How to evaluate $\int_0^1 \frac{\log x}{x-1}dx$

Question

$$I = \int_0^1 \frac{\log x}{x-1}dx$$ I tried to let $\log x = t \implies dx = e^t dt$ $$I = \int_0^1 \frac{te^t}{e^t-1}dt = \frac12+\int_0^1 \frac{t}{e^t-1}dt$$ I could not go further than this step, other than prove that $I > \frac12$.

Answer 1

$$I = \int_0^1\frac{\log x}{x-1} dx= -\int_0^1\log x(1+x+x^2+...)dx \\ = -\int_0^1 \log x \sum_{r=0}^\infty x^r dx \\ = -\sum_{r=0}^\infty\int_0^1x^r \log x dx$$ let $\log x = t \implies dx = e^t dt$: $$I = -\sum_{r=0}^\infty \int_{-\infty}^0te^{(r+1)t}dt\\ =-\sum_{r=0}^\infty \left[\frac{te^{(r+1)t}}{(r+1)} - \frac{e^{(r+1)t}}{(r+1)^2}\right]_{-\infty}^0 \\ = \sum_{r=0}^\infty \left[\frac{1}{(r+1)^2} + \lim_{t \to -\infty} \frac{(t(r+1)-1)e^{(r+1)t}}{(r+1)^2} \right]\\ = \sum_{r=0}^\infty \left[\frac{1}{(r+1)^2} - \lim_{t \to \infty} \frac{(t(r+1)+1)}{e^{(r+1)t}(r+1)^2} \right] \\ = \sum_{r=0}^\infty \left[\frac{1}{(r+1)^2} - \lim_{t \to \infty} \frac{r+1}{e^{(r+1)t}(r+1)^3} \right] \\ = \sum_{r=0}^\infty \frac{1}{(r+1)^2}\\ = \sum_{k=1}^\infty \frac{1}{k^2}\\ = \frac{\pi^2}{6}$$

The computation of the series $\sum \frac 1 {k^2}$ is also known as the Basel problem and has many good proofs here and here

The general form of this integral is known as Spence's Function or Dilogarithm. It's a transcendental function which has the values $Li_2(0)=0$ and $Li_2(1)=\zeta(2)=\frac{\pi^2}{6}$. If you know these values beforehand, you can cut out the lengthy calculation.

EDIT: Another easy way to calculate the integral is by using the taylor series expansion of $\log(1-x)$ $$I = -\int_0^1 \frac{\log(1-x)}{x}dx = \int_0^1 1 + \frac x2 + \frac{x^2}{3}+ \frac{x^3}{4} + ... dx\\ = \left[x + \frac{x^2}{2^2} + \frac{x^3}{3^2} + ... \right]_0^1 \\ = 1 + \frac1 {2^2} + \frac1 {3^2} + \frac 1 {4^2} + ...\\ = \frac{\pi^2}{6}$$