Integrating $\ln(x)\times\ln(1-x)$

Question

Is there a way I can derive the value of the integral

$ \int_0^1 \ln(x)\ln(1-x)dx$

using the fact that

$\displaystyle\sum_{i=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$

? (the actual value of the integral is $2-\frac{\pi^2}{6}$)

Thanks in advance

Answer 1

HINT

Replace $\log(1-y)$ by the corresponding infinite Taylor series. Reverse summation and integration. Each integrand will look as $y^i \log(y)$ you can very simply integrate by parts.

I am sure you can take from here. If not, just post.

Answer 2

Hint:

In addition to Leibovici's comment, recall that:

• $\ln(1-x) = -\sum^\infty_{n=1} \frac{x^n}{n}, \ |x|<1$, and:

• $\lim_{x \to 0} x \ln{x} = 0. $

Cheers!

Answer 3

You could be interested by the fact that the result of $ \int_0^a \ln(x)\ln(a-x)dx$ is given by

$$-\frac{1}{6} a \left(-6 (\log (a)-2) \log (a)+\pi ^2-12\right)$$

which is not much more difficult to establish than in the case where $a=1$.

Answer 4

Using $\ln(1-x)=-=-\sum_{k=1}^{\infty} \frac{x^k}{k}$ for $|x|<1$, we have $$ \begin{aligned} \int_0^1 \ln x \ln (1-x) d x = & -\sum_{k=1}^{\infty} \frac{1}{k} \int_0^1 x^k \ln x d x \\ = & \sum_{k=1}^{\infty} \frac{1}{k(k+1)^2} \\ = & \sum_{k=1}^{\infty} \frac{(k+1)-k}{k(k+1)^2} \\ = & \sum_{k=1}^{\infty}\left[\frac{1}{k(k+1)}-\frac{1}{(k+1)^2}\right] \\ = & \sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)-\sum_{k=1}^{\infty} \frac{1}{(k+1)^2} \\ = & 2-\frac{\pi^2}{6} \end{aligned} $$

Answer 5

Hint: In addition to previous comments: $$ \ln (1-x)=-\sum_{i=1}\frac{x^i}{i} $$ and $$ \int x^i\ln(x)dx=\frac {-1}{(i+1)^2} $$ and $$ \sum_{i=1}\frac{1}{i(i+1)^2}=\sum_{i=1}\frac{1}{i(i+1)}-\frac{1}{(i+1)^2}\\ $$