Evaluate $\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$ By accident, I find this summation when I pursue the particular value of $-\operatorname{Li_2}(\tfrac1{2})$, which equals to integral $\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x}$.
Notice this observation
$$\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x} = \int_{0}^{1} {\frac{\ln(1-x^{2})}{1+x} \mathrm{d}x} - \frac{(\ln2)^{2}}{2}$$
And using the Taylor series of $\ln(1-x^{2})$, I found this summation $\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$, where $H_{n}$ is the harmonic-numbers.
If the value of $\operatorname{Li_2}(\tfrac1{2})=\tfrac1{2}(\zeta(2)-(\ln2)^{2})$ is given, the result can be easily deduced, which is

$$\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)} = -\frac{\zeta(2)}{2}+(\ln2)^{2}$$

For the original goal is to calculate $\operatorname{Li_2}(\tfrac1{2})$, I expect some other approaches to the summation without using the value of $\operatorname{Li_2}(\tfrac1{2})$. I already knew some famous problem like Euler's Sum, which holds very similar form to this summation, but still in trouble finding the appropriate path.
 A: Well, ignoring the dilogarithm reflection formula, we still have
$$ \sum_{n=1}^{N}\frac{\log(2)}{n}=\log(2)H_N,\qquad \sum_{n=1}^{N}\frac{H_n}{n}\stackrel{\text{sym}}{=}\frac{H_n^2+H_n^{(2)}}{2} $$
and 
$$ \sum_{n=1}^{N}\frac{H_{2n}}{n}\stackrel{\text{SBP}}{=}H_N H_{2N}-\sum_{n=1}^{N-1}H_n\left(\frac{1}{2n+2}+\frac{1}{2n+1}\right) $$
can be reduced (up to known terms) to
$$ \sum_{n=1}^{N}\left[\frac{1}{n}\sum_{k=1}^{n}\frac{1}{2k-1}+\frac{1}{2n-1}\sum_{k=1}^{n}\frac{1}{k}\right]=\sum_{n=1}^{N}\frac{1}{n}\sum_{n=1}^{N}\frac{1}{2n-1}+\sum_{n=1}^{N}\frac{1}{n(2n-1)}. $$
Exploiting $H_n^{(2)}=\zeta(2)+o(1)$ and $H_n = \log(n)+\gamma+o(1)$ for $n\to +\infty$ we end up with the explicit value of $\text{Li}_2\left(\frac{1}{2}\right)$. Nice exercise!
A: With this answer I show an indirect method to the wished result of the integral $~\int\limits_0^1\frac{\ln(1-x)}{1+x}dx~$, 
and indirect means here: It’s used  $~\text{Li}_2\left(\frac{1}{2}\right)~$ without knowing it’s value, only as a catalyst. 
First a general formula. It’s not difficult to find out, that formally holds:
$$-\frac{d}{dx}(x+z)^y \sum\limits_{k=1}^\infty\frac{\left(\frac{x+z}{a+z}\right)^k}{k+y} = \frac{(x+z)^y}{x-a}$$ 
With the integration to $x$ and using Taylor series for $~y~$ around $~0~$ we get:

$$\frac{1}{n!}\int\frac{(\ln(x+z))^n}{x-a}dx = \sum\limits_{k=0}^n (-1)^{n-k+1}\frac{(\ln(x+z))^k}{k!}\text{Li}_{n-k+1}\left(\frac{x+z}{a+z}\right) + C$$

First we use $~(n;z;a):=(1;1;1)~$ :
$\displaystyle\int\limits_0^1\frac{\ln(1-x)}{1+x}dx = -\int\limits_{-1}^0\frac{\ln(1+x)}{1-x}dx =$
$\displaystyle = -\text{Li}_2\left(\frac{x+1}{2}\right)|_{-1}^0 + \ln(x+1)\text{Li}_1\left(\frac{x+1}{2}\right)|_{-1}^0 = -\text{Li}_2\left(\frac{1}{2}\right) $
Our next step is to transform the integral by partial integration :
$\displaystyle\int\limits_0^1\frac{\ln(1-x)}{1+x}dx = (\ln(1-x))(\ln(1+x) - \ln 2)|_0^1 + \int\limits_0^1\frac{\ln(1+x) - \ln 2}{1-x}dx = $
$\displaystyle = 0 + 2\int\limits_0^{1/2}\frac{\ln(1+2x) - \ln 2}{1-2x}dx = -\int\limits_0^{1/2}\frac{\ln(x+1/2)}{x-1/2}dx$
Now we use $~(n;z;a):=(1;\frac{1}{2};\frac{1}{2})~$ and $~\text{Li}_1\left(\frac{1}{2}\right)=\ln 2~$ :
$\displaystyle -\int\limits_0^{1/2}\frac{\ln(x+1/2)}{x-1/2}dx = -\text{Li}_2\left(x+\frac{1}{2}\right)|_0^{1/2} + \ln\left(x+\frac{1}{2}\right) \text{Li}_1\left(x+\frac{1}{2}\right)|_0^{1/2}$
$\displaystyle = -\frac{\pi^2}{6} + \text{Li}_2\left(\frac{1}{2}\right) + (\ln 2)^2 \enspace\enspace$ which is, as found before, the same as $~\displaystyle -\text{Li}_2\left(\frac{1}{2}\right)~$ . 
Comparing both results we get the wished formula.
Note: $~$ Here we see very well that the partial integration leads to the second representation of the result and both representations have as a common base the (yellow marked) general formula.

