Quadirlogarithm value $\operatorname{Li}_4 \left( \frac{1}{2}\right)$

Solution 1:

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{\ln^{3}\pars{x} \over 2 - x}\,\dd x:\ {\large ?}}$

\begin{align}&\overbrace{\color{#c00000}{\int_{0}^{1} {\ln^{3}\pars{x} \over 2 - x}\,\dd x}} ^{\ds{\mbox{Set}\ x \equiv \expo{-t}\ \imp\ t = -\ln\pars{x}}}\ =\ \half\int_{\infty}^{0}{-t^{3} \over 1 - \expo{-t}/2}\,\pars{-\expo{-t}\,\dd t} \\[3mm]&=-\,\half\int_{0}^{\infty} t^{3}\expo{-t}\sum_{n = 0}^{\infty}\pars{\half}^{n}\expo{-nt}\,\dd t =-\,\half\sum_{n = 0}^{\infty}\pars{\half}^{n} \int_{0}^{\infty}t^{3}\expo{-\pars{n + 1}t}\,\dd t \\[3mm]&=-\,\half\sum_{n = 0}^{\infty}{\pars{1/2}^{n} \over \pars{n + 1}^{4}}\ \overbrace{\int_{0}^{\infty}t^{3}\expo{-t}\,\dd t}^{\ds{=\ 3!\ = 6}}\ =\ -6\sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n^{4}} \end{align}

$$ \color{#66f}{\large% \int_{0}^{1}{\ln^{3}\pars{x} \over 2 - x}\,\dd x =-6\,{\rm Li}_{4}\pars{1 \over 2}} \approx -3.1049 $$

$\ds{{\rm Li_{s}}\pars{z}}$ is a PolyLogarithm Function.

Solution 2:

Wolfram page on polylogarithms says that no closed formula is known for $\mathrm{Li}_n\left(\frac12\right)$ for $n\geq4$, see the remark after their formula (17).

Hence, as I said answering your other question, I would be rather surprised if somebody comes with an answer.