# Convergence of an improper integral

When is $\displaystyle \int_{0}^{\frac{1}{2}}\left(x\log^{2}x\right)^{-r}dx$ convergent? And how to prove it?

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Let $I_r(\varepsilon):=\int_{\varepsilon}^{\frac 12}(x\log^2x)^{-r}dx$ and $t:=\log x$. We have $x=e^t$ hence $dx=e^tdt$, which gives $$I_r(\varepsilon)=\int_{\log\varepsilon}^{-\log 2}e^{-rt}t^{-r}e^tdt=\int_{\log\varepsilon}^{-\log 2}e^{(1-r)t}(t^2)^{-r}dt=\int_{\log 2}^{-\log \varepsilon}e^{s(r-1)}s^{-2r}ds.$$ So we have to see whether the integral $\int_1^{+\infty}e^{s(r-1)}s^{-2r}ds$ is convergent.
When $r>1$, it's easy to see it diverges, as the integrand goes to $+\infty$ as $s\to +\infty$, and when $r=1$, we use the fact that $\int_1^{+\infty}\frac 1{t^2}dt$ is convergent. When $r<1$, we have convergence using the fact that $\int_1^{+\infty}e^{-t}dt$ is convergent.