# How to give the order error term in Laplace asymptotic methods about integral?

I don't know why the order of error term is $O(n^{-1})$ or more high in asymptotically computing integral using Laplace's method? For instance, the following examples:

Suppose that $h(\theta)$ is a real function, has a unique manimum at $\hat{\theta}$ and has
continuously second derivative, then we get $\int_{-\infty}^{+\infty}e^{-nh(\theta)}d\theta=e^{-nh(\hat{\theta})}(\frac{2\pi}{nh^{''}(\hat{\theta})})^{\frac{1}{2}}(1+O(n^{-1})),$ where $h^{''}(\hat{\theta})\neq 0.$

But, I have computed this asymptotic expression many times, I still can not get the order of error term is $O(n^{-1}).$ On contrary, I can only get the order is $O(n).$ I do not know why.

There is another question on this expression, which is what is the $n$ that is arbitrary nature number. A lot of literature only use this result give some illustrations and do not give some proofs.

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Thank very much to Zarrax and Wesley for their unusually helpful answers! – user21535 Dec 20 '11 at 14:23

In 1989, Tierney, Kass, and Kadane answered this question far better than I could:

http://www.stat.cmu.edu/~kass/papers/nonlinearFunctions.pdf

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This is the basic idea... I think I'm assuming $h(\theta)$ is $C^4$, but anyhow:

First note you can factor out $e^{-nh(\hat{\theta})}$, so you can replace $h(\theta)$ by $h(\theta) - h(\hat{\theta})$ and assume $h(\hat{\theta}) = 0$. Then pick $\epsilon > 0$ such that $h(\theta)$ is of the form $c(\theta - \hat{\theta})^2 + O((\theta- \hat{\theta})^3)$ for $|\theta - \hat{\theta}| <\epsilon$, where ${\displaystyle c = {h''(\hat{\theta}) \over 2}}$. You only have to worry about the integral for $|\theta - \hat{\theta}| <\epsilon$ since the rest of the integral decays much faster than the inside.

You can always make a change of variables to $y = (\theta - \hat{\theta})+ O(\theta - \hat{\theta})^2$ so that the integral becomes that of $e^{-ncy^2}$. However you get a Jacobian factor of the form $1 + ay+ O(y^2)$. So your integral is $$\int_{-\epsilon}^{\epsilon} e^{-ncy^2}(1 + ay + O(y^2))\,dy$$ The first term tends to the main asymptotic term as $n$ goes to infinity. The second is $$a\int_{-\epsilon}^{\epsilon} ye^{-ncy^2}\,dy$$ Being the integral of an odd function, you just get 0. The final term is bounded in absolute value by $$C\int_{-\epsilon}^{\epsilon} y^2e^{-ncy^2}\,dy$$

If you change variables to $z = \sqrt{n}y$, you get a term asymptotic to ${C' \over n^{3 \over 2}}$ for some $C'$ as $n \rightarrow \infty$.

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