Transformation of $\log(X)+\log(1-X)$, where $X$ is uniform.

Question

I am trying to calculate the variance of: $$\log(X)+\log(1-X)$$ where $X \sim \mathrm{Unif}(0,1)$.

So far, I have tried to use a random variable transformation, i.e. define $Y=\log(X)+\log(1-X)$, then: $$X_1=\frac{1+\sqrt{1-4e^Y}}{2}$$ $$X_2=\frac{1-\sqrt{1-4e^Y}}{2}$$ Then I got the density, which is (same for both $X_1$ and $X_2$: $$f_Y(y)=\frac{e^y}{\sqrt{1-4e^y}}$$ And now, I can calculate the variance of that through the integral: $$Var[Y]=\int^{-2\log(2)}_{-\infty}dy\ y^2\frac{e^y}{\sqrt{1-4e^y}}-\bigg(\int^{-2\log(2)}_{-\infty}dy\ y\frac{e^y}{\sqrt{1-4e^y}}\bigg)^2$$

Answer 1

$$ Y=\log(X)+\log(1-X)\ . $$ By definition $$ \langle Y^2\rangle=\int_0^1 dx\, (\log(x)+\log(1-x))^2=8-\pi^2/3 $$ and $$ \langle Y\rangle^2=\left[\int_0^1 dx \, (\log(x)+\log(1-x))\right]^2=4\ . $$ Therefore, $\operatorname{Var}(Y)=4-\pi^2/3$.