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?

Answer 1

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 $x \to \infty$.