Asymptotic of a complex integral Consider the following integral
$$f(x):=\int_x^{+\infty}re^{-(r+ir^2)}dr$$
I want to understand the asymptotic behavior of $f(x)$ as $x\rightarrow +\infty$ 
Thank you for any suggestion.

I would start by trying to make the substitution $r+ir^2 = s$ to get
$$
r\,dr = \left(-\frac{i}{2} \pm \frac{i}{2\sqrt{1+4is}}\right) ds,
$$
where either $+$ or $-$ is chosen to pick the right branch of the square root.  In either case we have
$$
r\,dr \sim -\frac{i}{2}\,ds
$$
for large $r$ and $s$, so to first order we should have
$$
f(x) \approx -\frac{i}{2}\int_{x+ix^2}^\infty e^{-s}\,ds = -\frac{i}{2} e^{-(x+ix^2)}
$$
as $x \to \infty$.
This agrees with the numerics.  Below is a plot of $\operatorname{Re} \left( e^x f(x) \right)$ in blue vs. $\operatorname{Re} \left(-\frac{i}{2} e^{-ix^2}\right) = -\frac{1}{2} \sin x^2$ in red.

The imaginary parts agree as well.
