Showing $\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$

Question

Is there way to show $$\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$$ without using the Riemann zeta function?

Answer 1

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.

$$\frac{\ln(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3}...$$

$$\int\frac{\ln(1+x)}{x}dx = x - \frac{x^2}{4} + \frac{x^3}{9}...+C$$

$$\int\limits^{0}_{1}\frac{\ln(1+x)}{x}dx = -1 + \frac{1}{4} - \frac{1}{9}...$$

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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 $$\int\limits^{0}_{1}\frac{\ln(1-x)}{x}dx = 1 + \frac{1}{4} + \frac{1}{9}...$$

$$=\frac{\pi ^2}{6}$$

Answer 2

(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.)

Answer 3

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: $$\lim \limits_{t \nearrow 1}\sum \limits_{n = 1}^\infty \frac{t^n}{n^2} = \sum \limits_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$

Answer 4

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 $\pi^2 / 6$.