I am trying to compute an integral of the form
$I=\int_0^{x_\alpha}dx\,\frac{x^k}{(1-x^m)^n}\,,$
where $x_\alpha=\exp(2\pi i\alpha/m)$, i.e. some m-th root of unity and $m>0$ a positive integer. It looks a lot like Euler's beta function $B(a,b)$, and makes me want to do a substitution
$t=\frac{x^m}{x_\alpha^m}\,,\qquad x=x_\alpha t^\frac{1}{m}\,,\qquad dx=\frac{x_\alpha}{m}t^{\frac{1}{m}-1}dt\,.$
Plugging back one gets
$I=\frac{x_\alpha^{k-1}}{m}\int_0^1 t^{\frac{k+1}{m}-1}(1-t)^{(1-n)-1}=\frac{x_\alpha^{k-1}}{m}B(\frac{k+1}{m},1-n)\,.$
However, I could have defined $t$ such that $x=x_\beta t^\frac{1}{m}$, with $x_\beta$ another root of unity. The computation carries the same, as the upper integration boundary is still $t(x_\alpha)=(x_\alpha/x_\beta)^m=1$, so I get
$I=\frac{x_\beta^{k-1}}{m}B(\frac{k+1}{m},1-n)\,.$
For arbitrary $k$ these are obsiously not the same. Am I forced to choose a particular root when doing the change of variables? If so, what is the reason?
You are talking about complex integrals. You are integrating on a different path, so it's no wonder that the integrals are different