I am interested in the asymptotic behavior of sequences $(I_n)$ and $(J_n)$ as $n \rightarrow \infty$, where
$$I_n = \int_{1}^{\infty}\frac{e^{-nx^2}}{x^2}\, dx,$$
and
$$J_n = \int_{0}^{\infty}\frac{(1-e^{-x^2})^n}{x^2}\, dx.$$
In the first case, I integrated by parts to obtain
$$I_n = e^{-n} - 2\sqrt{n}\int_{\sqrt{n}}^{\infty}e^{-x^2}\,dx= e^{-n}-\sqrt{\pi n} \,\text{erfc}(\sqrt{n}). $$
The asymptotic behavior of the complementary error function is, of course, well-known (using repeated integration by parts) and I found
$$I_n \sim \frac{e^{-n}}{2n}.$$
Now I am having trouble with $J_n$ -- proceeding in the same manner or otherwise.
Making the change of variables $x = y\sqrt{\log n}$ transforms the integral into
$$ J_n = \frac{1}{\sqrt{\log n}} \int_0^\infty \frac{(1-e^{-y^2\log n})^n}{y^2}\,dy. $$
For $n > 1$ it can be shown that
$$ \left|\frac{(1-e^{-y^2\log n})^n}{y^2}\right| \leq \begin{cases} 1 & \text{if } 0 \leq y < 1 \\ 1/y^2 & \text{if } y \geq 1 \end{cases} \in L_1([0,\infty)) $$
and
$$ \lim_{n \to \infty} \frac{(1-e^{-y^2\log n})^n}{y^2} = \begin{cases} 0 & \text{if } 0 \leq y < 1 \\ 1/e & \text{if } y = 1 \\ 1/y^2 & \text{if } y > 1, \end{cases} $$
so by the dominated convergence theorem we have
$$ \lim_{n \to \infty} \int_0^\infty \frac{(1-e^{-y^2\log n})^n}{y^2}\,dy = \int_1^\infty \frac{dy}{y^2} = 1. $$
Thus
$$ J_n \sim \frac{1}{\sqrt{\log n}} $$
as $n \to \infty$.