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!

  • $\begingroup$ 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 $\sqrt{\pi}$). $\endgroup$ – metamorphy Oct 28 '18 at 10:59
  • $\begingroup$ Your integral $$\int_{0}^{1}\frac{dx}{\sqrt{\log(\frac{1}{x}})}=\sqrt{\pi}$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 28 '18 at 11:00
  • $\begingroup$ @metamorphy What does removable singularity mean and what does your curly O stand for? I don't think I've encountered those terms before $\endgroup$ – Yip Jung Hon Oct 28 '18 at 11:25
  • $\begingroup$ @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. $\endgroup$ – Parcly Taxel Oct 28 '18 at 11:26
  • $\begingroup$ I see, thank you, is there a link you can provide me to show me why the whole integral equals root of pi? $\endgroup$ – Yip Jung Hon Oct 28 '18 at 11:28


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.

  • $\begingroup$ Very nice way too! $\endgroup$ – gimusi Oct 28 '18 at 12:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.