laplace method on this integral

Question

How to get the leading asymptotic expansion for this integral $\int_{0}^{\pi/2}\sqrt{\sin(t)}\exp(-x\sin^4(t))dt $ in the limit $x\rightarrow\infty$ ? Because the maximum of the exponent is at $t=0$ for which the function in front of the exponential goes to 0.

Answer 1

Since the contribution of every interval $(\varepsilon,\pi/2)$ is at most $\mathrm e^{-Kx}$, one can concentrate on $t$ in a neighborhood of $0$ and use the change of variable $s=x\sin^4(t)$, which yields, when $x\to+\infty$, the equivalent $$\int_0^\infty (s/x)^{1/8}\mathrm e^{-s}\frac{\mathrm ds}{4x^{1/4}s^{3/4}}=\frac{1}{4x^{3/8}}\int_0^\infty s^{-5/8}\mathrm e^{-s}\mathrm ds=\frac{\Gamma(3/8)}{4x^{3/8}}.$$ More generally, if $u(t)\sim t^a$ and $v(t)\sim t^b$ when $t\to0$, and if $u$ and $v$ are uniformly bounded below by some positive constants on every interval $(\varepsilon,T)$ for some fixed $T$, then, when $x\to+\infty$, $$\int_0^Tu(t)\mathrm e^{-xv(t)}\mathrm dt\sim\frac{\Gamma(c)}{bx^c},\qquad c=\frac{a+1}b.$$