For what parameter $\alpha \in \mathbb{R}$ is this function integrable? 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.
 A: We assume $ x \in (0,1)$. There is no problem near the bound $1^-$.


*

*Assume $0<\alpha<1$. For $0<x<1/e$, we have $$
   0<-\frac{1}{x^\alpha\ln x}\le \frac1{x^{\alpha}} $$ and$$
   0<\frac{x-1}{x^\alpha\ln x}\le \frac{1-x}{x^{\alpha}} $$ giving $$
   0<\int_0^{1/e}\frac{x-1}{x^\alpha\ln x}\:dx\le
   \int_0^{1/e}\frac{1-x}{x^{\alpha}}\:dx=\left[x^{1-\alpha}
   \left(\frac{1}{1-\alpha}-\frac{x}{2-\alpha}\right)\right]_0^{1/e}<\infty
   $$ and the given integral is convergent.

*Assume $\alpha=1$. Then $$ \int_0^{1/e}\frac{-1}{x^\alpha\ln
   x}\:dx=\lim_{t \to 0^+} \left[-\ln(-\ln
   x)\frac{}{}\right]_t^{1/e}=\infty $$ and the given integral is
divergent.

*Assume $\alpha>1$. By the change of variable $u=-\ln x$, we have $$
   \int_0^{1/e}\frac{-1}{x^\alpha\ln
   x}\:dx=\int_1^{\infty}\frac{e^{(\alpha-1)u}}{u}\:du=\infty $$ and
the given integral is divergent.

A: Let
$$f(x)=\frac{x-1}{x^\alpha\ln(x)}$$ for
$0<x<1$ and $\alpha\in\mathbb R$.
we have

at $ x=1,$  it converges since 

$\lim_{x\to 1^-}f(x)=1$.

at$\;\;x=0$



*

*$\alpha>1$


Let $\alpha>a>1$
$lim_{x\to 0^+}x^af(x)=\infty \implies \int_0f(x)dx$ diverges.


*

*$\alpha<1$


let $\alpha<b<1$
$\lim_{x\to 0^+}x^bf(x)=0 \implies \int_0 f(x)dx$ converges.


*

*$\alpha=1$


$F(x)=\int_x f(t)dt=-(\ln(|\ln(x)|))$ 
goes to $\infty$.
