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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?

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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.

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