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I am using these notes as my reference, but I am running into some questions.

Say I am trying to find, for large $\lambda$ $$I(\lambda)=\int_0^{\pi/2}dxe^{-\lambda\sin^2(x)}$$ This has our maximum at $c=0$, where g(c)=0 and g'(c)$\neq$0. So when I pull out the $e^{\lambda g(c)}$, that is just 1, so I continue to expand g(x) about x = 0, and then change the bounds from -Infinity to Infinity. $$I(\lambda)\approx\int_{-\infty}^\infty dx\exp[-\lambda(x^2-x^4/3+...)]$$ To first order, this is just $\sqrt{(\pi/x)}$, but the second order diverges.

The reason this stumps me is because I can find an exact solution that depends on $e^x$ so I am not sure where i am going wrong in computing the behavior.

yankeefan11
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  • Instead of $$I(\lambda)\approx\int_{-\infty}^\infty dx\exp[-\lambda(x^2-x^4/3+...)]$$ you actually have $$I(\lambda)\approx\int_{0}^\infty dx\exp[-\lambda(x^2-x^4/3+...)]$$ because of the one-sided maximum at $x=0$. – Antonio Vargas Oct 12 '15 at 17:57
  • Still blows up at infinity – yankeefan11 Oct 12 '15 at 18:13
  • Perhaps you misunderstood: the idea is that the series in the exponent in the expression I wrote can be truncated at any point. You are not meant to sum the series then integrate from $0$ to $\infty$. You are meant to truncate the series, say just at $x^2$, then integrate. – Antonio Vargas Oct 12 '15 at 18:16
  • True. However, I am looking for an expression good through second order, so wouldn't I need to keep that term, OR, do I expand the results, (which would be 1/sqrt(x)) to 2 terms? – yankeefan11 Oct 12 '15 at 18:24
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    To get the higher-order corrections, or even a full asymptotic expansion, you can follow the suggestion given by Chappers. Try setting $x = \arcsin \sqrt{y}$ in the original integral, then applying Watson's lemma. – Antonio Vargas Oct 12 '15 at 18:30
  • You might like to look at these similar questions: one, two, and three. – Antonio Vargas Oct 12 '15 at 18:36
  • So I do the substitution to get

    $$I = \int_0^1 \frac{dy e^{-\lambda y}}{2\sqrt y\sqrt{1-y}}$$

    But can Watsons be used, because the g(y) blows up to infinity as I get to 0?

     – yankeefan11 Oct 12 '15 at 18:54
  • Ignore that last comment, I was able to work through it with Watsons lemma. Thanks for the help! – yankeefan11 Oct 12 '15 at 19:22
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    Great, glad to help. :) I suggest adding an answer to this question yourself to give the details on your calculation. – Antonio Vargas Oct 12 '15 at 19:23

3 Answers3

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Note that you have an endpoint maximum, and hence the integral should only be one-sided; this is the sort of case where Watson's lemma applies.

Chappers
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You can take advantage of the symmetry of your integrand to make the maximum point an interior point: $I(\lambda)=\int_0^{\pi/2} e^{-\lambda\sin^2(x)}\,dx ={1\over 2}\int_{-\pi/2}^{\pi/2}e^{-\lambda\sin^2(x)}\,dx$. Laplace's Method then yields $I(\lambda)\sim \sqrt{\pi\over 4\lambda}$. This is consistent with your exact evaluation of $I(\lambda)$ and the known large-$\lambda$ asymptototics for the Bessel function $I_0$: $I_0(x)={e^x\over\sqrt{2\pi x}}+e^x\cdot O(x^{-3/2})$.

John Dawkins
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That integral can be computed in a explicit way by using modified Bessel functions of the first kind:

$$ \int_{0}^{\pi/2} e^{-\lambda\sin(x)^2}\,dx = \frac{\pi}{2}e^{-\lambda/2}\,I_0(\lambda/2) \tag{1}$$ then the asymptotic form follows from Hankel's expansion: $$ \int_{0}^{\pi/2} e^{-\lambda\sin(x)^2}\,dx \approx \sqrt{\frac{\pi}{4\lambda }}\left(1+\frac{1}{4\lambda}\right).\tag{2} $$

Jack D'Aurizio
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  • So from Hankel's expansion, do you need to know that it can be explicitly calculated first? – yankeefan11 Oct 12 '15 at 17:22
  • Well, Hankel's expansion applies to Bessel functions. If you do not know you are dealing with a Bessel function, you may still prove $(2)$ through the usual Laplace method / Watson lemma. – Jack D'Aurizio Oct 12 '15 at 17:25