Question: I am trying to solve the following integral: $$\int\limits_{-x_0}^{x_0} \frac{|x|^{\alpha}}{x-c}\; dx$$
with $c\in \mathbb{C}$, $x_0\in \mathbb{R}^+$, $\alpha \in \mathbb{R}$.
Attempts: I split the absolute value and work on $$\int\limits_{0}^{x_0} \frac{x^{\alpha}}{x-c}\; dx,$$ I substitute $x'=x+c$ and then apply Newton's generalized binomial theorem. $$\int\limits_{0}^{x_0}\frac{x^{\alpha}}{x-c}\; dx=\int\limits_{-c}^{x_0-c} \frac{\left(x+c\right)^{\alpha}}{x}dx \;=\sum\limits_{k=0}^{\infty}\binom{\alpha}{k}\int\limits_{-c}^{x_0-c}dx\;\left\{\begin{array}{r}x^{\alpha-k-1}c^k,&|x|>|c|\\ x^{k-1}c^{\alpha-k},&|x|<|c|\end{array} \right. .$$ Then I can integrate over $x$.
This looks rather complicated though, as I have to consider different cases depending on the relative size of the absolute value of $x$ and $c$ to apply the binomial theorem correctly, thus breaking the integral and so on.
Any suggestion to solve this in a better way?