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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 !!!

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Notice that, $ \frac{1}{x^\alpha (1+x)^\beta } \sim \frac{1}{x^\alpha x^\beta }= \frac{1}{x^{\alpha+\beta} },$ as $x\to \infty.$ –  Mhenni Benghorbal Jan 21 '13 at 4:27

2 Answers 2

up vote 0 down vote accepted

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, $\beta \in (0,1)$.

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Thanks a lot ! !! –  theMissingIngredient Jan 18 '13 at 22:05

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.

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thanks, but what if we haven't learn this theorem and I want to solve the question without it? Got any idea? Thanks ! –  theMissingIngredient Jan 18 '13 at 17:46
    
The idea behind this theorem is exactly the same with using the Quotient test. In fact we compare our integral with $\int_a^{\infty}\frac{dx}{x^{\alpha}}$. –  Babak S. Jan 18 '13 at 17:50
    
but when comparing the integral using a limit to the integral of $1/x^\alpha$ , we get that the limit is 0 ... which isn't part of the quotient test assumptions... –  theMissingIngredient Jan 18 '13 at 18:00
    
Yes it is. When you know $\int_a^{\infty}g(x)dx$ converges and know that $\lim\frac{f(x)}{g(x)}=0$ then $\int_a^{\infty} f(x)dx$ converges. –  Babak S. Jan 18 '13 at 18:03
    
Very nice theorem here! +1 –  amWhy Feb 14 '13 at 0:09

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