Hint:
$$\frac{1}{n!}\int\frac{(\ln(x+z))^n}{(x-a)^{m+1}}dx =\\ =\frac{(-1)^m}{m!(a+z)^m}\sum\limits_{k=0}^n (-1)^{n-k+1}\frac{(\ln(x+z))^k}{k!}\sum\limits_{j=0}^m\begin{bmatrix}m~\\j~\end{bmatrix}\text{Li}_{n-k+1-j}\left(\frac{x+z}{a+z}\right) + C$$
for $~m\in\mathbb{N}_0~$ and with 
the Stirling numbers of the first kind $\begin{bmatrix}n~\\k~\end{bmatrix}~$ defined by $~\displaystyle\sum\limits_{k=0}^n\begin{bmatrix}n~\\k~\end{bmatrix}x^k:=\prod\limits_{k=0}^{n-1}(x+k)~$ 
A simple example: $\enspace\displaystyle\int\limits_0^1\frac{\ln(1-x)}{(1+x)^3}dx = \displaystyle -\int\limits_{-1}^0\frac{\ln(x+1)}{(x-1)^3}dx =-\frac{1+\ln 2}{8}$
A: Here is an approach that avoids knowing the value of $\operatorname{Li}_2 (\frac{1}{2})$. 
Let
$$S = \sum_{n = 1}^\infty \frac{1}{n} \left (H_{2n} - H_n - \ln 2 \right ).$$
Observing that
$$\int_0^1 \frac{x^{2n}}{1 + x} \, dx = H_n - H_{2n} + \ln 2,$$
your sum can be rewritten as
\begin{align}
S &= -\int_0^1 \frac{1}{1 + x} \sum_{n = 1}^\infty \frac{x^{2n}}{n} \, dx\\
&= \int_0^1 \frac{\ln (1 - x^2)}{1 + x} \, dx\\
&= \int_0^1 \frac{\ln (1 + x)}{1 + x} \, dx + \int_0^1 \frac{\ln (1 - x)}{1 + x} \, dx\\
&= I + J.
\end{align}
For the first of the integrals $I$, one has
$$I = \frac{1}{2} \ln^2 2.$$
Now consider $J - I$. Then
$$J - I = \int_0^1 \ln \left (\frac{1 - x}{1 + x} \right ) \frac{dx}{1 + x}.$$
Employing a self-similar substitution of $t = (1-x)/(1+x)$ leads to
\begin{align}
J - I &= \int_0^1 \frac{\ln t}{1 + t} \, dt\\
&= \sum_{n = 0}^\infty (-1)^n \int_0^1 t^n \ln t \, dt\\
&= \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\int_0^1 t^{n + s} \, dt \right ]_{s = 0}\\
&= \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\frac{1}{n + s + 1} \right ]_{s = 0}\\
&= -\underbrace{\sum_{n = 0}^\infty \frac{(-1)^n}{(n + 1)^2}}_{n \, \mapsto \, n - 1}\\
&= -\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n^2}\\
&= -\sum_{n = 1}^\infty \frac{1}{n^2} + \frac{1}{2} \sum_{n = 1}^\infty \frac{1}{n^2}\\
&= -\frac{1}{2} \zeta (2).
\end{align} 
Thus
$$J = I - \frac{1}{2} \zeta (2) = \frac{1}{2} \ln^2 2 - \frac{1}{2} \zeta (2).$$
Since $S = I + J$, we immediately see that
$$\sum_{n = 1}^\infty \frac{1}{n} \left (H_{2n} - H_n - \ln 2 \right ) = \ln^2 2 - \frac{1}{2} \zeta (2),$$
as desired. 
