 Solution

ID: past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2010/a/2/iii/solution

Fix

0 < ϵ < π /2

. On

[ϵ, π /2]

sin^{2} t \geq sin^{2} ϵ > 0

, so that part of the integral is bounded by

(π /2) e^{- x s i n^{2} ϵ}

, exponentially smaller than

x^{- 1/2}

By the assumed local equivalence, the remaining part is asymptotic to

\int_{0}^{ϵ} e^{- x t^{2}} d t

. Substitute

s = x t

\int_{0}^{ϵ} e^{- x t^{2}} d t = x^{- 1/2} \int_{0}^{ϵ x} e^{- s^{2}} d s \sim \frac{π}{2 x} .

The last step uses the Gaussian integral and monotone convergence of the upper limit to infinity. Hence

\int_{0}^{π /2} e^{- x s i n^{2} t} d t \sim \frac{1}{2} π / x .

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