Leading behavior of the integral $\int_0^{\pi/2}\sqrt{\sin t}e^{-x\sin^4t}dt$ as $x\to\infty$.

Question

$$\displaystyle\lim_{x\to\infty}\int_0^{\pi/2}\sqrt{\sin t}e^{-x\sin^4t}dt$$

I want to apply Laplace's method. And since the maximal of $-\sin^4t$ achieves at $t=0$, the contribution to the integral is at 0, but at the same time $\sqrt{\sin t}$ attains zero. So does that mean the leading behavior of the integral is just $0$? If so, how can we rigorously(kind of) show that?

If you're interested in learning about the justifications for approximations such as these you might be interested to read the appendix of my thesis. It's intended to be an instructive introduction to the asymptotic analysis of integrals such as these. You can download it here: https://arxiv.org/abs/1610.03387 — Antonio Vargas, Oct 12 '16 at 14:01

score 2 · Answer 1 · answered Oct 02 '16 at 21:03

2

Near $t=0$ we have

$$ \sin^4 t \sim t^4 \qquad \text{and} \qquad \sqrt{\sin t} \sim \sqrt{t}, $$

so by the Laplace method

$$ \int_0^{\pi/2} \sqrt{\sin t} e^{-x \sin^4 t}\,dt \sim \int_0^\infty \sqrt{t} e^{-xt^4}\,dt = \frac{1}{4} \Gamma\!\left(\frac{3}{8}\right) x^{-3/8} $$

as $x \to \infty$.

answered Oct 02 '16 at 21:03

Antonio Vargas

25,581

The justification for this is standard, see any text which goes into detail on the Laplace method. – Antonio Vargas Oct 02 '16 at 21:04

Leading behavior of the integral $\int_0^{\pi/2}\sqrt{\sin t}e^{-x\sin^4t}dt$ as $x\to\infty$.

1 Answers1