Improper integrals with parameter I have to find out for which $b$, the improper integral exist. $$\int_e^\infty \\! \frac{(\ln (x))^b}{x}\ \, \mathrm{d}x$$ I know that the improper integral exists if the limit exist. Is there a general methodic to find these $b$ or just trying? Thx in advance

One can first solve the integral with the substitution $u=\ln(x)$, which gives us

$$\int u^b\ du=\frac{u^{b+1}}{b+1}\color{#888888}{+c}$$

Placing in bounds, we then get

$$\left.\int_e^a\frac{(\ln(x))^b}x\ dx=\frac{(\ln(x))^{b+1}}{b+1}\right|_{x=e}^{x=a}$$

as $a\to\infty$, this integral exists if $(\ln(x))^{b+1}$ goes to $0$, i.e. for $b<-1$.

For $b=-1$, the integral turns out to be $\ln(\ln(x))$, which diverges.