Behavior of the following function at $x=0$ singularity

Question

I am trying to do the following integral: \begin{equation} \int\frac{1}{x^{2p}(x-1)^{2q}}\,\mathrm dx \end{equation} for positive $2p$ and $2q$. I want to understand how does this function blow up (the $x$ behavior) at $x=0$. In Mathematica e.g., if I do this integral with unspecified $2p$, the result is quite different from when I do it with a specific value of $2p$, like say $10$. I am not sure how Mathematica is doing it.

Answer 1

I suggest changing the second factor to $(1-x)^{2q}$, to get the minus sign out of the way, and thus to allow non-integer $q$. Then in some neighborhood of $0$ we have $$\frac1{2 x^{2p}} \le \frac{1}{x^p(1-x)^{2q}} \le \frac3{2 x^{2p}} $$ Integration preserves the inequality (up to additive term coming from $+C$) $$\frac1{2|2p-1| x^{2p-1}}-C \le \int\frac{1}{x^p(1-x)^{2q}} \le \frac3{2|2p-1| x^{2p-1}}+C $$ Simply put, the integral blows up like $1/x^{2p-1}$. An exceptional case is $p=1/2$; then the singularity is logarithmic.