I'm trying to prove that $$\int_0^{+\infty} e^{-x}\ln^2 x \, \text{d}x$$ Is convergent. My approach is $$\int_0^{+\infty} e^{-x}\ln^2 x \, \text{d}x = \int_0^1 e^{-x}\ln^2 x \, \text{d}x + \int_1^{+\infty} e^{-x}\ln^2 x \, \text{d}x$$ For the second integral I've used the inequality $\ln x \leq x-1 < x$, so being $x\in[1,+\infty)$, it follows that $\ln^2 x < x^2$; so $$\int_1^{+\infty} e^{-x}\ln^2 x \, \text{d}x < \int_1^{+\infty} e^{-x}x^2 \, \text{d}x = 5e^{-1}$$
So my questions are:
(1) Is the estimation correct? I'm not so sure of $\ln^2 x < x^2$, I'm pretty sure it works in $[1,+\infty)$ but I would like to know if there is a way to estimate $\ln^2 x <x^2$.
(2) I've noticed that $$\lim_{x \to +\infty} e^{-x}\ln^2 x=0$$ Is this enough to say that the integral in the interval $[1,+\infty)$ is convergent? If yes, how can I prove it? I'm suspicious because if the limit was $+\infty$ I could use comparison test, but in this case it doesn't give information so maybe the limit being $0$ isn't enough to prove the convergence.
Thanks for your time.
For the first integral, $\int \log^2(x) dx = x (\log^2(x) - 2 \log(x) + 2) $ so $\int_0^1 \log^2(x) dx = 2 $.
Since $\dfrac1{e} \le e^{-x} \le 1 $ for $0 \le x \le 1$, we have $\dfrac{2}{e} \le \int_0^1 e^{-x}\log^2(x) dx \le 2 $.
For the second integral, since $\log(x) < \log(1+x) \le x$ for $x \ge 1$, and $\int x^2 e^{-x} dx = -e^{-x} (x^2 + 2 x + 2) $, we have $\int_1^{+\infty} e^{-x}\ln^2 x dx \lt \int_1^{+\infty} e^{-x}x^2 dx =\dfrac{5}{e} $.