\(\displaystyle \displaystyle \int^{1}_{0} \frac{x^{p - 1}}{1 - x} log \Bigl( \frac{1}{x} \Bigr) dx = \displaystyle \int^{1}_{0} x^{p - 1} \sum^{\infty}_{n = 0} x^{n} log \Bigl( \frac{1}{x} \Bigr) dx = \sum^{\infty}_{n = 0} \int^{1}_{0} x^{n + p - 1} log \Bigl( \frac{1}{x} \Bigr) dx\). Now the obvious calculus thing to do is integrate by parts and I'm done. But to do that, assuming I'm treating this like a Riemann integral, I have to evaluate \(\displaystyle log \Bigl( \frac{1}{x} \Bigr) \frac{x^{n + p}}{n + p}\) from 0 to 1. Well okay, 1 is easy. Log of 0 isn't defined, though, and I'm not sure how to do a proofy demonstration that this is whole thing is zero. So that leads me to think that maybe I shouldn't be treating this like a Riemann integral and instead should find some sequence of step functions which approaches what I'm trying to integrate, i.e. to argue it as a Lebesgue integral. But finding a sequence of step functions which approaches \(\displaystyle x^{n + p - 1} log(1/x)\) seems daunting. Ideas?