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

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!

• Well, the integrand is $\mathcal{O}\big((1-x)^{-1/2}\big)$ when $x\to 1$, and has a removable singularity at $x=0$ (actually, the integral equals $\sqrt{\pi}$). – metamorphy Oct 28 '18 at 10:59
• Your integral $$\int_{0}^{1}\frac{dx}{\sqrt{\log(\frac{1}{x}})}=\sqrt{\pi}$$ – Dr. Sonnhard Graubner Oct 28 '18 at 11:00
• @metamorphy What does removable singularity mean and what does your curly O stand for? I don't think I've encountered those terms before – Yip Jung Hon Oct 28 '18 at 11:25
• @YipJungHon big-O notation: when $x$ is near 1, the function "looks like" $(1-x)^{-1/2}$. Removable singularity: we can define a value at $x=0$ so that the function remains continuous there. – Parcly Taxel Oct 28 '18 at 11:26
• I see, thank you, is there a link you can provide me to show me why the whole integral equals root of pi? – Yip Jung Hon Oct 28 '18 at 11:28

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.

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.

• Very nice way too! – user Oct 28 '18 at 12:01