Prove that ${\sum\limits_{n=1}^{\infty}}(-1)^{n-1} \frac{H_n}{n} = \frac{\pi^2}{12} - \frac12\ln^2 2$ We know that $H_n = \sum_{j=1}^{n}{1 \over j}$. Article in The Sum of Certain Series Related To Harmonic Numbers of Omran Kolba, we have proof of this identity which involves some advanced concepts. 
I tried to turn the sum into a definite integral and could not. I appreciate any help.
$$
\sum_{n=1}^{\infty}(-1)^{n-1} \dfrac{H_n}{n} = \sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}\sum_{i=2}^{n+1}\dfrac{1}{i+1} = \sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}\sum_{i=2}^{n+1}\int_{0}^{1}x^{i}dx = ?
$$
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Note that
\begin{align}
H_{n}&=\int_{0}^{1}{1 - t^{n} \over 1 - t}\,\dd t
=-n\int_{0}^{1}\ln\pars{1 - t}t^{n - 1}\,\dd t
\end{align}
where we integrated by parts.

Then,
\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}
\pars{-1}^{n - 1}\,{H_{n} \over n}} =
-\int_{0}^{1}\ln\pars{1 - t}
\sum_{n = 1}^{\infty}\pars{-t}^{n - 1}\,\dd t
\\[5mm] = &\
-\int_{0}^{1}{\ln\pars{1 - t} \over 1 + t}\,\dd t =
-\int_{0}^{1}{\ln\pars{t} \over 2 - t}\,\dd t
=-\int_{0}^{1/2}{\ln\pars{2t} \over 1 - t}\,\dd t
\\[5mm] = &\
-\int_{0}^{1/2}{\ln\pars{1 - t} \over t}\,\dd t =
\int_{0}^{1/2}{{\rm Li}_{1}\pars{t} \over t}\,\dd t
=\int_{0}^{1/2}{\rm Li}_{2}'\pars{t}\,\dd t
\\[5mm] = &\ {\rm Li}_{2}\pars{\half} =
\bbox[10px,border:1px groove navy]
{{\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}}
\approx 0.5822
\end{align}
$\large\mbox{See this link}$.
A: You may consider the standard identity $$\sum_{n=1}^{\infty}H_n x^{n-1} = -\dfrac{\ln(1-x)}{x(1-x)} \quad -1 < x<1,\,x\neq0.$$ Then integrate from $x=-1$ to $x=0$ to obtain easily$$\sum_{n=1}^{\infty}(-1)^{n-1} \dfrac{H_n}{n} \!= -\! \int_{-1}^{0}\dfrac{\ln(1-x)}{x(1-x)} dx = -\!\int_{-1}^{0}\left(\dfrac{\ln(1-x)}{x}\! + \!\dfrac{\ln(1-x)}{1-x}\right) \! dx=\dfrac{\pi^2}{12} - \dfrac{1}{2}\ln^2 2.$$
