It's claimed that $$\int_0^1 \frac{\log(1-x)}{x}dx=-\frac{\pi^2}{6}$$ by first expanding $\frac{\log(1-x)}{x}$ into a power series and then doing term-by-term integration. I want to justify this by Levi's Theorem for Series, that is, if $(f_n)$ is a sequence of $L^1$ functions such that $\sum_{n=0}^{\infty} \int_X \lvert f_n\rvert \,\text{d}\mu \lt \infty$. Then $\sum_{n=0}^{\infty} f_n$ converges a.e. to an $L^1$ function $f$ such that $$\int_X \sum_{j=1}^{\infty} f_k\text{d}\mu=\sum_{k=0}^{\infty}\int_Xf_k\,\text{d}\mu.$$ Note that we are focusing on Lebesgue integration, since the integrand is not defined at $x=0$ or $x=1$. I'm not sure how to show this, possibly by Taylor expansion? Could someone help to show to derive this? Thanks.
[Math] Show $\int_0^1 \frac{\log(1-x)}{x}dx=-\frac{\pi^2}{6}$
integrationlebesgue-integralsequences-and-seriestaylor expansion
Best Answer
$$ \begin{eqnarray} \int_0^1\frac{\ln(1-x)}{x}\ \mathrm{d}x&=&\int_0^1\frac{1}{x}\sum_{n=1}^\infty\frac{(-1)^{n-1}(-x)^n}{n} & \textrm{(Taylor series of }\ln(1+z))\\ &=& \int_0^1\sum_{n=1}^\infty\frac{1}{x}\frac{(-1)(x^n)}{n}\\ &=& -\sum_{n=1}^\infty\int_0^1\frac{x^{n-1}}{n}\ \mathrm{d}x \\ &=&-\sum_{n=1}^\infty\left[\frac{x^n}{n^2}\right]_0^1 \\ &=&-\sum_{n=1}^\infty\frac{1}{n^2} \\ &=&-\frac{\pi^2}{6} \end{eqnarray} $$
The last equality is the Basel problem and requires a more complex approach.