Convergence of two Improper Integrals

Question

1) $\quad \displaystyle \int_1^\infty \frac{1}{x^\alpha (1+x)^\beta }dx $

2) $\quad\displaystyle \int _0 ^1 \frac{\sin^\beta x } { x(1-x)^\beta } dx $

In the first one- I can't find any suitable candidate for comparison at $\infty$ . In the second one- I need to split this integral into $ \int_0 ^{0.5} + \int_{0.5}^1 $ . The first one converges iff the integral from $0 $ to $0.5 $ of $x^{\beta-1} $ converges. But how can I check the second one?

Thanks everyone !!!

Answer 1

1) $\alpha + \beta > 1$ for integrability at $\infty$.

2) $\beta > 0$ for integrability at $x=0$, and $\beta < 1$ for integrability at $x=1$. Therefore, $\beta \in (0,1)$.

Answer 2

For some cases of $\alpha,\beta$ we can have the result.

Let $$\lim_{x\to\infty}x^{\alpha}\frac{1}{x^{\alpha}(1+x)^{\beta}}=A$$

There is an applicable theorem as follows:

Theorem: If $\lim_{x\to\infty}x^{\alpha}f(x)=A$ then $\int_a^{\infty}f(x)dx$ converges if $\alpha>1$ and $A<\infty$. It diverges if $\alpha\leq1$ and $A\neq0$ ($A$ may be infinte).

For exampe,for $\alpha>1$ and $\beta\geq0$ then above limit is finite when $x\to+\infty$. Or for any $\alpha\leq 1$ and $\beta\leq 0$ the improper integral (1) diverges.