Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2010 / A / 2 / iii / Solution

Fix $0<ϵ<\pi /2$. On $[ϵ,\pi /2]$, ${\sin }^{2}t\ge {\sin }^2ϵ>0$, so that part of the integral is bounded by $(\pi /2)e^{-x{\sin }^2ϵ}$, exponentially smaller than $x^{-1/2}$.

By the assumed local equivalence, the remaining part is asymptotic to ${\int }_0^{ϵ}{e}^{-x{t}^2}dt$. Substitute $s=\sqrt{x}t$:The last step uses the Gaussian integral and monotone convergence of the upper limit to infinity. Hence