Rate of Divergence of an Infinite Integral

Question

Let $$G(n)=\int_{1}^{\infty}x^{-x}x^{n}dx.$$ Clearly, for each fixed $n$, $G(n)<\infty$ and $\lim_{n\rightarrow \infty}G(n)=\infty$.

Question: At what rate of $n$ does $G(n)$ go to infinity?

Answer 1

Disclaimer. This is not a complete answer.

---

Let $W$ be the Lambert W-function. Then the maximum of $x^{n-x}$ occurs at $x = b$, where we set

$$a = \frac{\sqrt{n}}{W(en)} \qquad \text{and} \qquad b = \frac{n}{W(en)}. $$

Here, we suppress the dependence on $n$ from the notation for simplicity. Then with the substitution $x = b + at$,

$$ G(n) = a \cdot b^{n-b} \int_{\frac{1}{a}-\sqrt{n}}^{\infty} b^{-at}\left(1 +\frac{t}{\sqrt{n}} \right)^{n-b-at} \, dt $$

Although I have no formal proof, it seems that the integrand converges to $e^{-t^2/2}$ and the dominated convergence theorem is applicable. Assuming that this is true, we obtain the following asymptotics

$$ G(n) \sim a \cdot b^{n-n} \cdot \sqrt{2\pi} = \frac{\sqrt{2\pi n}}{W(en)} \exp\left( n \frac{(W(en) - 1)^2}{W(en)} \right). $$