Convergence of the improper integral $\int_{0+}^{1-} \frac{\log x}{1-x} dx$

Question

Let $0 < t_{1} \leq t_{2} < 1.$ Then $$\int_{t_{1}}^{t_{2}} \frac{\log x}{1-x} dx = \int_{1/t_{2}}^{1/t_{1}} \frac{\log u^{-1}}{1 - u^{-1}}(-u^{-2}) du = \int_{1/t_{2}}^{1/t_{1}} \frac{\log u}{u^{2} - u} du = \int_{1/t_{2}}^{1/t_{1}} \frac{u^{1/2}}{u^{2}-u} du = l + o(1)\bigg|_{t_{1} \to 0+}$$ for some $l \in \mathbb{R},$ so $\int_{0+}^{t_{2}} \frac{\log x}{1-x} dx$ converges.

However, I am not sure how to show the remaining part, $i.e.$ the part where $t_{2} \to 1-.$ (Though an answer is certainly appreciated, I would like to see if the present approach can be continued.)

Answer 1

$$\int \frac{\log x}{1-x} dx=\text{Li}_2(1-x)$$ So, $$\int_{\epsilon}^{1-\epsilon} \frac{\log x}{1-x} dx=\text{Li}_2(\epsilon )-\text{Li}_2(1-\epsilon )$$ and the limit is $-\frac{\pi ^2}{6}$ when $\epsilon$ goes to $0$ as Lucian commented.

Answer 2

Near $1$, $\frac{\log(x)}{1-x} = -\frac{\log(x)-\log(1)}{x-1} = -1 + O(|x-1|)$ by Taylor's theorem. In particular there is a small interval around $1$ where $\left | \frac{\log(x)}{1-x} \right | \leq 2$. That should get you the convergence you need.