Convergence/Divergence of $\int_{0}^{1/e} \frac{\log \left(\frac{1}{x}\right)}{(\log^2 (x)-1)^{3/2}} \mathrm{dx}$ Initially I wanted to compute 
$$\int_{0}^{1/e} \frac{\log \left(\frac{1}{x}\right)}{(\log^2 (x)-1)^{3/2}} \mathrm{dx}$$
but it seems that Mathematica says that the integral diverges. I thought of
some variable change, but I also wonder if there is something easy to prove
it diverges. Any hint / suggestion here would be precious to me. Thanks!
 A: Making the change of variables $ x=e^{-(t+1)} $ gives
$$ \int _{0}^{\infty }\!{\frac { \left( t+1 \right) {{\rm e}^{-t-1}}}{{
t}^{3/2} \left( t+2 \right) ^{3/2}}}{dt}. $$
The integrand behaves as $c\,t^{-3/2}$ as $t \to 0 $.
A: Try substituting $x=e^{-u-1}$ to get
$$
\int_0^\infty\frac{u+1}{(u^2+2u)^{3/2}}e^{-u-1}\mathrm{d}u
=\int_0^\infty\color{#00A000}{\frac{u+1}{(u+2)^{3/2}e}}\color{#C00000}{u^{-3/2}}\color{#0000FF}{e^{-u}}\mathrm{d}u
$$
The part in green is bounded over the positive reals. $e^{-u}$ would be integrable at $\infty$, but the factor of $u^{-3/2}$ is not integrable at $0$.
Therefore, the integral does not converge.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{1/\expo{}}{\ln\pars{1/x} \over \bracks{\ln^{2}\pars{x} - 1}^{3/2}}
     \,\dd x:\ {\large ?}}$

With $\ds{\pars{~x \equiv \expo{-t}\quad\imp\quad t = -\ln\pars{x}~}\quad}$ and $\ds{\quad\Lambda > \expo{}}$:
  \begin{align}&\color{#c00000}{%
\int_{0}^{1/\Lambda}{\ln\pars{1/x} \over \bracks{\ln^{2}\pars{x} - 1}^{3/2}}
\,\dd x}
=\int_{\infty}^{\ln\pars{\Lambda}}{t \over \pars{t^{2} - 1}^{3/2}}\,\pars{-\expo{-t}\,\dd t}
\\[3mm]&=-\int_{t\ =\ \ln\pars{\Lambda}}^{t\ \to\ \infty}
\expo{-t}\dd\bracks{\pars{t^{2} - 1}^{-1/2}}
={1 \over \Lambda\root{\ln^{2}\pars{\Lambda} - 1}}
-\int_{\ln\pars{\Lambda}}^{\infty}{\expo{-t} \over \root{t^{2} - 1}}\,\dd t
\\[3mm]&={1 \over \Lambda\root{\ln^{2}\pars{\Lambda} - 1}}
-\int_{\ln\pars{\Lambda}}^{1}{\expo{-t} \over \root{t^{2} - 1}}\,\dd t
-\
\underbrace{\int_{1}^{\infty}{\expo{-t} \over \root{t^{2} - 1}}\,\dd t}
_{\ds{=\ {\rm K}_{0}\pars{1}}}
\end{align}
  where $\ds{{\rm K}_{\nu}\pars{z}}$ is a
  Modiffied Bessel Function.
  See ${\bf\mbox{9.6.23}}$.

The first term, in the right hand side, shows clearly the divergence:
$$
{1 \over \Lambda\root{\ln^{2}\pars{\Lambda} - 1}}
\sim {1 \over \root{2\expo{}}}\,{1 \over \pars{\Lambda - \expo{}}^{1/2}}\,,
\qquad \Lambda \gtrsim \expo{}
$$
