Improper integral and its convergence. Is this procedure correct? Any quicker and simpler ways? 
Find the values of $\alpha,\beta\in\mathbb{R}$ for which $$\int_0^1\frac{x^{\alpha-1}}{\sqrt{\lvert1-x^\alpha\rvert}}\lvert\ln(x)\rvert^\beta\ dx$$ converges.


$\large \text{Procedure}:$ 
The possible singularities of the integrand occur at $x=\{0,1\}$, thus, I will treat two seperate casses. 

$1.$ Starting with$\ x=0$, the possible values that $\alpha$ can assume are $\alpha>0$ and $\alpha<0$. If $\alpha=0$ the denominator of the fraction vanishes. 
- Case for $\alpha>0$: 
Since: $$\frac{x^{\alpha-1}}{\sqrt{\lvert1-x^\alpha\rvert}}\lvert\ln(x)\rvert^\beta=O\left(x^{\alpha-1}\lvert\ln(x)\rvert^\beta\right) \quad \text{as $x\to0^+$}$$
We are able to use the Limit Comparison Test with $x^{\alpha-1}\lvert\ln(x)\rvert^\beta$. 
Considering:
$$x^{\alpha-1}\lvert\ln(x)\rvert^\beta=o\left(\frac1{x^\gamma}\right) \quad \text{as $x\to0^+$}$$
with $1-\alpha<\gamma<1$ (which can be shown by using the subsitution $x=e^{-t}$). 
The integral $$\int_0^1\frac1{x^\gamma}\ dx$$
converges for  $\gamma<1$. Thus,$$\int_0^1\frac1{x^\gamma} \ dx \quad \text{converges} \iff \int_0^1x^{\alpha-1}\lvert\ln(x)\rvert^\beta \ dx \quad \text{converges} \iff \int_0^1\frac{x^{\alpha-1}}{\sqrt{\lvert1-x^\alpha\rvert}}\lvert\ln(x)\rvert^\beta\ dx \quad \text{converges for $\alpha>0$ and $\forall\beta$} $$
- Case for $\alpha<0$: 
Since: $$\frac{x^{\alpha-1}}{\sqrt{\lvert1-x^\alpha\rvert}}\lvert\ln(x)\rvert^\beta=O\left(\frac{x^{\alpha-1}}{x^{\frac\alpha2}}\lvert\ln(x)\rvert^\beta\right) \quad \text{as $x\to0^+$}$$
We are able to use the Limit Comparison Test with $\lvert\ln(x)\rvert^\beta/x^{1-\alpha/2}$. 
Knowing that $1-\alpha/2$ is always positive (since $\alpha<0$), we can compare the latter function with $1/x$: $$\displaystyle \lim_{x\to0^+}\frac{\lvert\ln(x)\rvert^\beta/x^{1-\alpha/2}}{1/x}=\lim_{x\to0^+}\frac{\lvert\ln(x)\rvert^\beta}{x^{\lvert\alpha\rvert/2}}=+\infty$$
Therefore,$$\int_0^1\frac1x \ dx \quad \text{diverges} \iff \int_0^1\frac{\lvert\ln(x)\rvert^\beta}{x^{1-\alpha/2}} \ dx \quad \text{diverges} \iff \int_0^1\frac{x^{\alpha-1}}{\sqrt{\lvert1-x^\alpha\rvert}}\lvert\ln(x)\rvert^\beta\ dx \quad \text{diverges for $\alpha<0$} $$ $$$$
$2.$ Finishing with$\ x=1$: 
Since:
$$\lvert\ln(x)\rvert^\beta=O\left(\lvert1-x\rvert^\beta\right) \quad \text{as $x\to1$}$$
and $$\lvert1-x^\alpha\rvert = O\left(\lvert1-x\rvert\right) \quad \text{as $x\to1$}$$ 
We obtain:
$$\frac{x^{\alpha-1}}{\sqrt{\lvert1-x^\alpha\rvert}}\lvert\ln(x)\rvert^\beta\sim \frac{\lvert1-x\rvert^\beta}{\sqrt{\lvert1-x\rvert}}=\frac1{(1-x)^{1/2-\beta}}$$
The integral $$\int_0^1\frac1{(1-x)^{1/2-\beta}}\ dx$$
converges for $1/2-\beta<1 \iff \beta>-1/2$ and by the Limit Comparison Test so does the initial one. 
By $1$ and $2$, we can conclude that the improper integral converges for $\boxed{\alpha>0\ \text{and} \ \beta>-1/2}$.
$$$$ Does this seem correct? Since it took me a long time to do this, is there any quicker way to find the values of $\alpha$ and $\beta$? Are there any tips or tricks to these type of exercises? 
Thank you for any help or suggestion.
 A: Hint. Alternatively, one may perform the change of variable 
$$
u=x^\alpha, \qquad du=\alpha x^{\alpha-1}dx.
$$ 
Case 1. Assume $\alpha>0$. Then
$$
\begin{align}
\int_0^1\frac{x^{\alpha-1}}{\sqrt{\lvert1-x^\alpha\rvert}}\lvert\ln(x)\rvert^\beta\ dx&=\frac1{\alpha}\int_0^1\frac{\alpha x^{\alpha-1}}{\sqrt{\lvert1-x^\alpha\rvert}}\left|\frac1\alpha\ln(x^\alpha)\right|^\beta\ dx
\\\\&=\frac1{\alpha^{\beta+1}}\int_0^1\frac{|\ln u|^\beta}{\sqrt{1-u}}\ du
\\\\&=\frac2{\alpha^{\beta+1}}\int_0^1(-\ln (1-t^2))^\beta\ dt \quad (t=\sqrt{1-u}).
\end{align}
$$ As $t \to 0^+$, one has
$$
(-\ln (1-t^2))^\beta \sim t^{2\beta}
$$ which gives a convergent integral iff $-2\beta<1$.
As $t \to 1^-$, one has
$$
(-\ln (1-t^2))^\beta \sim (-\ln (1-t))^\beta
$$ which gives a convergent integral iff $\beta>-1$.
Case 2. Assume $\alpha<0$. Then
$$
\begin{align}
\int_0^1\frac{x^{\alpha-1}}{\sqrt{\lvert1-x^\alpha\rvert}}\lvert\ln(x)\rvert^\beta\ dx&=\int_0^1\frac{x^{-|\alpha|-1}}{\sqrt{\lvert1-x^{-|\alpha|}\rvert}}\left|\frac1\alpha\ln(x^{-|\alpha|})\right|^\beta\ dx
\\\\&=\frac1{|\alpha|^{\beta+1}}\int_1^\infty\frac{(\ln u)^\beta}{\sqrt{u-1}}\ du
\\\\&=\frac2{|\alpha|^{\beta+1}}\int_1^\infty(\ln (1+t^2))^\beta\ dt \quad (t=\sqrt{1-u}).
\end{align}
$$ As $t \to \infty$, one has
$$
(\ln (1+t^2))^\beta \sim 2^{\beta}\ln^\beta t
$$ which gives a divergent integral for all values of $\beta$.
In conclusion, the given integral is convergent if and only if

$$
\alpha > 0, \quad \beta>-\frac12.
$$

