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This was inspired by Oksana's post. Let, $$a = \ln 2 \quad\quad\\ b = \ln 3\quad\quad\\ c = \ln 5\quad\quad$$

then the following,

\begin{align} A &= \text{Li}_3\left(\frac12\right)\\ B &= \text{Li}_3\left(\frac13\right)+\text{Li}_3\left(\frac23\right)\\ C &= \text{Li}_3\left(\frac14\right)+\text{Li}_3\left(\frac34\right)\\ D &= \text{Li}_3\left(\frac15\right)+\text{Li}_3\left(\frac25\right)+\text{Li}_3\left(\frac35\right)+\text{Li}_3\left(\frac45\right)\\ E &= \text{Li}_3\left(\frac16\right)+\text{Li}_3\left(\frac56\right)\\ \end{align}

have known expression in terms of $a,b,c,$ plus other polylogs. Namely,

\begin{align} 12A &= 2a^3-6a\zeta(2)+\tfrac{21}2\zeta(3)\\ 12B &=-3\,\text{Li}_3\left(\tfrac14\right)+4a^3+4b^3-6ab^2+6(a-2b)\zeta(2)+\tfrac{45}2\zeta(3)\\ 6C &= -12\,\text{Li}_3\left(\tfrac13\right)+16a^3+2b^3-12a^2b-12a\zeta(2)+19\zeta(3)\\ 6D &=\, D_1+D_2- 6 (a - 4 b + 2 c)\zeta(2)-\tfrac{19}2\zeta(3)\\ 12E &=\, E_1+E_2- 3 (a - 8 b + 4 c) \zeta(2)-\tfrac{63}4\zeta(3)\\ \end{align}

where,

\begin{align} D_1&=18\,\text{Li}_3\left(\tfrac13\right)+21\,\text{Li}_3\left(\tfrac23\right)-3\,\text{Li}_3\left(\tfrac16\right)-3\,\text{Li}_3\left(\color{blue}{\tfrac38}\right)\\ D_2&=14 a^3 - 6 b^3 + 4 c^3 - 12 a^2 b + 12 a b^2 - 9 a c^2 - 3 b c^2 + 6 a b c\\ \end{align}

and,

\begin{align} E_1&= 36\,\text{Li}_3\left(\tfrac13\right)+36\,\text{Li}_3\left(\tfrac23\right)-15\,\text{Li}_3\left(\tfrac15\right)-18\,\text{Li}_3\left(\tfrac25\right)-9\,\text{Li}_3\left(\tfrac45\right)+6\,\text{Li}_3\left(\color{blue}{\tfrac58}\right)\\ E_2&=-23 a^3 - 8 b^3 + 7 c^3 + 12 a^2 b + 21 a^2c + 30 a b^2 - 6 b^2 c - 18 a c^2 - 12 a b c \\ \end{align}

Eliminating $\zeta(3)$ between $D$ and $E$, then moving all the polylogs to the LHS and the logarithms to the RHS gives the simplification sought by Oksana in her post.

Note: The relation for $D$ and $E$ were found by Kirill (though I had to fix some of the typos).

Question: Can we find a closed-form for, $$F = \text{Li}_3\left(\frac18\right)+\text{Li}_3\left(\frac38\right)+\text{Li}_3\left(\frac58\right)+\text{Li}_3\left(\frac78\right)$$ similar to the others?

Tito Piezas III
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  • If you arrange these in a matrix you get $$ \begin{matrix} A \ B \ C \ D \ E \end{matrix} \sim \begin{matrix} 1 \ 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 1 & 1 \ 1 & 0 & 0 & 0 & 1 \ \end{matrix} $$ Searching "1,1,1,1,0,1,1,1,1,1,1,0,0,0,1" in the OEIS gives A047999 Sierpiński's [Sierpinski's] triangle (or gasket): Therefore you might also want to investigate $$ F = \text{Li}_3\left(\frac17\right)+\text{Li}_3\left(\frac27\right)+\text{Li}_3\left(\frac57\right)+\text{Li}_3\left(\frac67\right) $$ – Benedict W. J. Irwin Jul 15 '19 at 12:40
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    @BenedictW.J.Irwin: I'm afraid it may not be the case. Technically, all the entries should be $1$, but since $\text{Li}_3(\frac24)$, $\text{Li}_3(\frac26)+\text{Li}_3(\frac46)$, etc have closed-forms, they are redundant. But I was thinking of including them earlier. – Tito Piezas III Jul 15 '19 at 12:56

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