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**ragnar** I am asked to show that $\displaystyle \displaystyle \int^{1}_{0} \frac{x^{p - 1}}{1 - x} log \Bigl( \frac{1}{x} \Bigr) dx = \sum^{\infty}_{n = 0} \frac{1}{(n + p)^{2}}$ for p > 0. I have a lot of it so far, and am basically on the cusp.

$\displaystyle \displaystyle \int^{1}_{0} \frac{x^{p - 1}}{1 - x} log \Bigl( \frac{1}{x} \Bigr) dx = \displaystyle \int^{1}_{0} x^{p - 1} \sum^{\infty}_{n = 0} x^{n} log \Bigl( \frac{1}{x} \Bigr) dx = \sum^{\infty}_{n = 0} \int^{1}_{0} x^{n + p - 1} log \Bigl( \frac{1}{x} \Bigr) dx$. 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?