Divergence of $\int_{0}^{1/2} 1/(|\sqrt{x}\ln(x)|)^{p} dx$ The matter of interest is
$$\int_{0}^{1/2} \frac{1}{|\sqrt{x}\ln(x)|^p}\, dx$$
I am aware that this integral converges for $p=2$ (that's not too hard to show). I also believe that this integral diverges for $p>2$...but how can I show that using elementary calculus and related techniques (comparison test etc)? 

By enforcing the substitution $x=e^{-z}$ we get
$$ \int_{0}^{1/2}\frac{dx}{\left(-\sqrt{x}\log x\right)^p} = \int_{\log 2}^{+\infty}\exp\left[\left(\frac{p}{2}-1\right)z\right]\frac{dz}{z^p} $$
and we clearly need $p\leq 2$ to ensure the (improperly-Riemann or Lebesgue)-integrability of $\exp\left[\left(\frac{p}{2}-1\right)z\right]\frac{1}{z^p}$ over $(\log 2,+\infty)$.
