Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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

2 Answers 2

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

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

share|improve this answer
Américo: this question is tagged homework. I would appreciate it if you did not give complete answers. –  Qiaochu Yuan Dec 3 '10 at 23:22
@Qiaochu Yuan: Would it be acceptable to transcribe my last two comments only? If you deem necessary I may delete my answer rightaway, or replace it by those comments, mentioning your hint. –  Américo Tavares Dec 3 '10 at 23:30
@Qiaochu Yuan: in the meantime I have shorten my answer. –  Américo Tavares Dec 3 '10 at 23:38
Undeleted today. –  Américo Tavares May 2 at 13:57
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 $\beta-1 \lt \alpha \lt 1$.

share|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.