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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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For the behavior near 0 you can ignore the 1/(1 - x)^a term. Then the problem is to compare the logarithm in the numerator with the power function in the denominator. Are you familiar with the first few terms in the Taylor expansion of ln(1 + x)? –  Qiaochu Yuan Dec 3 '10 at 17:01
    
I made that observation regarding 1/(1 - x)^a aswell. We just started learning about Taylor expansions but I think we're expected to solve this without that material. –  daniel.jackson Dec 3 '10 at 17:22
    
If I am not mistaken, for $0<\beta <1$, since $\displaystyle\int_{0}^{1/2}\dfrac{1}{x^\beta}dx$ converges, $\dfrac{\ln ^{\alpha }(1+x)}{x^{\beta }(1-x)^{\alpha }}/(1/x^{\beta })$ tends to $0$ with $x$ and your integral converges. –  Américo Tavares Dec 3 '10 at 17:55
    
Why? If $\alpha \lt 0, \ln^\alpha(1+x) \underset{x\to 0}{\to} \infty$. And still if $0 \leq \alpha \lt 1$, what about $\beta \geq 1$? The LCT gives no conclusion when $g(x)$ doesn't converge and $\frac{f(x)}{g(x)}\to 0$ –  daniel.jackson Dec 3 '10 at 18:07
    
The condition is only $\beta<1$ –  Américo Tavares Dec 3 '10 at 18:13
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1 Answer

up vote 1 down vote accepted

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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