TITLE: Evaluating $\int_0^{1/2} \log(1-x) \log(1-2x) \ dx$.
QUESTION [6 upvotes]: While messing around with Wolfram Alpha, I discovered that 

$$\int_0^{\frac{1}2} \log(1-x) \log(1-2x) \ dx = 1 - \frac{\pi^2}{24} - \frac{\log(2)}2.$$

I've tried all sorts of standard tricks, but I cannot seem to prove it. Can someone prove this beautiful identity?

REPLY [3 votes]: More generally, $\ln(1-x) \ln(1-tx)$ has a rather messy antiderivative involving logarithms and dilog, leading to (for $t > 1$)
$$ \int_0^{1/t} \ln(1-x) \ln(1-t x)\; dx = \frac{2}{t} + \frac{1-t}{t} \left(\ln \left(\frac{t}{t-1}\right) - \text{dilog}\left(\frac{t}{t-1}\right)\right)$$
Yours is the case $t=2$, where $\text{dilog}(2) = -\pi^2/12$.