Laplace's Method for Integral asymptotes when g(c) = 0

Question

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.

Answer 1

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.

Answer 2

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: $$ \int_{0}^{\pi/2} e^{-\lambda\sin(x)^2}\,dx \approx \sqrt{\frac{\pi}{4\lambda }}\left(1+\frac{1}{4\lambda}\right).\tag{2} $$

Answer 3

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 $I_0$: $I_0(x)={e^x\over\sqrt{2\pi x}}+e^x\cdot O(x^{-3/2})$.