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There are 2 improper points: $0, 1$.

I found that $\displaystyle \int_{\frac{1}{2}}^{1}\frac{\ln^\alpha(1+x)}{x^\beta(1-x)^\alpha}\mathrm dx$ exists for $\alpha < 1$ (by using the limit comparison test, with $g(x)=\frac{1}{(1-x)^\alpha}$).

But I'm having trouble picking $g(x)$ for the other interval. If $\alpha \lt 0$ then $\ln^\alpha(1+x) \underset{x\to 0}{\to} \infty$, and everything I try gives me $\frac{f(x)}{g(x)}\to 0$ or $\frac{f(x)}{g(x)}\to \infty$ which doesn't give me all possible solutions.

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    I don't know why I thought ln(1+x)/x approaches inifinity near 0...2010-12-04

2 Answers 2

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We can pick $g(x)=\frac{1}{x^(\beta-\alpha)}$ and then $\frac{f(x)}{g(x)}\to 1$ as $x\to 0$. By the LCT we get $\int_{0}^{\frac{1}{2}}\frac{\ln^\alpha(1+x)}{x^\beta(1-x)^\alpha}$ exists iif $\beta-\alpha \lt 1$.

So overall $\int_{0}^{1}\frac{\ln^\alpha(1+x)}{x^\beta(1-x)^\alpha}$ exists iif $\beta-1 \lt \alpha \lt 1$.

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You may analyze the convergence of the improper integral

$I=\int_{0}^{1/2}f(x)dx=\int_{0}^{1/2}\frac{\ln ^{\alpha }(1+x)}{x^{\beta }(1-x)^{\alpha }}dx$

by LCT with

$J=\int_{0}^{1/2}g(x)dx=\int_{0}^{1/2}\frac{1}{x^{\beta -\alpha }}dx<\infty \qquad \beta -\alpha <1\text{.}$