**Problem:** For what values of the parameter $\alpha \in \mathbb{R}$ is the function $$ f: (0,1) \rightarrow \mathbb{R}: x \mapsto \frac{x-1}{x^{\alpha} \ln(x)} $$ integrable? 

**My attempt:** Notice that $$\lim\_{x \to 1} f(x) x^{\alpha} = \lim\_{x \to 1} \frac{x-1}{\ln(x)} = 1 $$ by the use of l'Hopital. This means that $f(x) = \Theta(\frac{1}{x^{\alpha}})$ for $x \to 1$, i.e. $f(x)$ is asymptotically bounded by this function. So for $x \to 1$, we know that we must have $\alpha > 1$ for $f$ to be integrable. 

Now I wanted to determine the behavior of this function near $x = 0$. I let $\alpha = 1$ first. Let $\epsilon > 0$ and consider the interval $(\epsilon, 1/e)$. Is $f$ integrable on this interval?

We have $$ \lim\_{\epsilon \to 0} \int\_{\epsilon}^{1/e} f(x) dx = - \lim\_{\epsilon \to 0} \int\_{\epsilon}^{1/e} \frac{1-x}{x\ln(x)} dx \leq - \lim\_{\epsilon \to 0} \int\_{\epsilon}^{1/e} \frac{1}{x \ln(x)} dx. $$ Now if I let $u = \ln(x)$, then the limits of integration change to $\ln(\epsilon)$ and $-1$, but I would get $$ - \lim\_{\epsilon \to 0} \big[ \ln(u) \big]\_{\ln(\epsilon)}^{-1} $$ which is not defined.

I'm not sure how to determine the whether this function is integrable around zero, and for what value of the parameter. Any help/suggestions is appreciated.