TITLE: Integrate $(ax)^{-x}$
QUESTION [5 upvotes]: I need to find the asymptotics of the following expression as $r\to\infty$:
$$\int_{r^2+r}^\infty e^{-r^2}\Big(\frac{er^2}{x}\Big)^x \,dx.$$
Clearly,
$$\int_{3r^2}^\infty \Big(\frac{er^2}{x}\Big)^x \,dx \to 0,$$
so the main contribution comes from the part near the lower bound. Estimating
$$\int_{r^2+r}^{r^2+Cr} \Big(\frac{er^2}{x}\Big)^x\, dx$$
yields a lower bound of order $Cr$.
My guess is that the above integral behaves like $r\log r$ but I have no idea how to show this.

REPLY [2 votes]: Let $x = r^2 t, f(t) = r^2 t (1 - \ln t)$. The integral of $e^{f(t)}$ is determined by the maximum at $t = 1$, but we need to subtract the contribution from the interval $[1, 1 + 1/r]$:
$$f(t) = r^2 - \frac {r^2} 2 (t - 1)^2 + O((t - 1)^3), \\
e^{-r^2} \int_{r^2 + r}^\infty \left( \frac {e r^2} x \right)^x dx =
e^{-r^2} r^2 \int_{1 + 1/r}^\infty e^{f(t)} dt \sim \\
r^2 \int_{1 + 1/r}^\infty e^{-r^2 (t - 1)^2/2} dt =
r \sqrt \frac \pi 2 \operatorname{erfc} \frac 1 {\sqrt 2}.$$