Asymptotic of a complex integral

Question

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.

Answer 1

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.