How do I get the asymptotic expansion for the integral
$$ I(R) = \int_0^1 \frac{e^{- R x}}{\sqrt{1-x}}dx,$$
when $R \to +\infty$?
I was able to relate this integral to the Dawson function $F(x)$, so that
$$ I(R) = \frac 2 {\sqrt R} F(\sqrt R) ,$$
and from there I was able to get the
$$ I(R) = \frac{1}{R} + \frac{1}{2R^2} + \ldots, ~~~ {\rm as}~~~ R\to +\infty.$$
But what is the simplest way to get this directly from the original integral?
Simple integration by parts does not seem to do the trick here.