Prove that $\int_{0}^{1/2} x^{-\alpha}|\log(x)|^{-2\alpha} d \, x$ diverges for $\alpha>1$ As a part of a larger proof I need to show that $\int_{0}^{1/2} x^{-\alpha}|\log(x)|^{-2\alpha} d \, x$ diverges for $\alpha>1$ (which I know to be true from numerical simulation with Maple). 
However, I have tried to find a lower bound for this integral that diverges all day and just cannot figure it out. I am desperate for help, so any suggestions will be useful!
 A: By the change of variable
$$
x=e^{-u},\quad u=-\ln x,\quad dx=-e^{-u}du,
$$ one obtains
$$
\int_{0}^{1/2} x^{-\alpha}|\log(x)|^{-2\alpha} d \, x=\int_{\ln 2}^{+\infty} \frac{e^{(\alpha-1)u}}{u^{2\alpha}}du
$$ A potential problem of convergence is as $u \to +\infty$, in which case: 

  
*
  
*if $\alpha -1<0$ then
  

$$
\frac{e^{(\alpha-1)u}}{u^{2\alpha}}<e^{(\alpha-1)u}
$$  and the latter function is convergent over $[\ln 2,+\infty)$ giving the convergence of your initial integral.

  
*
  
*if $\alpha -1=0$ then
  

$$
\int_{0}^{1/2} x^{-\alpha}|\log(x)|^{-2\alpha} d \, x=\int_{\ln 2}^{+\infty} \frac{1}{u^{2}}du=\frac1{\ln 2}<\infty
$$ giving the convergence of your initial integral.

  
*
  
*if $\alpha -1>0$ then
  

$$
\frac{e^{(\alpha-1)u}}{u^{2\alpha}} \to +\infty, \quad u \to +\infty
$$ giving the divergence of your initial integral.
A: Hint: For all $p,q> 0, \lim_{x\to 0^+} x^p|\log x|^q = 0.$
Added later: The hint gives us a computation free way to show divergence. Write
$$\tag 1 x^{-\alpha}|\log x|^{-2\alpha} = \frac{1}{x}\cdot \frac{1}{x^{\alpha - 1}|\log x|^{2\alpha}}.$$
By the hint, the second fraction $\to \infty$ as $x \to 0^+.$ It follows easily that the left side of $(1)$ is $> 1/x$ in some $(0,\delta),$ and since $\int_0^\delta (1/x)\,dx = \infty,$ we're done.
