Show that
$$\int_1^\infty e^{-\frac{x t^2}{2}}\log(t) dt \sim \frac{e^{-\frac{x}{2}}}{x^2}$$
for large positive $x$.
I tried Taylor expanding the log. Each of the resulting terms can be integrated separately and yields a hypergeometric function upon integration. However, even after using the asymptotic form of these hypergeometric functions, the sum can't be done in closed form.
Recognizing that this is dominated by what happens near the lower limit and Taylor expanding $\log(t)\approx t-1$, this is asymptotic to $$ \int_1^\infty (t-1)e^{-\frac{1}{2}xt^2}dt = \int_0^\infty te^{-\frac{1}{2}x(t+1)^2}dt \\= \frac{e^{-x/2}}{x^2}\int_0^\infty te^{-t-\frac{1}{2}t^2/x}dt \\\sim \frac{e^{-x/2}}{x^2}\int_0^\infty te^{-t}dt \\= \frac{e^{-x/2}}{x^2}.$$