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?
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$\begingroup$ I've used Laplace's Method on a single function $q$ rather than a family varying with $c$. If you let $q$ depend on $c$, don't you need some sort of uniform bound on the rest of the function that is automatic when $q$ doesn't depend on $c$? By the way, I think that should be $\exp(\frac{27}{256}c^4)$ in the last line, as $\exp(cq(x_0)) \ne\exp(cx_0).$ $\endgroup$– Douglas ZareSep 21, 2012 at 10:54
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2$\begingroup$ You can transform this to a problem with a fixed function q(x) by first substituting $x=c^3y$ and then applying Laplace's method to the transformed problem. $\endgroup$– Michael RenardySep 21, 2012 at 14:19
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$\begingroup$ @Douglas - corrected, thank you! @Michael - true. I just ordered Bleistein and Handelsman's Asymptotic Expansions of Integrals, which will hopefully lead to a fuller understanding of this technique. $\endgroup$– Michael TinkerSep 21, 2012 at 17:38
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$\begingroup$ @Douglas - the issue of uniformity bothers me a lot as well, since I also need to estimate things like $\int_{\mathbb{R}}e^{\eta x -\sigma x^{2} - \tau x^{4}}dx$ where $\eta,\sigma \in \mathbb{R}$ and $\tau \in \mathbb{R}^{+}$. $\endgroup$– Michael TinkerSep 21, 2012 at 17:52
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$\begingroup$ In addition to @Michael's comment, this seems to be a standard Watson's lemma integral. $\endgroup$– Igor RivinOct 6, 2012 at 1:16
1 Answer
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 $f(x,c) = c q(x,c).$
alt text http://www.freeimagehosting.net/newuploads/txpgu.png
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.
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$\begingroup$ @Douglas - appreciated. My professor, when recommending Bleistein's book, said he had vague memories of such a uniformity condition. (I.e. a condition for when $\int_{\mathbb{R}}e^{f(x,c)}dx$ admits an asymptotic estimate in the parameter $c$.) In some days or weeks I will report back with a verdict on the accuracy of his recollection. :) $\endgroup$ Sep 22, 2012 at 6:10
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