ do appear in the sum.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.