Evaluating improper integrals with $\ln(x) $ Can someone help me understand how to use the limit version of comparison test in order to check whether or not the following integrals converges:
$$\int _ {1}^{\infty} \frac{\ln^2(x) }{x^2} dx,\,\int _ {0}^{1} \frac{\ln(x) }{x-1} dx? $$
I know I need to calculate the limit of $\frac{\frac{\ln^2(x)}{x^2 }}{g(x)}$ when $x\to 0$ and $x\to \infty$ , but I also need the limit to be a finite number , different than $0$. Can someone help me understand which function $g(x)$ should I choose in either case ? 
Thanks in advance 
 A: Looking for the limit to be nonzero is a heavy and unnecessary restriction. If the limit is zero, with $g$ at the bottom and $g$ is integrable, then the integral converges.
For the first one, you only consider the limit at infinity. You need a power greater than one to guarantee convergence, and you can use the rest to kill the logarithm.  So a natural choice is $g(x)=x^{-3/2}$. This choice comes from splitting $x^2$ as $x^{3/2}x^{1/2}$, where the first factor makes the integral convergent while the second one kills the logarithm. 
For the second the idea is similar. At zero you have no issues because only the logarithm makes it improper, and it is integrable (or you can use $g(x)=1/\sqrt x$). At $1$,  the limit exists, so the integral is not improper (or use $g(x)=1$).
A: $\newcommand{\+}{^{\dagger}}%
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With $\mu < -1$:
\begin{align}
\color{#00f}{\large\int_{1}^{\infty}{\ln^{2}\pars{x} \over x^{2}}\,\dd x}&=
\lim_{\mu \to -2}\partiald[2]{}{\mu}\int_{1}^{\infty}x^{\mu}\,\dd x
=
\lim_{\mu \to -2}\partiald[2]{}{\mu}{-1 \over \mu + 1}
=
\lim_{\mu \to -2}{-2 \over \pars{\mu + 1}^{3}} = \color{#00f}{\Large 2}
\end{align}

\begin{align}
\int_{0}^{1}{\ln\pars{x} \over x - 1}\,{\rm d}x&=
\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1}{1 - x^{\mu} \over 1 - x}\,{\rm d}x
=
\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1}\ln\pars{1 - x}\pars{-\mu x^{\mu -1}}\,{\rm d}x
\\[3mm]&=
-\lim_{\mu \to 0}\partiald{}{\mu}\mu\lim_{\nu \to 0}
\partiald{}{\nu}\int_{0}^{1}\pars{1 - x}^{\nu}x^{\mu -1}\,{\rm d}x
\\[3mm]&=
-\lim_{\mu \to 0}\partiald{}{\mu}\mu\lim_{\nu \to 0}
\partiald{{\rm B}\pars{\nu + 1,\mu}}{\nu}
\end{align}
where ${\rm B}\pars{x,y}$ is the Beta Function which satisfies
$\ds{{\rm B}\pars{x,y} = {\Gamma\pars{x}\Gamma\pars{y} \over \Gamma\pars{x + y}}}$.
$\Gamma\pars{z}$ and $\Psi\pars{z}$ are the Gamma and Digamma functions, respectively.
\begin{align}
\color{#00f}{\large\int_{0}^{1}{\ln\pars{x} \over x - 1}\,{\rm d}x}&=
-\lim_{\mu \to 0}\partiald{}{\mu}\mu\lim_{\nu \to 0}
\partiald{}{\nu}
\bracks{\Gamma\pars{\nu + 1}\Gamma\pars{\mu} \over \Gamma\pars{\nu + 1 + \mu}}
\\[3mm]&=
-\lim_{\mu \to 0}\partiald{}{\mu}\Gamma\pars{\mu + 1}\lim_{\nu \to 0}
\partiald{}{\nu}
\bracks{\Gamma\pars{\nu + 1} \over \Gamma\pars{\nu + 1 + \mu}}
\\[3mm]&=
-\lim_{\mu \to 0}\partiald{}{\mu}\Gamma\pars{\mu + 1}
\bracks{\Psi\pars{1} - \Psi\pars{1 + \mu} \over \Gamma\pars{1 + \mu}}
=\Psi'\pars{1} = \color{#00f}{\large{\pi^{2} \over 6}}
\end{align}
Also, this integral can be solved with the change of variables $x = \expo{-t}$ and expanding $\pars{1 - \expo{-t}}^{-1}$ in powers of $\expo{-t}$.

A: Do you have to use the limit comparison test? What about the Condensation Test, for example?:
$$a_n:=\frac{\log^2n}{n^2}\implies 2^na_{2^n}=\frac{2^n\log^2(2^n)}{(2^n)^2}=\frac{n^2\log^2n}{2^n}$$
and now you can apply the $\;n-$th root test:
$$\sqrt[n]{2^na_{2^n}}=\frac{\sqrt[n]{n^2\log^2n}}2\xrightarrow[n\to\infty]{}\frac12<1$$
