Integral involving Dilogarithm $\int_{1/2}^{1} \frac{\mathrm{Li}_{2}\left(x\right)\ln\left(2x - 1\right)}{x}\,\mathrm{d}x$ I need your help in evaluating the following integral in closed form.  $$\displaystyle\int\limits_{0.5}^{1}
\frac{\mathrm{Li}_{2}\left(x\right)\ln\left(2x - 1\right)}{x}\,\mathrm{d}x$$
Since the function is singular at $x = 0.5$, we are looking for Principal Value. The integral is finite and was evaluated numerically.
I expect the closed form result to contain $\,\mathrm{Li}_{3}$ and $\,\mathrm{Li}_{2}$.
Thanks
 A: Note that
\begin{align}
I_1=&\int_0^1 \frac{\ln^2(1-x)\ln(1+x)}{1+x}dx =\int_0^1 \frac{\ln^2x \ln(2-x)}{2-x}dx\\
=& \>\ln2 \int_0^1 \frac{\ln^2x }{2-x}dx +\int_0^1 \frac{\ln^2x }{2-x} \left( -\int_0^1 \frac x{2-xy}dy\right) dx\\
=&\>2\ln2 Li_3(\frac12)+\int_0^1 \int_0^1 \frac{\ln^2x }{1-y}\left(\frac1{2-xy}-\frac1{2-x}\right)dy\>dx\\
= &\>2\ln2 Li_3(\frac12)+ 2\int_0^1\frac{Li_3(\frac y2)}ydy +2 \int_0^1 \frac{Li_3(\frac y2)-Li_3(\frac12)}{1-y}\>\overset{ibp}{dy}\\
= & \> 2\ln2 Li_3(\frac12)+2Li_4(\frac12)+ 
2\int_0^{\frac12}\underset{=J}{\frac{\ln(1-2y)Li_2(y)}y}dy\tag1
\end{align}
A similar procedure establishes
\begin{align}
I_2=\int_0^1 \frac{\ln^2(1-x)\ln x}{1+x}dx
= \>2Li_4(1)+ 2J
+ 2\int^1_{\frac12}\frac{\ln(2y-1)Li_2(y)}ydy\tag2
\end{align}
Combine (1) and (2) to express the original integral as
\begin{align}
\int^1_{\frac12}\frac{\ln(2y-1)Li_2(y)}ydy
=\frac12 (I_2-I_1)+\ln2 Li_3(\frac12)+Li_4(\frac12)-Li_4(1)\tag3
\end{align}
where the integrals $I_1$ and $I_2$ are known, given by
\begin{align}
&I_1 =-\frac{\pi^4}{360} +2\ln2 \zeta(3)-\frac{\pi^2}6\ln^22+\frac14 \ln^42\\
&I_2 = -6Li_4(\frac12)+\frac{11\pi^4}{360}-\frac14\ln^42
\end{align}
Substitute into (3) to obtain the close-form
\begin{align}\int^1_{\frac12}\frac{\ln(2y-1)Li_2(y)}ydy
=& -2{Li}_4\left(\frac{1}{2}\right)-\frac{1}{8}\ln2\zeta(3) + \frac{\pi^4}{180}- \frac{1}{12}\ln^42
\end{align}
A: $\displaystyle \int\limits_{0.5}^1 \frac{Li_2(x)\ln(2x-1)}{x}dx=$
$\displaystyle =\sum\limits_{k=1}^\infty \frac{1}{k^2 2^k}\sum\limits_{v=0}^{k-1} {\binom {k-1} v} \lim\limits_{h\to 0}\frac{1}{h}\left(\frac{(2x-1)^{v+h+1}}{v+h+1}-\frac{(2x-1)^{v+1}}{v+1}\right)|_{0.5}^1$
$\displaystyle =-\sum\limits_{k=1}^\infty \frac{1}{k^3 2^k}\sum\limits_{v=1}^k {\binom k v} \frac{1}{v} = -\int\limits_0^1 \frac{Li_3(\frac{x+1}{2})-Li_3(\frac{1}{2})}{x}dx $
First note: 
Be $\,\displaystyle H_k(x):=x\int\limits_0^1 \frac{(xt)^k-1}{xt-1}dt=\sum\limits_{v=1}^k \frac{x^v}{v}$ . $\,$ It's $\enspace\displaystyle \sum\limits_{v=1}^k {\binom k v} \frac{1}{v}=H_k(2)-H_k(1)$ .
Second note: 
We can define e.g. $\,\displaystyle Fi_n(x):=\int\limits_0^{1-x}\frac{Li_n(t+x)-Li_n(x)}{t}dt\,$ for $\,|x|\leq 1\,$ . 
Then it's $\,\displaystyle \int\limits_{0.5}^1 \frac{Li_2(x)\ln(2x-1)}{x}dx=-Fi_3(\frac{1}{2})\,$ .
