How to calculate the closed form of the Euler Sums We know that the closed form of the series
$$\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}\left( {\begin{array}{*{20}{c}}
   {2n}  \\
   n  \\
\end{array}} \right)}}}  = \frac{1}{3}\zeta \left( 2 \right).$$
but how to evaluate the following series
$$\sum\limits_{n = 1}^\infty  {\frac{{{H_n}}}{{n\left( {\begin{array}{*{20}{c}}
   {2n}  \\
   n  \\
\end{array}} \right)}}} ,\sum\limits_{n = 1}^\infty  {\frac{{{H_n}}}{{{n^2}\left( {\begin{array}{*{20}{c}}
   {2n}  \\
   n  \\
\end{array}} \right)}}} .$$
 A: Using the main result proved in this question,
$$ \sum_{n\geq 1}\frac{x^n}{n^2 \binom{2n}{n}}=2 \arcsin^2\left(\frac{\sqrt{x}}{2}\right)\tag{1}$$
it follows that:
$$ \sum_{n\geq 1}\frac{H_n}{n^2 \binom{2n}{n}}=\int_{0}^{1}\frac{2 \arcsin^2\left(\frac{\sqrt{x}}{2}\right)-\frac{\pi^2}{18}}{x-1}\,dx \tag{2}$$
and:
$$ \sum_{n\geq 1}\frac{H_n}{n\binom{2n}{n}}=\int_{0}^{1}\frac{\frac{2\sqrt{x}\arcsin\left(\frac{\sqrt{x}}{2}\right)}{\sqrt{4-x}}-\frac{\pi}{3\sqrt{3}}}{x-1}.\tag{3}$$
The RHS of $(2)$ equals:
$$ \int_{0}^{\pi/6}\frac{8\sin(t)\cos(t)}{4\sin^2(t)-1}\left(2t^2-\frac{\pi^2}{18}\right)\,dt=-\int_{0}^{\pi/6}4t\cdot\log(1-4\sin^2(t))\,\tag{4}$$
that is computable in terms of dilogarithms and $\zeta(3)$. In fact, we have:
$$ \sum_{n\geq 1}\frac{H_n}{n^2\binom{2n}{n}}=\frac{\pi}{18\sqrt{3}}\left(\psi'\left(\frac{1}{3}\right)-\psi'\left(\frac{2}{3}\right)\right)-\frac{\zeta(3)}{9},\tag{5}$$
where $\psi'\left(\frac{1}{3}\right)-\psi'\left(\frac{2}{3}\right)=9\cdot L(\chi_3,2)$ with $\chi_3$ being the non-principal character $\!\!\pmod{3}$.
Now, I am working on $(3)$ that should be similar. It boils down to:
$$-16\int_{0}^{\pi/6}\log(1-4\sin^2(t))\left(\frac{t}{\cos^2(t)}+\tan(t)\right)\,dt=\\=\frac{2\pi^2}{3}+2\log^2(3)-2\log^2(4)-4\,\text{Li}_2\left(\frac{3}{4}\right)-16\int_{0}^{\pi/6}\log(1-4\sin^2(t))\left(\frac{t}{\cos^2(t)}\right)\,dt$$
and the last integral depends on
$$ \int_{0}^{\frac{1}{\sqrt{3}}}\arctan(z)\log(1-3z^2)\,dz,\qquad \int_{0}^{\frac{1}{\sqrt{3}}}\arctan(z)\log(1+z^2)\,dz,$$
so $(2)$ is a complicated expression involving $\log(2),\log(3),\pi$ and their squares, the previous $L(\chi_3,2)$ and $\text{Li}_2\left(\frac{3}{4}\right)$.
