Asymptotic Expansion of Integrals in general form

Question

How to understand asymptotic expansion of $$ I(s) = \int_0^1 \frac{dx}{\sqrt{1+s f(x)}}$$ where $f$ is periodic of period 1, $f(x) \geq 0$ as $s\longrightarrow \infty$?

Answer 1

just a small hint

Let $$J (s)=I (\frac {1}{s}) $$

$$=\int_0^1\sqrt {\frac{s}{f (x)}}\frac {dx}{(1+\frac{s}{f (x)})^\frac12} $$

use that near $t=0$,

$$(1+t)^\frac12=1+\frac {t}{2}+... $$