Convergence of $\int_{2}^{+\infty} \frac1{x \ln^\alpha x}dx$

Question

I can't analyse the convergence of this integral:

$$\int_{2}^{+\infty} \frac1{x \ln^\alpha x}dx$$ with $\alpha \in R$.

I have tried to find some functions and use comparison theorem, but I haven't succeeded. Can you help me?

Answer 1

Let use $u = \ln x$ then we have $$ \int_{\ln 2}^\infty \frac{1}{u^{\alpha}}du $$ what are the convergence properties of the above?

Answer 2

Another approach: for $\;\alpha=0\;$ the question is almost trivial, so assume $\;\alpha\neq 0\;$ , and then

$$\int\frac{dx}{x\log^\alpha x}=\int\left(\log x\right)'\log^{-\alpha}x\;dx=\begin{cases}\log\log x&,\;\;\alpha=1\\{}\\\frac{\log^{-\alpha+1}}{1-\alpha}&,\;\;\alpha\neq 1\end{cases}$$

It follows at once from the above that we have convergence iff $\;-\alpha+1<0\;$ ...