Show $\int_0^1 \frac{\log(1-x)}{x}dx=-\frac{\pi^2}{6}$ 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.
 A: $$
\begin{eqnarray}
\int_0^1\frac{\ln(1-x)}{x}\ \mathrm{d}x&=&\int_0^1\frac{1}{x}\sum_{n=1}^\infty\frac{(-1)^{n-1}(-x)^n}{n} & \textrm{(Taylor series of }\ln(1+z))\\
&=& \int_0^1\sum_{n=1}^\infty\frac{1}{x}\frac{(-1)(x^n)}{n}\\
&=& -\sum_{n=1}^\infty\int_0^1\frac{x^{n-1}}{n}\ \mathrm{d}x \\
&=&-\sum_{n=1}^\infty\left[\frac{x^n}{n^2}\right]_0^1 \\
&=&-\sum_{n=1}^\infty\frac{1}{n^2} \\
&=&-\frac{\pi^2}{6}
\end{eqnarray}
$$
The last equality is the Basel problem and requires a more complex approach.
A: If you want to be formal, the Taylor Expansion (Wikipedia) $\log(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}$ converges uniformly on sets of the form $[\epsilon,0]$.
So you can calculate
$$\int_\epsilon^1\frac{\log(1-x)}{x}dx=-\sum_{n=1}^\infty\int_\epsilon^1\frac{x^{n-1}}{n}=-\sum_{n=1}^\infty\frac{1-\epsilon^n}{n^2}=-\frac{\pi^2}{6}+\sum_{n=1}^\infty\frac{\epsilon^n}{n^2}.$$
where the last equality is well-known Basel problem.
Now the integral on the LHS is simply the integral of $\frac{\log(1-x)}{x}\chi_{[\epsilon,1]}$ on $[0,1]$, where $\chi_{A}$ denotes the characteristic function of a set $A$. This converges pointwise, and in a decreasing way, to $\frac{\log(1-x)}{x}$ on $(0,1]$, so (and all these functions are negative), so by Monotone Convergence, 
$$\int_0^1\frac{\log(1-x)}{x}dx=\lim_{\epsilon\to 0^+}\int_\epsilon^1\frac{\log(1-x)}{x}dx=\lim_{\epsilon\to 0^+}-\frac{\pi^2}{6}+\sum_{n=1}^\infty\frac{\epsilon^n}{n^2}=-\frac{\pi^2}{6}$$
where the limit $\lim_{\epsilon\to 0}\sum_{n=1}^\infty\frac{\epsilon^n}{n^2}=0$ can be calculated directly (for $\epsilon<1$, the terms in the sum are always smaller than $\epsilon/n^2$, so you can use the Basel problem again).
A: This can be generalized as
$$
\begin{align}
\int_0^1\frac{\log(1-x)^{n-1}}x\,\mathrm{d}x
&=\int_0^1\frac{\log(x)^{n-1}}{1-x}\,\mathrm{d}x\\
&=(-1)^{n-1}\int_0^\infty\frac{x^{n-1}}{1-e^{-x}}\,e^{-x}\,\mathrm{d}x\\
&=(-1)^{n-1}\sum_{k=1}^\infty\int_0^\infty x^{n-1}e^{-kx}\,\mathrm{d}x\\
&=(-1)^{n-1}\sum_{k=1}^\infty\frac1{k^n}\int_0^\infty x^{n-1}e^{-x}\,\mathrm{d}x\\[6pt]
&=(-1)^{n-1}\zeta(n)\,\Gamma(n)\tag{1}
\end{align}
$$
For $n=2$, this gives
$$
\begin{align}
\int_0^1\frac{\log(1-x)}x\,\mathrm{d}x
&=-\zeta(2)\Gamma(2)\\
&=-\frac{\pi^2}6\tag{2}
\end{align}
$$
A: A more intuitive explanation, and the way to actually derive this integral (something I tried messing around with a while back) is to try long division of $\frac{1}{1-y}$. The motivation for this is to attempt a new solution at the Basel problem. 
$$\frac{1}{1-y} = 1 + y + y^2 + y^3 \dots$$
You get an infinite series of $1 + y + y^2 + y^3 \dots$ onto infinity. You integrate both sides to get $$-\ln(1-y) = y + \frac{y^2}{2} + \frac{y^3}{3} + \frac{y^4}{4}\dots \text{ on to infinity }$$ 
Because they are equivalent functions, the constants of integration ($C$) cancel out on both sides. You want the denominators of the right hand side to match the denominators of the series $\frac{1}{n^2}$, so that you can solve the Basel problem. So you divide both sides by $y$ and integrate once more from $0$ to $1$ to get the series $$\int_0^1\frac{-\ln(1-y)}{y}dy= 1 + 1/4 + 1/9 + 1/16\dots$$ You reverse the bounds of integration to account for the negative sign from earlier, and you get $$\int_1^0\frac{\ln(1-y)}{y}dy= 1 + 1/4 + 1/9 + 1/16\dots=\frac{\pi^2}{6}$$
The same integral with reversed bounds is the negative answer.
$$\int_0^1\frac{\ln(1-y)}{y}dy= -\frac{\pi^2}{6}$$
