What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$ Is there closed form for 
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$$
where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.
My approach,
In this paper page $95$ Eq $(5)$ we have 
$$\sum_{n=1}^\infty \overline{H}_n\frac{x^n}{n}=\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2(-x)-\ln2\ln(1-x)-\operatorname{Li}_2\left(\frac12\right)$$
Divide both sides by $x$ then integrate we get
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n=\int\frac{\operatorname{Li}_2\left(\frac{1-x}{2}\right)}{x}\ dx-\operatorname{Li}_3(-x)+\ln2\operatorname{Li}_2(x)-\operatorname{Li}_2\left(\frac12\right)\ln x$$
and my question is how to find the remaining integral? Thanks
Maybe you wonder why I have it as an indefinite integral, I meant so as I am planning to plug $x=0$ to find the constant after we find the closed form of the integral if possible.
I tried Mathematica, it gave


Edit
With help of $Mathematica$ I was able to find
\begin{align}
\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n&=-\frac13\ln^3(2)+\frac12\ln^2(2)\ln(1-x)-\frac12\zeta(2)\ln(x)+\frac32\ln^2(2)\ln(x)\\
&\quad-\ln(2)\ln(x)\ln(1-x)-\frac12\ln(2)\ln^2(x)-\frac12\ln^2(2)\ln(1-x)\\
&\quad-\ln^2(2)\left(\frac{x}{1+x}\right)+\ln(2)\ln\left(\frac{x}{1+x}\right)[\ln(1-x)+\ln(x)]\\
&\quad+\ln(x)\ln(1-x)\ln(1+x)+\ln(x)\operatorname{Li}_2\left(\frac{1-x}{2}\right)+\ln\left(\frac{x}{1+x}\right)\operatorname{Li}_2(x)\\
&\quad+\ln(1+x)\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{x}{1+x}\right)\ln\left(\frac{2x}{1+x}\right)-\operatorname{Li}_2\left(\frac{2x}{1+x}\right)\ln\left(\frac{2x}{1+x}\right)\\
&\quad+\operatorname{Li}_2\left(\frac{1+x}{2}\right)\ln\left(\frac{x}{2}\right)-\ln\left(\frac{x}{1+x}\right)\operatorname{Li}_2\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(x)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)\\
&\quad+\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(-x)+\ln(2)\operatorname{Li}_2(x)+\frac{7}{8}\zeta(3)
\end{align}
 A: This is a long comment to https://math.stackexchange.com/a/3523732/198592 which just provides my result for comparison.
Let $\overline{H}_n=\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k}$ be the alternating harmonic sum and define the generating function of order $q=0,1,2,...$ as
$$g_{q}(x) = \sum_{n=1}^\infty\frac{\overline{H}_n}{n^q}x^n\tag{1}$$
For $q=2$ I have obtained 
$$\begin{align}
{g}_2(x)& =-\operatorname{Li}_3\left(\frac{x+1}{2}\right)-\operatorname{Li}_3(-x)-\operatorname{Li}_3(x)-\operatorname{Li}_3\left(\frac{x}{x+1}\right)+\operatorname{Li}_3\left(\frac{2 x}{x+1}\right)\\
& +\log (2) \operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{x+1}{2}\right) \left(\log (x)-\log \left(\frac{2 x}{x+1}\right)\right)\\
& + \operatorname{Li}_2\left(\frac{1}{2}-\frac{x}{2}\right) \log (x)+\operatorname{Li}_2(x) \left(\log \left(\frac{x}{x+1}\right)+\log (x+1)\right)\\
& +\left(\operatorname{Li}_2\left(\frac{x}{x+1}\right)
-\operatorname{Li}_2\left(\frac{2 x}{x+1}\right)\right) \log \left(\frac{2 x}{x+1}\right)+\frac{1}{2} \log ^2(2) \log (x)\\
& +\frac{1}{2} \log ^2\left(\frac{2 x}{x+1}\right) \left(\log \left(\frac{1-x}{2}\right)+\log \left(\frac{1}{x+1}\right)-\log \left(-\frac{x-1}{x+1}\right)\right)\\
& +\log (2) \log (x) \log \left(\frac{2 x}{x+1}\right)-\frac{1}{2} \log (2) \log (x) (\log (x)-2 \log (x+1)+\log (4))\\
& -\frac{1}{12} \pi ^2 \log (x)+\log \left(\frac{1-x}{2}\right) \log \left(\frac{x+1}{2}\right) \log (x)+\frac{7 \zeta (3)}{8}+\frac{\log ^3(2)}{6}
\end {align}\tag{2}$$
Notice the appearance of $\zeta(3)$ which I don't see in your expression.
Here is the graph of the g.f.

