Log with Dilog

We start by proving some identities involving Harmonic Numbers :

$\displaystyle H_1(x) = \sum_{n=1}^\infty \dfrac{H_n}{n}x^n = {\rm Li}_2(x)+\dfrac{\ln^2 (1-x)}{2}$

Proof: We start by the generating function of the harmonic numbers,

$\displaystyle \begin{aligned} & \sum_{n=1}^\infty {\rm H}_{n} x^n = -\dfrac{\ln(1-x)}{1-x} \\ & \int_0^x \sum_{n=1}^\infty {\rm H}_{n} x^{n-1} = -\int_0^x \dfrac{\ln(1-x)}{x}\; dx -\int_0^x \dfrac{\ln(1-x)}{1-x}\; dx \\ & \sum_{n=1}^\infty \dfrac{{\rm H}_{n}}{n} x^n = {\rm Li}_2(x)+\dfrac{\ln^2 (1-x)}{2} \end{aligned}$

as claimed.

$\displaystyle H_2(x)=\sum_{n=1}^\infty \dfrac{{\rm H}_n}{n^2}x^n = {\rm Li}_3(x)-{\rm Li}_3(1-x)+\ln(1-x){\rm Li}_2(1-x)+\dfrac{1}{2}\ln x\ln^2 (1-x)+\zeta(3)$

Proof: It is easy to see that $\displaystyle H_2(x)=\int_0^x \dfrac{H_1(x)}{x}\; dx$ , which we use to derive :

$\displaystyle \begin{aligned} &\int_0^x \sum_{n=1}^\infty \dfrac{{\rm H}_n}{n^2}x^{n-1}\; dx = \int_{0}^x \dfrac{{\rm Li}_2(x)}{x}\; dx +\dfrac{1}{2}\int_0^x \dfrac{\ln^2 (1-x)}{2}\; dx \\ & H_2(x) = {\rm Li}_3(x)+\dfrac{I}{2}\end{aligned}$

where $\displaystyle I =\int_0^x \dfrac{\ln^2 (1-x)}{2}\; dx$ for which we evaluate the indefinite as,

$\displaystyle \begin{aligned} \int \dfrac{\ln^2(1-x)}{x}\; dx & = -\int \dfrac{\ln^2 t}{1-t} \quad x\to 1-t \\ &= \ln^2 t\ln(1-t)-2\int \dfrac{\ln t\ln(1-t)}{t}\; dt \\ &=\ln^2 t\ln(1-t)+2\ln t {\rm Li}_2(t)-2{\rm Li}_3(t)+C \end{aligned}$

Now after substituting back $t=1-x$ , the constant $C=2\zeta(3)$ can be determined by setting $x=0$ and hence substituting back we have our final result as desired.

It is now easy to see by partial fractions that ,

$\displaystyle \sum_{n=1}^\infty \dfrac{{\rm H}_n}{n^2 (n+1)}x^n = H_2(x)-H_1(x)+\sum_{n=1}^\infty\dfrac{{\rm H}_n}{n+1}x^n$

where the last sum can be obtained as $\displaystyle \begin{aligned}\sum_{n=1}^\infty\dfrac{{\rm H}_n}{n+1}x^n &= \dfrac{1}{x}\int_0^x \sum_{n=1}^{\infty} {\rm H_n}x^n &= \dfrac{\ln^2 (1-x)}{2x}\end{aligned}$

Also it is easy to verify that,

$\displaystyle \sum_{n=1}^\infty \dfrac{x^n}{(n^2+n)^2}=2\log(1-x)+{\rm Li}_2(x)-3-\dfrac{2\log(1-x)}{x}+\dfrac{{\rm Li}_2(x)}{x}$

Now coming back to the problem, for any real $0<x<1$ ,

$\displaystyle \begin{aligned} \int_0^1 {\rm Li}_2 (xy)\log(1-y)\; dy &= \sum_{n=1}^\infty \dfrac{x^n}{n^2} \int_0^1 y^n \ln(1-y)\; dy \\ &= \sum_{n=1}^\infty \dfrac{x^n}{n^2} \dfrac{-{\rm H}_{n+1}}{n+1} \\ &= \sum_{n=1}^\infty \dfrac{{\rm H}_{n+1}}{n^2(n+1)}x^n \\ &= \sum_{n=1}^\infty \dfrac{{\rm H}_{n}}{n^2(n+1)}x^n+\sum_{n=1}^\infty \dfrac{x^n}{n^2(n+1)^2}x^n \\ & = H_2(x)-H_1(x)+\dfrac{\ln^2 (1-x)}{2x}+2\log(1-x)+{\rm Li}_2(x)-3-\dfrac{2\log(1-x)}{x}+\dfrac{{\rm Li}_2(x)}{x}\end{aligned}$

Now putting $x=\dfrac12$ we have ,

$\displaystyle \int_0^1 {\rm Li}_2\left(\dfrac{x}{2}\right)\ln(1-x)\; dx = 3+\dfrac{\log^2 2}{2}+\dfrac{\zeta(2)\log 2}{2}-\zeta(3)-\zeta(2)-2\log 2\approx -0.422968$ which makes the answer $\boxed{24}$ .