Convergence and absolute convergence of $\int_0^\infty \frac{ \sin \pi x}{x^\alpha|1-x|^\beta}\text{d}x$ Please. I need help.
I have to study the convergence and the absolute convergence of $$\int_0^\infty \frac{ \sin \pi x}{x^\alpha|1-x|^\beta}\text{d}x,$$
where $\alpha$, $\beta>0$.
EDIT: Thanks to a comment below, I've already solved the absolute convergence.
Now, for the conditional convergence I tried to do it using integration by parts, choosing $$u=\frac{1}{x^\alpha|1-x|^\beta}$$ and $dv=\sin \pi x$ but I'm not sure that this can help. Any hint?
 A: You need to look at the points $x=0,1,\infty$ separately. At $x=0$ the integrand behaviour is $x^{1-\alpha}$, at $x=1$ we get $x^{1-\beta}$, and as $x\to\infty$, it is like $x^{-\alpha-\beta}$. This means that the integral is absolutely convergent if and only if
$$
\max\{\alpha,\beta\}<2\quad\textrm{and}\quad \alpha+\beta>1.
$$
On the other hand, for conditional convergence, nothing will change for the singularities at $x=0$ and $x=1$, because the integrand does not oscillate around these points. Hence the condition $\max\{\alpha,\beta\}<2$ is necessary. However, as $x\to\infty$ the integrand oscillates, meaning that the integral might converge conditionally but not absolutely in some cases. The integral behaves like
$$
\int_1^\infty\frac{\sin x}{x^{\alpha+\beta}}dx
$$
and an integration by part argument shows that this is conditionally convergent if and only if $\alpha+\beta>0$. To conclude, the original integral is conditionally convergent if and only if
$$
\max\{\alpha,\beta\}<2,
$$
since $\min\{\alpha,\beta\}>0$ is given.