I find the following boundary values
$$g_2(+1) = \frac{1}{4} \pi ^2 \log (2)-\frac{\zeta (3)}{4}\simeq 1.40976$$
$$g_2(-1) = -\frac{1}{4} \pi ^2 \log (2)+\frac{5 \zeta (3)}{8}\simeq -0.958987$$
The values at $\pm \frac{1}{2}$ are somewhat too long to be provided here at the moment.
To facilitate comparison here is the Mathematica statement
g2[x_]:=Log[2]^3/6 - 1/12 \[Pi]^2 Log[x] + 1/2 Log[2]^2 Log[x] + 
 Log[2] Log[x] Log[(2 x)/(1 + x)] + 
 1/2 (Log[(1 - x)/2] + Log[1/(1 + x)] - 
    Log[-((-1 + x)/(1 + x))]) Log[(2 x)/(1 + x)]^2 + 
 Log[(1 - x)/2] Log[x] Log[(1 + x)/2] - 
 1/2 Log[2] Log[x] (Log[4] + Log[x] - 2 Log[1 + x]) + 
 Log[x] PolyLog[2, 1/2 - x/2] + 
 Log[2] PolyLog[2, x] + (Log[x/(1 + x)] + Log[1 + x]) PolyLog[2, x] + 
 Log[(2 x)/(
   1 + x)] (PolyLog[2, x/(1 + x)] - 
    PolyLog[2, (2 x)/(1 + x)]) + (Log[x] - 
    Log[(2 x)/(1 + x)]) PolyLog[2, (1 + x)/2] - PolyLog[3, -x] - 
 PolyLog[3, x] - PolyLog[3, x/(1 + x)] + PolyLog[3, (2 x)/(1 + x)] - 
 PolyLog[3, (1 + x)/2] + (7 Zeta[3])/8

A: From this paper page $101$ we have 
$$\sum_{n=1}^\infty\overline{H}_n\frac{x^{n+1}}{(n+1)^2}=\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(x)$$
$$+\ln(1+x)\left[\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\frac12\ln 2\ln(1+x)\right]+\operatorname{Li}_3\left(\frac{1}{2}\right)$$
but
$$\sum_{n=1}^\infty\overline{H}_n\frac{x^{n+1}}{(n+1)^2}=\sum_{n=0}^\infty\overline{H}_n\frac{x^{n+1}}{(n+1)^2}=\sum_{n=1}^\infty\overline{H}_{n-1}\frac{x^n}{n^2},\quad \overline{H}_{n-1}=\overline{H}_n+\frac{(-1)^n}{n}$$
$$=\sum_{n=1}^\infty\overline{H}_{n}\frac{x^n}{n^2}+\operatorname{Li}_3(-x)$$
Thus

$$\sum_{n=1}^\infty\overline{H}_{n}\frac{x^n}{n^2}=\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(-x)-\operatorname{Li}_3(x)$$
$$+\ln(1+x)\left[\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\frac12\ln 2\ln(1+x)\right]+\operatorname{Li}_3\left(\frac{1}{2}\right)$$

