Proving that $\int^1_0 \frac{1}{\sqrt{\ln(\frac{1}{x})}}dx$ converges?

Question

How do you prove that $\int^1_0 \frac{1}{\sqrt{\ln(\frac{1}{x})}}dx$ converges? I've tried more or less everything I can think of and still can't get the answer. Any hints will be appreciated!

Answer 1

HINT

We have that

$$\int^1_0 \frac{1}{\sqrt{\ln(\frac{1}{x})}}dx =\int_1^\infty \frac{1}{x^2\sqrt{\ln x}}dx =\int_1^2 \frac{1}{x^2\sqrt{\ln x}}dx+\int_2^\infty \frac{1}{x^2\sqrt{\ln x}}dx$$

and then refer to limit comparison test.

Answer 2

Substitute $x = e^{-u^2}$ (or equivalently, $u = \sqrt{\log(1/x)}$) and notice that for $0 < a < b < 1$,

$$ \int_{a}^{b} \frac{dx}{\sqrt{\log(1/x)}} = \int_{\log^{1/2}(1/b)}^{\log^{1/2}(1/a)} 2e^{-u^2} \, du. $$

So, as $a \to 0^+$ and $b \to 1^-$,

$$ \lim_{\substack{a \to 0^+ \\ b \to 1^-}} \int_{a}^{b} \frac{dx}{\sqrt{\log(1/x)}} = \int_{0}^{\infty} 2e^{-u^2} \, du = \sqrt{\pi} $$

Of course, even without knowing the value of $\int_{0}^{\infty} 2e^{-u^2} \, du$, an easy comparison tells that this integral converges.