Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm going through some old notes I took on harmonic analysis some time ago and came across the claim that $$\int_0^1 \frac{dx}{x\log \frac{1}{x}}$$ is unbounded. I know almost nothing about singular integrals, so I have no idea if this is easy or hard. Could anyone help me out?


share|cite|improve this question
up vote 4 down vote accepted

$-\int_0^1 d/(x\log x)=-\int_{-\infty}^0 dt/t$, where $t=\log x$, and the latter integral surely diverges (at both ends) (as the primitive function is $\log |t|$)

the divergence at $x=0$ probably needs this substitution. The divergence at $1$ is clear without any substitution - your function there behaves as $c/(x-1)$.

share|cite|improve this answer
Wow! I'm pretty slow tonight :). Cheers. – Glen Wheeler Apr 9 '11 at 18:11
Isn't the question about $1/(x\text{log}(1/x))$, not $1/(x\text{log}(x))$? – Alex Becker Apr 9 '11 at 18:15
@Alex Yeah, but $\log 1/x = -\log x$. – Glen Wheeler Apr 9 '11 at 18:16
@Glen: I feel so stupid right now. – Alex Becker Apr 9 '11 at 18:17
@Alex I know exactly what you mean :D. – Glen Wheeler Apr 9 '11 at 18:25

Start by changing variables with $u = \text{log}\frac{1}{x}$, so that $du = -\frac{1}{x}dx$ and the indefinite integral becomes $\int -\frac{1}{u}du = -\text{log}(|u|) + C$, and substituting back then gives $\int_0^1 \frac{dx}{x\log \frac{1}{x}} = -\text{log}(\text{log}(\frac{1}{1})) + \text{log}(\text{log}(\frac{1}{0})) = -\text{log}(0) + \text{log}(\infty) = \infty + \infty = \infty$. In order to make this rigorous one would have to use limits, but the result is the same.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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