Integral $\int_{0}^{1} \dfrac{\ln(1-x)}{x}~dx$

Question

The integal is $$\int_{0}^{1} \dfrac{\ln(1-x)}{x}~dx$$

My opnion is use fundamental theorem of calculus, calculate $F(x)$ and then use $F(1^-)-F(0^+)$.

But i cannot find the undetermined integral of $\frac{\ln(1-x)}{x}$.

Answer 1

Use the maclaurin series of $\ln{(1-x)}$:

$$I=\int_0^1 \sum_{n=1}^{\infty} -\frac{x^n}{n} \cdot \frac{1}{x} \; dx$$

By Fubini theorem, we can interchange the integral and summation: $$I=-\sum_{n=1}^{\infty} \int_0^1 \frac{x^{n-1}}{n} \; dx$$ $$I= -\sum_{n=1}^{\infty} \frac{x^n}{n^2} \bigg \rvert_0^1$$ $$I= -\sum_{n=1}^{\infty} \frac{1}{n^2}$$ $$I=-\frac{\pi^2}{6}$$