Find:

$\displaystyle \lim_{r \rightarrow 0^+} \frac{1}{r|\ln r|^{\alpha}} \int^{r}_{0} |\ln x|^{\alpha} e^{-x^2} dx$ for $\displaystyle \alpha \in (0;1)$.

My intuition says that would be 0 so I have tried to use squeeze theorem. This limit is bounded below by 0.

I've thought I could bound it above using Hölder's inequality with $\displaystyle p=\frac{1}{\alpha}$ and $\displaystyle q=\frac{1}{1-\alpha}$ now what I have got:

$\displaystyle \frac{1}{r|\ln r|^{\alpha}} \int^{r}_{0} |\ln x|^{\alpha} e^{-x^2} dx \leq \left( \frac{\int^{r}_{0} |\ln x|dx}{|\ln x| \int^{r}_{0} e^{\frac{-x^2}{1-\alpha}}dx} \right)^{\alpha} \cdot \frac{\int^{r}_{0}e^{\frac{-x^2}{1-\alpha}}dx}{r} $. Does it get me anywhere?