find for which $\alpha, \beta$ does $\int\limits_{100}^{\infty}\frac{1}{xln(x)^{\alpha}ln(ln(x))^{\beta}}dx$ converge

Question

find for which $\alpha, \beta$ the following integral converges:

$\int\limits_{100}^{\infty}\frac{1}{x\ln(x)^{\alpha}\ln(\ln(x))^{\beta}}\,dx$

I thought about trying substitution and got to $\int\limits_{\ln(100)}^{\infty}\frac{1}{u^{\alpha}\ln(u)^{\beta}}\,du$

and I thought about separating to different cases here but it seemed too complicated.

also thought about another substitution which gets $\int\limits_{\ln(\ln(100))}^{\infty}\frac{1}{e^{v\alpha -v}v^{\beta}}\,dv$

but this seems even more confusing.

So I want to ask:

1. are my substitutions not correct by any chance?
2. tips on how to approach this question?

Answer 1

Hint : Since find the values of $\alpha,\beta$ necessary to apply the Integral test and use Cauchy's test to determine convergence of the series.

Answer 2

You were right by trying to start with a substitution. After you get

$$\int^{\infty}_{\ln(100)} \frac{1}{u^\alpha\ln(u)^\beta}$$ From this, we can do a p-test and find that the integral will converge for all $$\alpha \in (1, \infty)$$ or $$\alpha = 1 \text{ and } \beta \in (1, \infty)$$