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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!

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up vote 13 down vote accepted

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.

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Nice answer (+1) – user 1618033 Dec 26 '12 at 18:14

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 $.

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Nice answer (+1) – user 1618033 Dec 26 '12 at 18:15
@Chris'ssister: You are welcome. – Mhenni Benghorbal Dec 26 '12 at 18:25

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1/\expo{}}{\ln\pars{1/x} \over \bracks{\ln^{2}\pars{x} - 1}^{3/2}} \,\dd x:\ {\large ?}}$

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: $$ {1 \over \Lambda\root{\ln^{2}\pars{\Lambda} - 1}} \sim {1 \over \root{2\expo{}}}\,{1 \over \pars{\Lambda - \expo{}}^{1/2}}\,, \qquad \Lambda \gtrsim \expo{} $$

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