How to evaluate $\sum _{n=1}^{\infty } \frac{(-1)^{n+1} H_{2 n}^{(2)}}{n} = 2\zeta(3) - \frac \pi 2 G- \frac {\pi }{48}\ln 2$? What is the best way to calculate the following sum?$$S=\sum _{n=1}^{\infty } \frac{(-1)^{n+1} H_{2 n}^{(2)}}{n} = 2\zeta(3) - \frac \pi 2 G- \frac {\pi^2}{48}\ln 2$$
I tried putting $$f(z) = \frac{\pi \csc(\pi z/2)}{z^3}\psi'(-z)$$
and integrating around an infinitely large contour. The residues in the complex plane sum up to zero.
While this eventually gives the right answer, it involves tedious computations. I would be very happy with a non-contour solution.
 A: Denote the sum as $\mathcal{S}$. Then
\begin{align}
\small\mathcal{S}
&\small\ =-2\mathcal{Re}\sum^\infty_{n=1}\frac{H_n^{(2)}}{n}i^n\\
&\small\ =-2\mathcal{Re}\left[\mathcal{Li}_3(z)+2\mathcal{Li}_3(1-z)-\mathcal{Li}_2(z)\ln(1-z)-2\mathcal{Li}_2(1-z)\ln(1-z)-\ln{z}\ln^2(1-z)-2\zeta(3)\vphantom{\frac{}{}}\right]\Bigg{|}_{z=i}\\
&\small\ =\boxed{\displaystyle2\zeta(3)-\frac{\pi\mathbf{G}}{2}-\frac{\pi^\color{red}{2}}{48}\ln{2}}
\end{align}
The second equality can be derived by computing the integral $\small\displaystyle\int\frac{\mathcal{Li}_2(z)}{z(1-z)}dz$.
To arrive at the third equality from the second, it suffices to know that
\begin{align}
\small\mathcal{Li}_\nu(i)&\small\ =(2^{1-2\nu}-2^{-\nu})\zeta(\nu)+i\beta(\nu)\\
\small\mathcal{Re}\ \mathcal{Li}_3(1\pm i)&\small\ =\frac{35}{64}\zeta(3)+\frac{\pi^2}{32}\ln{2} 
\end{align}
The first identity follows from the series representation of the polylogarithm. For the second, setting $z=\pm i$ in Landen's trilogarithm identity yields
$$\small\mathcal{Re}\ \mathcal{Li}_3(1\pm i)=-\mathcal{Re}\ \mathcal{Li}_3\left(\frac{1\mp i}{2}\right)+\frac{35}{32}\zeta(3)+\frac{\pi^2}{192}\ln{2}+\frac{\ln^3{2}}{48}$$
Setting $z=1\pm i$ in the inversion formula,
$$\small\mathcal{Re}\ \mathcal{Li}_3(1\pm i)=\mathcal{Re}\ \mathcal{Li}_3\left(\frac{1\mp i}{2}\right)+\frac{11\pi^2}{192}\ln{2}-\frac{\ln^3{2}}{48}$$
Averaging these two equalities gives us the closed form for $\small\mathcal{Re}\ \mathcal{Li}_3(1\pm i)$.
A: We use the fact, $$I=\int_0^1\frac{log(1+x^2)\log{x}}{1-x}dx=-\frac{1}{2}{\pi}{G}+2\zeta(3)-\frac{3}{16}{\pi^2}ln2$$
 $$Log(1+x^2)=\sum_{n=1}^\infty\frac{(-1)^{n-1}{x^{2n}}}{n}$$
$$I=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\int_0^1\frac{x^{2n}}{1-x}log{\,x}\,dx$$
 But $$\int_0^1\frac{x^{p-1}}{1-x^{q}}{log^{n}x}\,dx=-\frac{1}{q^{n+1}}\operatorname\psi^{n}(\frac{p}{q})$$
We have,$$\int_0^1\frac{x^{2n}}{1-x}{log\,x}\,dx=-\operatorname\psi^{'}{(2n+1)}=\frac{\pi^2}{6}-H_{2n}^{(2)}$$
And also deduce the result.
