The following problem is proposed by Cornel Valean:
$$\sum_{n=1}^{\infty} \frac{H_n}{2n+1}\left(\eta(2)- \overline{H}_n^{(2)}\right)$$ $$=2 G^2-2\ln(2) \pi G+\ln^2(2)\frac{\pi ^2}{6} +\frac{53}{1440}\pi ^4-\frac{1}{6} \ln ^4(2)-4 \operatorname{Li}_4\left(\frac{1}{2}\right),$$ where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}, \ m\ge1$, is the $n$th generalized harmonic number of order $m$, $\overline{H}_n^{(m)}=1-\frac{1}{2^m}+\cdots+(-1)^{n-1}\frac{1}{n^m}, \ m\ge1$, represents the $n$th generalized skew-harmonic number of order $m$, $G$ denotes the Catalan's constant, $\zeta$ designates the Riemann zeta function, $\eta$ designates the Dirichlet eta function, and $\operatorname{Li}_n$ signifies the Polylogarithm.
Here is my approach: Using $$\eta(2)-\overline{H}_n^{(2)}=-\int_0^1\frac{(-x)^n\ln(x)}{1+x}dx$$
and $$\sum_{n=1}^\infty\frac{H_n}{2n+1}x^n=-\int_0^1\frac{\ln(1-y^2x)}{1-y^2x}dy,$$
the sum converts to
$$\int_0^1\int_0^1\frac{\ln(x)\ln(1+y^2x)}{(1+x)(1+y^2x)}dydx.$$ Any idea how to crack this integral or a different approach than mine?
Thank you in advance.
A solution may be found in this short presentation, Exploiting The Master Theorem of Series to Evaluate a Series with the Tail of the Dirichlet Eta Function Artistically, by C.I. Valean.
Another solution, as emphasized at the given link, may be obtained by exploiting \begin{equation*} \int_0^1 x^{2n-1}\operatorname{arctanh}^2(x)\textrm{d}x=\log(2)\frac{H_{2n}}{n}-\frac{1}{2}\log(2)\frac{H_n}{n}+\frac{1}{2}\frac{1}{n}\sum_{k=1}^{n-1}\frac{H_k}{2k+1}, \end{equation*} which is given and derived in More (Almost) Impossible Integrals, Sums, and Series, Springer, Cham, First Edition (2023), more precisely in Chapter 1, Section 1.22, p.29.
End of story (apart from this, I would love to see more real solutions using very different ideas)