Problem Statement
Prove that $$\sum_{k=1}^{\infty}\frac{1}{(k+p)^2}=-\int_0^1\frac{x^p}{1-x}\log x \ \mathrm{d}x$$
Background
I just learned the limit theorems (MCT, LDCT, Fatou's Lemma). This problem is the same as this one except that I wish for a solution using the tools I know, rather than complex analysis.
Attempt First, recalling the geometric series, \begin{align*} -\int_0^1\frac{x^p}{1-x}\log x \ \mathrm{d}x&=-\int_0^1 \lim_{n\to\infty} \sum_{k=0}^{n}x^{k+p} \log x \ \mathrm{d}x \\ &=-\lim_{n \to \infty}\int_0^1\sum_{k=0}^{n}x^{k+p} \log x \ \mathrm{d}x, \end{align*} where the limits were interchanged by invoking the "reverse" Monotone Convergence after realizing that $\require{cancel}\sum_{k=0}^{n}x^{k+p} \log x \geq \sum_{k=0}^{n+1}x^{k+p} \log x.$ Next, we use integration by parts as follows: \begin{align*} -\lim_{n \to \infty}\int_0^1\sum_{k=0}^{n}x^{k+p} \log x \ \mathrm{d}x&=-\lim_{n \to \infty}\int_0^1\log x \ \mathrm{d}\left(\sum_{k=1}^{n}\frac{x^{k+p}}{k+p} \right)\\ &=-\lim_{n\to\infty}\left(\cancelto{0}{\log(1)\sum_{k=1}^{n}\frac{x^{k+p}}{k+p}}-\cancelto{0}{\lim_{x\to 0}\log(x)\sum_{k=1}^{n}\frac{x^{k+p}}{k+p}}\\ -\int_0^1\sum_{k=1}^{n}\frac{1}{x}\frac{x^{k+p}}{k+p}\ \mathrm{d}x\right)\\ &=\lim_{n\to\infty}\int_0^1\sum_{k=1}^n \frac{x^{k+p-1}}{k+p} \\ &=\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{(k+p)^2}, \end{align*} as desired.
Question
Are my steps rigorous and justified enough for a first year graduate real analysis course?
You have $\sum_{k=0}^{n} x^{k+p} \leq \sum_{k=0}^{n+1} x^{k+p}$, but as $x\leq 1$, you have $\log(x)\leq 0$ so you actually have $\sum_{k=0}^{n} x^{k+p} \log(x) \geq \sum_{k=0}^{n+1} x^{k+p} \log(x)$. Anyway, as your functions have negative values, you can actually use their opposites, which will change the inequality and allows you to use MCT.
Except for that, it seems rigorous enough for me (although you have an innatention error in the second line of your second calculation, it should be the limit when $x$ goes to 0 not to $\infty$ I guess).