I would like to receive some help with checking the calculation result:
I was calculating the result of the following improper integral:
$$\int\limits_3^\infty \frac{dx}{x \cdot \ln x \cdot (\ln\ln x)^{1 + \alpha}} \, \,\,\text{where}\,\,\, \alpha > 0$$
I did it this way:
$$\int\limits_3^\infty \frac{dx}{x \cdot \ln x \cdot (\ln\ln x)^{1 + \alpha}} = \lim_{\beta \to \infty} \int\limits_3^{\beta} \frac{dx}{x \cdot \ln x \cdot (\ln\ln x)^{1 + \alpha}} = (*).$$
As i understand, now we should calculate the definite integral $\int\limits_3^{\beta} \frac{dx}{x \cdot \ln x \cdot (\ln \ln x)^{1 + \alpha}}$,using $\beta$ as constant. Now i used the next substitution: $u(x) = \ln(\ln x)$, $du(x) = \frac{dx}{x \ln x}$. I calculated the new boundaries: $u(3) = \ln(\ln3)$, $u(\beta) = \ln(\ln \beta)$.
$$(*) = \lim_{\beta \to \infty} \int\limits_3^{\beta} \ln(\ln x)^{-1-\alpha} \frac{dx}{x \ln x} = \lim_{\beta \to \infty} \int\limits_{\ln(\ln3)}^{\ln(\ln \beta)} u^{-1 - \alpha}du.$$
From here i got the next result :
$$ \lim_{\beta \to \infty} \left( \frac{1}{\alpha} \left( \frac{1}{\ln(\ln(\ln(\ln3)))^{\alpha}} - \frac{1}{\ln(\ln(\ln(\ln \beta)))^{\alpha}}\right) \right) = \frac{1}{\alpha} \cdot \frac{1}{\ln(\ln(\ln(\ln 3)))^{\alpha}}.$$
Please, could you tell me if my calculation was right?
Thank you, for your time and help!