3
$\begingroup$

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.

  • 1
    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)?2010-12-03
  • 0
    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.2010-12-03
  • 0
    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.2010-12-03
  • 0
    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$2010-12-03
  • 0
    The condition is only $\beta<1$2010-12-03
  • 0
    Yes, $\int_{0}^{1/2}\frac{1}{x^\beta}dx$ converges iif $\beta \lt 1$. So if $g(x)=\frac{1}{x^\beta}$ and $\alpha \gt 0$ then $\frac{f(x)}{g(x)}\to 0$ and $\int_{0}^{1/2} f(x)$ converges by the LCT. But like I said, this doesn't cover the case where $\beta \geq 1$ since the LCT in this form (limit is 0) isn't an if and only if.2010-12-03
  • 0
    Okay, so you don't need the full Taylor expansion. The main thing you need to prove is that ln(1 + x)/x approaches 1 as x approaches 0, and the rest should be clear.2010-12-03
  • 0
    The limit can be computed by L'Hôpital's rule http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule2010-12-03
  • 0
    ... and then see that the integrand behaves like $x^{\alpha -\beta }$ near $x=0$.2010-12-03
  • 0
    I don't know why I thought ln(1+x)/x approaches inifinity near 0...2010-12-04

2 Answers 2

1

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 $\beta-1 \lt \alpha \lt 1$.

1

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

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