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I'm trying to find out for which values of $\alpha$ and $\beta$ the integral $\int\limits_2^{\infty}\frac{dx}{x^{\alpha}\ln^{\beta}x}$ does converge. I know that when $\alpha=1$ then $\beta$ must be greater than $1$. I tried to use integration by parts but It didn't work, so I would appreciate some hints. Thanks in advance.

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    $\begingroup$ Have you tried u-sub, where u = ln(x) and du = 1/x? $\endgroup$
    – Sentient
    Dec 13, 2015 at 23:38

2 Answers 2

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Covergence:

(1) $\alpha>1$, and $\beta\in \mathbb R$;

(2)$\alpha=1$, and $\beta>1$.

All other cases are divergent.

You have already know the case (2), so let me explain case (1). The key point is to see $x^{\alpha}$ is always the dominate term.

If $\alpha>1$, then $\frac{\alpha+1}{2}>1$, and $\frac{\alpha-1}{2}>0$. So we have $$\frac{1}{x^{\alpha}ln^{\beta}x}=\frac{1}{x^{\frac{\alpha+1}{2}}}\frac{1}{x^{\frac{\alpha-1}{2}}ln^{\beta}x}.$$

But notice that $\int_2^{\infty}\frac{1}{x^{\frac{\alpha+1}{2}}}dx<\infty$, and the term $\frac{1}{x^{\frac{\alpha-1}{2}}ln^{\beta}x}$ is bounded as the limit $$\lim_{x\to \infty}\frac{1}{x^{\frac{\alpha-1}{2}}ln^{\beta}x}=0$$ for any $\beta$.

Therefore in this case the integral

$$\mid\int_2^{\infty}\frac{1}{x^{\alpha}ln^{\beta}x}dx\mid\le\int_2^{\infty}\mid \frac{1}{x^{\alpha}ln^{\beta}x}\mid dx=\int_2^{\infty}\mid \frac{1}{x^{\frac{\alpha+1}{2}}}\mid \mid \frac{1}{x^{\frac{\alpha-1}{2}}ln^{\beta}x}\mid dx\le \int_2^{\infty}\frac{M}{x^{\frac{\alpha+1}{2}}}dx\le \infty$$

And the case when $\alpha<1$ then divergence is similar.

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  • $\begingroup$ There is math processing error. Please, edit the answer. $\endgroup$ Sep 10, 2016 at 16:34
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Hint: what happens, when $ \alpha \leq 0 $?

For the other case, $ \alpha > 0 $, try to use the integral test to consider the convergence of the series: $ \sum\limits_{n = 2}^{\infty} \frac{1}{x^{\alpha}\ln^\beta(x)} $

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