Showing $\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$ Is there way to show $$\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$$ without using the Riemann zeta function?
 A: Note: As pointed out by Hans, this answer uses the fact that $\zeta(2) = \dfrac{\pi ^2}{6}$ and therefore isn't what the OP asked for.
$$\frac{\ln(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3}...$$
$$\int\frac{\ln(1+x)}{x}dx = x - \frac{x^2}{4} + \frac{x^3}{9}...+C$$
$$\int\limits^{0}_{1}\frac{\ln(1+x)}{x}dx = -1 + \frac{1}{4} - \frac{1}{9}...$$

After question edit:
We have $$\frac{\ln(1-x)}{x} = - 1 + \frac{x}{2} - \frac{x^2}{3}...$$
$$\int\frac{\ln(1-x)}{x}dx = -x + \frac{x^2}{4} - \frac{x^3}{9}...+C$$
Therefore $$\int\limits^{0}_{1}\frac{\ln(1-x)}{x}dx = 1 + \frac{1}{4} + \frac{1}{9}...$$
$$=\frac{\pi ^2}{6}$$
A: (I assume that what you mean is that you don't want to use the fact that $\sum 1/n^2=\pi^2/6$.)
Yes, there is a way of seeing this, and this is the basis of Mikael Passare's paper How to compute $\sum 1/n^2$ by solving triangles
(free preprint on arXiv,
or the published version on JSTOR for subscribers).
If you set $x=e^{-t}$, your integral becomes
$$
\int_0^{\infty} -\ln(1-e^{-t}) \, dt
,
$$
or
$$
\int_0^{\infty} -\ln(1-e^{-x}) \, dx
$$
if we call the variable $x$ again.
This is the area of a region in the first quadrant of the $xy$-plane, below the curve $y=-\ln(1-e^{-x})$ (or equivalently $e^{-x}+e^{-y}=1$). This is the region called $U_0$ i Passare's paper, and he shows that there are two other regions $U_1$ and $U_2$ with the same area, and then via a clever area-preserving change of variables that the combined area of $U_0$, $U_1$ and $U_2$ equals the area of a certain right triangle $T$ which is obviously $\pi^2/2$. Hence the area of $U_0$ (your integral) is $\pi^2/6$.
(See also this answer and this.)
A: To expand on Tim's solution:
The series expansion of the logarithm yields the following representation for all $x \in [-1, 1)$:
$$\log(1 - x) = -\sum \limits_{n = 1}^\infty \frac{x^n}{n}$$
This means that we need to calculate the following integral:
$$\int_0^1 \sum \limits_{n = 1}^\infty \frac{x^{n - 1}}{n} \; dx$$
Ususally you would apply the dominated convergence theorem to interchange the integral and the limit at this point. However, the series doesn't have an integrable majorant [I didn't check this rigorously, but I'm pretty sure this is the case].
We circumvent this by writing the integral as a limit:
$$\int_0^1 \sum \limits_{n = 1}^\infty \frac{x^{n - 1}}{n} \; dx = \lim \limits_{t \nearrow 1} \int_0^t \sum \limits_{n = 1}^\infty \frac{x^{n - 1}}{n} \; dx$$
Now since the Taylor series converges uniformly on every interval $[0, t]$ for $0 < t < 1$, we are allowed to interchange the sum and the integral:
$$\lim \limits_{t \nearrow 1} \int_0^t \sum \limits_{n = 1}^\infty \frac{x^{n - 1}}{n} \; dx = \lim \limits_{t \nearrow 1} \sum \limits_{n = 1}^\infty\int_0^t \frac{x^{n - 1}}{n} \; dx = \lim \limits_{t \nearrow 1}\sum \limits_{n = 1}^\infty \frac{t^n}{n^2}$$
Finally we can interchange the limit and the sum by Abel's theorem:
$$\lim \limits_{t \nearrow 1}\sum \limits_{n = 1}^\infty \frac{t^n}{n^2} = \sum \limits_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$
A: Using much the same method as above, for the required integral $\int_{1}^{0}\dfrac{\ln(1-x)}{x}dx$ it "feels reasonable" to first make the substitution $1-x = e^{-z}$ then $dx|_{x=1}^{x=0} = e^{-z}dz|_{z=\infty}^{z=0}$ so that the integral can be written as
$\int^{\infty}_{0}\dfrac{z e^{-z}}{1-e^{-z}}dz$.
After this expand $\dfrac{1}{1-e^{-z}} = \sum_{n=0}^{n=\infty} e^{-n z}$. Eventually have to deal with the sum $\sum_{n=0}^{n=\infty} \dfrac{1}{(n+1)^2}$ which is identified as $\pi^2 / 6$.
