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

Solution 1:

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.

Solution 2:

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

Solution 3:

By definition $$\sum_{k=1}^{\infty}(-1)^{k} \frac{(\beta-1)^{k}}{k^2 }=\text{Li}_2(1-\beta )$$ If $\beta$ is small $$\text{Li}_2(1-\beta )=\zeta(2)+\beta \log \left(\frac{\beta }{e}\right)+O\left(\beta ^2\right)$$