Convergence of $\int_2^\infty \frac{dx}{x^2 \cdot (ln(x))^{\alpha}}$?

Question

For which values of $\alpha > 0$ is the following improper integral convergent ?:

$$\int_{2}^{\infty}{{\rm d}x \over x^{2}\ \ln^{\alpha}\left(\, x\,\right)}$$

I tried to solve this problem by parts method but I am nowhere near to the answer.

Answer 1

The integrand is always dominated by $x^{-2}$ and hence the integral is convergent for all $\alpha$. Just evaluate the integral by the fundamental theorem.

Answer 2

We have for every $\alpha\in \Bbb R$ (not only for $\alpha>0$)

$$\frac{1}{x^2(\ln x)^\alpha}=_\infty o\left(\frac1{x^{3/2}}\right)$$ and since the integral

$$\int_2^\infty \frac{dx}{x^{3/2}}$$ is convergent since $\dfrac32>1$ then the given integral is also convergent.

Answer 3

For $\alpha\ge0$ $$ \begin{align} \int_2^\infty\frac{\mathrm{d}x}{x^2\log(x)^\alpha} &\le\frac1{\log(2)^\alpha}\int_2^\infty\frac{\mathrm{d}x}{x^2}\\ &=\frac1{2\log(2)^\alpha} \end{align} $$

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For $\alpha\lt0$, $$ \begin{align} \int_2^\infty\frac{\mathrm{d}x}{x^2\log(x)^\alpha} &\le\int_1^\infty\frac{\log(x)^{-\alpha}\mathrm{d}x}{x^2}\\ &=\int_0^\infty x^{-\alpha}\,e^{-x}\,\mathrm{d}x\\[4pt] &=\Gamma(1-\alpha) \end{align} $$