My task is to determine wetherthe integral $\int_e^\infty{\frac{\ln(x)^\alpha}{x}\,dx}$ does exist or not in depencence of $\alpha\in \mathbb{R}$.
For that I wrote $b$ instead of $\infty$ and then calculated the integral using substitution, which should be $\frac{1}{\alpha+1}((\ln(b))^{\alpha+1}-1)$. Now for the improper integral I get $\lim\limits_{b->\infty}(\int_e^b(\frac{\ln(x)^\alpha}{x}dx))=\frac{1}{\alpha+1}\cdot(\lim\limits_{b->\infty}\ln(b)^{\alpha+1}-1)$
And because $\lim\limits_{c->\infty}\ln(c)=\infty$ and $\lim\limits_{c->\infty} c^{\alpha+1} = \left\{ \begin{array}{ll} \infty & \alpha\geq -1 \\ 1 & \alpha=-1 \\ 0 & \alpha<-1 \end{array} \right. $
So the limit only exists for $\alpha<-1$, is this right?
$$I=\int_e^\infty\frac{\ln^a(x)}{x}dx$$ $$u=\ln(x),\,dx=xdu$$ $$I=\int_1^\infty u^adu=\left[\frac{u^{a+1}}{a+1}\right]_1^\infty$$ and this is clearly divergent