Happy New Year 2024 Romania!
Here is a question proposed by Cornel Ioan Valean, $$\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{2^{2n}}\binom{2n}{n}\sum_{k=1}^n (-1)^{k-1}\frac{H_k}{k}-\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{2^{2n}}\binom{2n}{n}\overline{H}_n^{(2)}$$ $$=\frac{\sqrt{2}}{4}\log ^2(2)+\frac{3 \sqrt{2}}{8}\pi ^2+\sqrt{2}\log (2) \log(\sqrt{2}-1) +6 \sqrt{2} \text{Li}_2\left(1-\sqrt{2}\right),$$ 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 $m$, and $\operatorname{Li}_2$ denotes the Dilogarithm.
The author derivation's flow is as follows, in very large steps: it starts with combining the Wallis' integral, $\displaystyle\int_0^{\pi/2} \sin^{2n}(x)\textrm{d}x=\frac{\pi}{2^{2n+1}}\binom{2n}{n}$, and the Cauchy product, $\displaystyle \frac{\log^2(1+x)}{1-x}=\sum_{n=1}^{\infty} x^n \left(2\overline{H}_n^{(2)} -2\sum_{k=1}^n (-1)^{k-1}\frac{H_k}{k}\right)$, then exploiting $\displaystyle \int_0^{\pi/2}\frac{\log^2(\cos(\theta))}{\cos^2(\theta)+y^2 \sin^2(\theta)}\textrm{d}\theta=\frac{1}{4}\int_0^{\infty} \frac{\log^2(1+x^2)}{1+y^2 x^2}\textrm{d}x=\frac{\pi^3}{24}\frac{1}{y}+\frac{\pi}{2}\frac{1}{y}\log^2\left(1+\frac{1}{y}\right)+\frac{\pi}{2}\frac{1}{y}\operatorname{Li}_2\left(\frac{1-y}{1+y}\right),\ y>0$, already found in More (Almost) Impossible Integrals, Sums, and Series (2023), page $73$, the sequel of (Almost) Impossible Integrals, Sums, and Series (2019), and finally, everything is finalized upon employing the identity, $\displaystyle \operatorname{Li}_2\left(-(1-\sqrt{2})^2\right)=\frac{7}{24}\pi^2-\log^2(\sqrt{2}-1)+6 \operatorname{Li}_2(1-\sqrt{2})$, found in the same book, page $299$.
I would love to see more different ways of attacking the problem.
