# Thread: Integration of the product of two logarithms

1. ## Integration of the product of two logarithms

Find $\displaystyle \int\limits_{0}^{1}ln(x)ln(1-x)\, dx$

According to Wolfram alpha, the answer is $\displaystyle 2 - \frac{\pi^2}{6}$

Is there a way of arriving at this result without using dilogarithms? I've tried using integration by parts and logarithmic series, but to no avail.

2. As in...

http://www.mathhelpforum.com/math-he...te-159304.html

$\displaystyle \displaystyle \int_{0}^{1} x^{m}\ \ln^{n} x\ dx = (-1)^{n}\ \frac{n!}{(m+1)^{n+1}}$ (1)

... and because for $\displaystyle |x|<1$...

$\displaystyle \displaystyle \ln (1-x)= - \sum_{m=1}^{\infty} \frac{x^{m}}{m}$ (2)

... combining (1) and (2) You have...

$\displaystyle \displaystyle \int_{0}^{1} \ln x\ \ln (1-x)\ dx = \sum_{m=1}^{\infty} -\frac{1}{m}\ \int_{0}^{1} x^{m}\ \ln x\ dx=$

$\displaystyle \displaystyle = \sum_{m=1}^{\infty} \frac{1}{m\ (m+1)^{2}} = \sum_{m=1}^{\infty} \frac{1}{m\ (m+1)} - \sum_{m=1}^{\infty} \frac{1}{(m+1)^{2}} = 1+1-\frac{\pi^{2}}{6}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$