# For which values of $\alpha$ and $\beta$ does the integral $\int\limits_2^{\infty}\frac{dx}{x^{\alpha}\ln^{\beta}x}$ converge?

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.

• Have you tried u-sub, where u = ln(x) and du = 1/x? – Sentient Dec 13 '15 at 23:38

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.

• There is math processing error. Please, edit the answer. – Egor Okhterov Sep 10 '16 at 16:34

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: $\sum\limits_{n = 2}^{\infty} \frac{1}{x^{\alpha}\ln^\beta(x)}$