# 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!

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## 2 Answers

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) –  Chris's sis Dec 26 '12 at 18:14
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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) –  Chris's sis Dec 26 '12 at 18:15
@Chris'ssister: You are welcome. –  Mhenni Benghorbal Dec 26 '12 at 18:25
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