On an generalized integral exercise: $ \int_{0}^{+\infty} \frac{dx}{\sqrt{x} | 1-x |^{\alpha}} $. I am asked to determine for which $\alpha > 0$ does the following generalized integral converge:
$$ \int_{0}^{+\infty} \frac{dx}{\sqrt{x} | 1-x |^{\alpha}} $$
I did the following $$ \int_{0}^{+\infty} \frac{dx}{\sqrt{x} ( 1-x )^{\alpha}} =  \int_{0}^{b} \frac{dx}{\sqrt{x} ( 1-x )^{\alpha}} +  \int_{b}^{1} \frac{dx}{\sqrt{x} ( 1-x )^{\alpha}} + \int_{1}^{c} \frac{dx}{\sqrt{x} ( -1+x )^{\alpha}}  + \int_{c}^{+\infty} \frac{dx}{\sqrt{x} ( -1+x )^{\alpha}}  $$
to get rid of the absolute values and have only one problematic point per integral but I am a bit dubious of my next move (how do I find the anti-derivative of the arguments of the integrals? Expanding everything out does not help right?). 
 A: Okay andre is absolutely right (took me like 2 hours to figure this out), you can see that $\alpha <1$ in the 2nd integral and see that $\alpha>1/2$.
$\alpha<1$
The idea is you want to create two functions $f$ and $g$ where $f(x)>\frac{1}{\sqrt{x}(1-x)^{\alpha}}>g(x)$ for all $x\in (b,1)$, $\int\limits_{b}^{1}g(x)dx=\infty$, and $\int\limits_{b}^{1}f(x)dx=\infty$. (clearly if $\int\limits_{b}^{1}\frac{dx}{\sqrt{x}(1-x)^{\alpha}}$ converges then this is impossible, but you want to see which $\alpha$ work )
So I choose a $0<k$ such that $f(x)=\frac{k}{(1-x)^{\alpha}}<\frac{1}{\sqrt{x}(1-x)^{\alpha}}$ for all $x\in (b,1)$, and choose an $0<m$, $g(x)=\frac{m}{(1-x)^{\alpha}}>\frac{1}{\sqrt{x}(1-x)^{\alpha}}$. (Such an $m$ and $k$ exist since $\frac{1}{\sqrt{x}}$ is bounded on $(b,1)$)
$\int\limits_{b}^{1}\frac{kdx}{(1-x)^{\alpha}}<\int\limits_{b}^{1}\frac{dx}{\sqrt{x}(1-x)^{\alpha}}<\int\limits_{b}^{1}\frac{mdx}{(1-x)^{\alpha}}=m\frac{1}{(1-\alpha)(1-x)^{\alpha-1}}\big|_{b}^{1}=m\frac{1}{(1-\alpha)(1-1)^{\alpha-1}}-m\frac{1}{(1-\alpha)(1-0)^{\alpha-1}}$.
Since $\frac{1}{(1-1)^{\alpha-1}}$ is only defined for $\alpha-1<0$ we have $\alpha<1$.
$\alpha>\frac{1}{2}$
Work with the last integral and choose an  $f(x)=\frac{k}{x^{\alpha +\frac{1}{2}}}$ and $g(x)=\frac{m}{x^{\alpha+\frac{1}{2}}}$, (I do this because we can choose $m,n\in \mathbb{R}$, such that $m<1<(\frac{x}{x-1})^{\alpha}<\frac{c}{c-1}<k$ since $(\frac{x}{x-1})^{\alpha}$ is bounded on $(c,\infty)$). After that similar steps in-sue.
