Approaching $\sum_{n=1}^\infty\frac{\overline{H}_n-H_{n/2}}{n^3}$ elegantly How to elegantly prove that $$\sum_{n=1}^\infty\frac{\overline{H}_n-H_{n/2}}{n^3}=2\text{Li}_4\left(\frac12\right)-\frac{49}{16}\zeta(4)+\frac72\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$
where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number, $H_{n/2}=\int_0^1\frac{1-x^{n/2}}{1-x}\ dx$ is the harmonic number, $\text{Li}_r$ is the polylogarithm function and $\zeta$ is the Riemann zeta function. 
What I mean by elegant solutions is solutions involving cancellation of challenging integrals/ sums , symmetry , manipulations and new ideas that save us tedious calculations. However, all solutions are appreciated. 
Thank you
 A: Here's a preliminary answer which boils the problem to find the sum
$$s = \sum_{n=1}^{\infty} \frac{1}{n^3}(\overline{H_{n}} - H_{n/2})\tag{1}$$
down to the tough (?) sum
$$s_1 = -\sum_{n=1}^{\infty} \frac{H_{n-\frac{1}{2}}}{(2n-1)^3}\tag{2}$$
Let us, just for information, look at the integral representation of the sum
$$s_i = \int_{0}^{1} \sum _{n=1}^{\infty } \frac{\frac{1-(-x)^n}{x+1}-\frac{1-x^{n/2}}{1-x}}{n^3}\,dx 
\\= \int_{0}^{1}\frac{-x \operatorname{Li}_3\left(\sqrt{x}\right)-\operatorname{Li}_3\left(\sqrt{x}\right)-x \text{Li}_3(-x)+\text{Li}_3(-x)+2 x \zeta (3)}{(x-1) (x+1)}\,dx
\\\simeq 0.260631\tag{3}$$ 
The main idea is to split the sum $(1)$ into even and odd parts and then use the well-known relations
$$\overline{H_{2k}} = H_{2k} - H_{k}, \overline{H_{2k+1}}=H_{2k+1} - H_{k}\tag{4a}$$
and
$$\overline{H_{2k-1}}=H_{2k-1}-H_{k}+\frac{1}{n}\tag{4b}$$
This gives
$$s = s_1 + s_2 + s_3+ s_4 + s_5 + s_6$$
Where
$\begin{align}
&s_2 = \sum_{n=1}^{\infty}\frac{H_{2n}}{(2n)^3}\\
&s_3 = \sum_{n=1}^{\infty}\frac{H_{2n-1}}{(2n-1)^3}\\
&s_4 = -2\sum_{n=1}^{\infty}\frac{H_{n}}{(2n)^3}\\
&s_5 =- \sum_{n=1}^{\infty}\frac{H_{n}}{(2n-1)^3}\\
&s_6 = \sum_{n=1}^{\infty}\frac{1}{n(2n-1)^3}
\end{align}$
Notice that 
$$s_2+s_3= \sum_{n=1}^{\infty}\frac{H_{n}}{n^3}$$
and 
$$s_A = s_2+s_3+s_4+s_5 = \sum _{n=1}^{\infty } \left(\frac{1}{n^3}-\frac{1}{(2 n)^3}-\frac{1}{(2 n-1)^3}\right) H_n\tag{5}$$
so that
$$s = s_1 + s_A + s_6\tag{6}$$
Mathematica gives
$$s_A =-\frac{7 \pi ^4 \zeta (3)}{720}+40 \zeta (3)-\frac{7 \pi ^2 \zeta (5)}{48}+\frac{7 \zeta (7)}{2}+14 \zeta (3) \log (2)
\\
+8 \pi ^2-\frac{\pi ^4}{9}+48 \log ^2(2)-6 \pi ^2 \log (2)-160 \log (2)\tag{7}$$
and
$$s_6 = \frac{7 \zeta (3)}{4}-\frac{\pi ^2}{4}+\log (4)\tag{8}$$
The result $(6)$ is numerically correct.
I am sure that someone around here has already calculated the sum $s_1$ which would then complete the result.
