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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?

Cheers!

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2 Answers

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 $c/(x-1)$.

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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
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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.

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