Proving an integral is equal to $\sum_{k=1}^\infty\frac{1}{(p+k)^2}$ for $p>0$. I'm studying for a qualifying exam and I'm stuck on this problem from Bass's "Real Analysis for Graduate Students" (Exercise 7.14). It asks us to prove that
$$\sum_{k=1}^\infty\frac{1}{(p+k)^2}=-\int_0^1\frac{x^p}{1-x}\log(x)dx$$ for $p>0$. 
Note that this exercise comes from the chapter that introduces the monotone convergence theorem, Fatou's lemma, and the dominated convergence theorem. My issue is probably that I don't immediately see how to apply any of these theorems to this particular problem. I have tried playing around with Feynman's trick and log series, but haven't made any notable progress. I've been stuck here for awhile so any help is appreciated.
 A: First, expand $1/(1-x)$ into its geometric series to get
$$-\int_0^1\sum_{k\ge 0} x^mx^p\log x\,dx$$
Now, consider the partial sums
$$S_N=\sum_{k=0}^Nx^k=\frac{1-x^{N+1}}{1-x}$$
We can easily show that $S_N\le S_{N+1}$, as $S_{N+1}-S_N=x^{N+1}\ge 0$ as $x\in[0,1]$.
By the monotone convergence theorem,
\begin{align}
-\int_0^1\sum_{k\ge 0} x^k x^p\log x\,dx&=-\int_0^1\lim_{N\to\infty}\sum_{k= 0}^N x^k x^p\log x\,dx \\
&=-\lim_{N\to\infty}\int_0^1\sum_{k= 0}^N x^k x^p\log x\,dx \tag{1} \\
&=-\lim_{N\to\infty}\sum_{k= 0}^N\int_0^1 x^k x^p\log x\,dx \\
&=-\sum_{k\ge 0}\int_0^1 x^k x^p\log x\,dx \\
&=\sum_{k\ge 0}\frac{1}{(k+p+1)^2} \tag{2}\\
&=\sum_{k\ge 1}\frac{1}{(k+p)^2} \\
\end{align}
Where the monotone convergence theorem was used in $(1)$ and integration by parts in $(2)$.
A: Proceeding naively,
$\begin{array}\\
\int_0^1\frac{x^p}{1-x}\log(x)dx
&=\int_0^1x^p\log(x)\sum_{n=0}^{\infty} x^ndx\\
&=\sum_{n=0}^{\infty} \int_0^1x^{p+n}\log(x)dx\\
&=\sum_{n=0}^{\infty} \dfrac{x^{n+p+1}((n+p+1)\ln(x)-1}{(n+p+1)^2}|_0^1
\qquad\text{(according to Wolfy)}\\
&=\sum_{n=0}^{\infty}\dfrac1{(n+p+1)^2} (x^{n+p+1}((n+p+1)\ln(x)-1)|_0^1\\
&=\sum_{n=0}^{\infty}\dfrac{-1}{(n+p+1)^2}\\
&=\sum_{n=1}^{\infty}\dfrac{-1}{(n+p)^2}\\
\end{array}
$
Looks OK to me.
