Evaluating the convergence of $\int_{(1, +\infty)}\frac{1}{x \log x \ e^x}$

Question

I'm reading a proof where the integral over $(1,+\infty)$ of the function $$(1, +\infty)\ni x \mapsto f(x)=\frac{1}{x \log x \ e^x}$$ is shown to be $+\infty$ by invoking the fact that if $f,g$ are continuous positive functions with $f \sim g$ (i.e. f is asymptotically equivalent to g) and the integral of $g$ diverges over the domain, the so does the integral of f.

Now, the proposed solution gives the estimates $$\frac{1}{x \log x \ e^x}= \frac{e^{-x}}{x \log x } \sim \frac{e^{-1}}{\log(1+x-1)} \sim \frac{e^{-1}}{x-1}=\frac{1}{e(x-1)}$$ and then $$\int_1^{+\infty}\frac{1}{e(x-1)}=\frac{1}{e}\int_1^{+\infty}\frac{1}{(x-1)}=\frac{1}{e}\log(x-1)_{|_1^{+\infty}}=+\infty$$

Question

I do not understand the estimates above.

Answer 1

For an improper integral to converge, the limit must exist at both endpoints. In this instance there is no problem with the endpoint at $\infty$. When $x>2$ say, the integrand positive and bounded above by $e^{-x}$, so the integral is less than $\int_2^\infty e^{-x}\mathrm{d}x=e^{-2}$. When $x\to1+$ the integrand goes to $\infty$. So the questions comes down to investigating the behavior of $\int_t^2f(x)\mathrm{d}x$ as $t\to1+$

The asysmptotic behavior the proof talks about is as $x\to1+$.