How to evaluate $\int_0^1 -\frac {\ln(1-x)}{x}dx$

Question

This integral came up while I was trying to evaluate $$\sum_{n=1}^\infty \frac {1}{n^2}.$$ The value should be $$\frac {\pi^2}{6},$$ but how do I solve the integral and evaluate it?

Answer 1

Substitute $t=1-x$ to reexpress the integral as

$$\int_0^1 -\frac {\ln(1-x)}{x}dx = \int_0^1 \frac {\ln t}{t-1}dt = \frac {\pi^2}6$$

where $\int_0^1 \frac {\ln x}{x-1}dx = \frac {\pi^2}6$ is used in the last step

Answer 2

$$I=\int_{0}^{1} \frac{\ln(1-x)}{x}dx= \int_{0}^{1} \sum_{k=1}^{\infty}\frac{x^{k-1}}{k}=\sum_{k=1}^{\infty} \frac{1}{k^2}=\zeta(2)=\frac{\pi^2}{6}.$$

Answer 3

Using the definition of polylogarithms $$I(a)=-\int_0^a \frac {\ln(1-x)}{x}dx=\text{Li}_2(a)$$ Expandinga series around $a=1$ gives $$I(a)=\frac {\pi^2}6-\frac{1}{4} (1-a) (5-a-2 (3-a) \log (1-a))+\cdots$$

If you use this approximation for $a=0.75$, it will give $0.989$ while the exact value is $0.978$.