TITLE: Uniform Convergence of $\sum_{n=1}^\infty -x^{2n} \ln x$
QUESTION [7 upvotes]: Last month, I read a proof of $\displaystyle \sum_{n=0}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ that used the following integral:
$$ \int_0^\infty \int_0^\infty \frac{1}{(1+y)(1+x^2y)} \, dx \, dy$$ It was in August-September's issue of the American Mathematical Monthly. Anyway, I was able to follow most of  it upto the last step where he switches integral and summation.
$$\int_0^1 \frac{\ln x}{x^2-1} dx = \sum_{n=0}^\infty \int_0^1 -x^{2n} \ln x\,dx$$
I think I grasp how uniform convergence of a series can allow you to switch infinite sums and integral signs. But from what I can see $\sum_{n=1}^\infty -x^{2n} \ln x$ doesn't uniformly converge on $[0,1]$. It may be uniformly convergent on $[0,k]$ if $k \in (0,1)$? Is that enough to justify the switch?
Alternatively, the partial sums are strictly increasing and bounded. But how do you prove they converge to the integral on the left side?
Edit: 
Is there a theorem that says: if $f_n \to f$ pointwise on $[a,b]$ and $f_n$ is uniformly bounded then, $\int_a^b f_n(x) \, dx \to \int_a^b f(x) \, dx$? Is the following a proper proof? I'm trying to avoid using Lebesgue Integrals as I haven't studied them.
Suppose that $\displaystyle \left| \int_0^1 f(x)\,dx - \lim_{n\to \infty} \int_0^1 f_n(x)\,dx \right| = \epsilon>0$. 
Since $\displaystyle \int_0^k f(x)\, dx = \lim_{n \to \infty} \int_0^k f_n(x)\,dx$ by uniform convergence on $[0,k]$. It remains to show that if $k$ is chosen properly then $\displaystyle \left| \int_k^1 f(x)\,dx-\lim_{n\to\infty} \int_k^1 f_n(x)\,dx \right| < \epsilon$, a contradiction. $f_n$ being uniformly bounded implies
$$ \left| \int_k^1 f(x)\,dx - \lim_{n\to\infty} \int_k^1 f_n(x)\,dx \right| < (1-k)M$$
So let $1-\frac{\epsilon}{M}<k<1$ so that $(1-k)M < \epsilon$.

REPLY [2 votes]: $\cal Theorem:$ If $\{f_n:X\rightarrow R\}_n$  converges to $f$ pointwise and $\{f_n\}_n, f$ be continuous and $f_{n+1}(x)\leq f_n(x)$ for all $x\in X$ and $X\subset R$ be a compact set, then it converges uniformly.
I'm not sure, but I think this theorem can help you.
And also check the first chapter of Wlter Rudin's book it is named "Real and Complex Analysis".This is the theorem you asked for it(Theorem 1.34).