Prove that $\int_{0}^{1} \frac{\ln(x)}{x-1} dx = \sum_{k=1}^{\infty} \frac{1}{k^2}$ Notice: I've seen questions alike these on MSE, however, from what I've found, none of them presents a solution like mine. That's why I hope some of you can confirm whether the use of my method produces a correct solution. Thanks.
The problem is stated as:

Prove that $\int_{0}^{1} \frac{\ln(x)}{x-1} dx = \sum_{k=1}^{\infty} \frac{1}{k^2}$

My attempt:
First of all, we use the fact that $\ln(x) = \ln(1+(x-1))$ which then can be written as $\sum_{k=1}^{\infty} \frac{(-1)^{k-1}(x-1)^k}{k}dx$, substituting this into the integrand, we have:
$$\int_{0}^{1} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}(x-1)^{k-1}}{k}dx$$
We would like to interchange the summation and the integral, since that'd simplify the problem alot. In order to do this, one has to show that the integrand converges uniformly on $[0,1]$. Since we have a power series, it's easy to see that this does indeed converge uniformly within our radius of convergence given by $R = 1$.
In other words, we have uniform convergence for $|x-1| < 1 \Leftrightarrow x \in (0,2)$. Therefore, it also converges uniformly on the interval $x \in [\beta, 1]$ for $\beta > 0$. Our original expression can therefore be rewritten to:
$$ \lim_{\beta \rightarrow 0^+} \int_{\beta}^{1} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}(x-1)^{k-1}}{k}dx$$
Now, we can interchange the summation with the integral, since we are integrating within our radius of convergence for the power series, for which we have uniform convergence, as previously stated. After some algebraic simplifications, we have:
$$ \lim_{\beta \rightarrow 0^+} \sum_{k=1}^{\infty} - \frac{(-1)^{k-1}(\beta-1)^{k}}{k^2 }$$
Once again, we can move our limit inside the summation, since we have that the sum converges uniformly on the interval $[\beta, 1]$, which can easily be verified by the Weierstrass M - test.
Finally, we have that:
$$ \sum_{k=1}^{\infty} \lim_{\beta \rightarrow 0^+} - \frac{(-1)^{k-1}(\beta-1)^{k}}{k^2} =\sum_{k=1}^{\infty} - \frac{(-1)^{k-1}(-1)^{k}}{k^2} = \sum_{k=1}^{\infty} \frac{1}{k^2} $$
QED.

Feel free to add any comments on my solution. I'm mostly unsure of the step where I add the limit for $\beta$.
Thanks.
 A: It's a nice approach. The only thing I would clean up a bit:
When applying the Weierstrass M-test, use the fact that
$ \sum_{k=1}^{\infty} \left|-\frac{(-1)^{k-1}(\beta-1)^{k}}{k^2 }\right|\leq \sum_{k=1}^{\infty} \frac{1}{k^2 }<\infty$
so that the sum is dominated by a sum that's independent of $\beta$. Then the limit as $\beta \to 0^+$ may proceed.
A: Substitute $x=1-t$
$$-\int_{0}^{1}\frac{\ln(1-t)}{t}dt=\int_{0}^{1}\sum_{r=1}^{\infty}\frac{t^{r}}{tr}dt=\\\int_{0}^{1}\sum_{r=1}^{\infty}\frac{t^{r-1}}{r}dt=\\\sum_{r=1}^{\infty}\int_{0}^{1}\frac{t^{r-1}}{r}dt=\sum_{r=1}^{\infty}\frac{1}{r^{2}}=\zeta(2)$$.
For justification of interchange of summation and integral see this
A: By definition
$$\sum_{k=1}^{\infty}(-1)^{k}  \frac{(\beta-1)^{k}}{k^2 }=\text{Li}_2(1-\beta )$$ If $\beta$ is small
$$\text{Li}_2(1-\beta )=\zeta(2)+\beta \log \left(\frac{\beta }{e}\right)+O\left(\beta ^2\right)$$
A: Just expand $\frac{1}{x-1}$ into a geometric series and do the elementary integral.
More precisely, letting $x=e^{-t}$ the integral becomes
$$i= \int_{0}^{\infty} \frac{t e^{-t}}{1-e^{-t}}\;dt$$
Expanding the integrand into a geometric series we obtain
$$i=\sum_{k=1}^{\infty} \int_{0}^{\infty} t e^{-k t}\;dt$$
and letting $t=\frac{s}{k}$ i.e. $dt=\frac{ds}{k}$ the integral becomes
$$\frac{1}{k^2} \int_{0}^{\infty} s e^{-s}\;ds=\frac{1}{k^2}$$
since the integral is equal to unity. Q.E.D.
