I have to prove as an excercise of my Measure Theory lessons that for $p > -1$ it is true that: $$\int_0^1 \frac{-\ln(x) x^p}{1-x} dx = \sum_{n \geq 1} \frac{1}{(n+p)^2}$$
As an advice, the excercise says that there is some series of part of the expression bellow the integral that helps to prove the result. I am trying to use that $x\in (0,1)$ so $\frac{1}{1-x} = \sum_{n \geq 0} x^n$. Applying to this excercise we have:
$$\int_0^1 \frac{-\ln(x) x^p}{1-x} dx = \int_0^1 -\ln(x) \cdot \sum_{n \geq 0} x^{n+p} dx = \sum_{n \geq 0} \int_0^1 -\ln(x) \cdot x^{n+p} dx $$
The thing is that the last integral diverges: $$ \int_0^1 -\ln(x) \cdot x^{n+p}=\left [\dfrac{{x}^{n+p+1}}{(n+p+1)^2}-\dfrac{\left(n+p+1\right)\,{x}^{n+p+1}\,\ln\left(x\right)}{(n+p+1)^2} \right]_{x \rightarrow 0} ^{x=1} $$ and $\lim_{x \rightarrow 0} \ \ln(x) = -\infty$, so I am getting stuck and do not know how to proceed.
Thanks in advance :)