For which $p,q$ does the $\int_0^{\infty} \frac{x^p}{\mid{1-x}\mid^q}dx$ exist? For which $p,q$ does the $\int_0^{\infty} \frac{x^p}{\mid{1-x}\mid^q}dx$ exist ?
Can you help me, I have been siting hours on this question .
I got that for $ q<1$ and $p>q+1$, but I am not sure if thats right
 A: The only correct bound you have is $$q<1.$$ $x^p|1-x|^{-q}$ has potential issues at $0,1,+\infty$. 


*

*At $0$, the function is like $x^p$. need $p>-1$.

*At $1$, the function is like $|1-x|^{-q}$. Need $q<1$.

*At $\infty$, the function is like $x^{p-q}$. Here we need $p-q<-1$. 


The result is the following triangle of admissible values: ($x$-axis is $p$, $y$-axis is $q$)

From this graph, we can verify that your answer is not right: your answer allows the values $(p,q) = (-5,-10)$. Indeed, $-10=q<1$ and $p=-5 > -10+1=q+1$. But this gives the integral
$$ \int_0^\infty \frac{|1-x|^{10}}{x^5} dx$$
This integrand has a non-integrable singularity at 0, is integrable around 1, and has polynomial growth at infinity.
A: I would try to get rid of the absolute value to simplify:
$$
\int_0^\infty \frac{x^pdx}{|1-x|^q}
 = \int_0^1 \frac{x^p dx}{(1-x)^q}
 + \int_1^\infty \frac{x^p dx}{(x-1)^q}
$$
Now the first integral has a problem at $x \to 1^-$, and the second as $x \to 1^+$, and also a potential problem at $x\to \infty$, and if $p < 0$ then both of those have a problem at $x \to 0$.
Can you take it from here?
