Find convergence of improper integral. Hello I have to find the convergence of this improper integral: 
$$\int_{e}^{\infty} \frac{1}{x\log^2x} dx$$
So I started by doing the following:
$\lim \limits_{x \to A} \int_{e}^{A} \frac{1}{x\log^2x} dx$, but I don't really know how to solve this integral:$\int_{e}^{A} \frac{1}{x\log^2x} dx$.
Any tips would be great, thank you.
 A: Hint: Make the substitution $u=\log x$. Then $du=\frac{1}{x}\,dx$.
A: Integration by parts gives us:
$$
\int_e^\infty \frac{dx}{x\ln ^2x} = \left[ \frac{1}{\ln x} \right]_e^\infty +2\int_e^\infty \frac{\ln x \,dx}{x \ln^3 x} = -1 +2\int_e^\infty \frac{dx}{x\ln^2 x}
$$
Thus:
$$
\int_e^\infty \frac{dx}{x\ln^2 x} = 1
$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\large\tt\mbox{With}\quad 0\ <\ \delta\ <\ 1\quad\mbox{and}\quad
     \pars{~x\ \equiv\ \expo{-t}\quad\imp\quad t\ =\ -\ln\pars{x}~}:}$

\begin{align}
&\color{#66f}{\large\lim_{\delta \to 0^{+}}\bracks{%
\int_{\epsilon}^{1 - \delta}{\dd x \over x\ln^{2}\pars{x}}
+\int_{1 + \delta}^{\infty}{\dd x \over x\ln^{2}\pars{x}}}}
\\[3mm]&=\lim_{\delta \to 0^{+}}\bracks{%
-\int_{-\ln\pars{\epsilon}}^{-\ln\pars{1 - \delta}}{\dd t \over t^{2}}
-\int_{-\ln\pars{1 + \delta}}^{-\infty}{\dd t \over t^{2}}}
\\[3mm]&=\lim_{\delta \to 0^{+}}\bracks{%
-\,{1 \over \ln\pars{1 - \delta}} +{1 \over \ln\pars{\epsilon}}
+{1 \over \ln\pars{1 + \delta}}} = \color{#66f}{\large +\infty}
\end{align}

