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

Is there a known closed form for the following

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

I know that we can derive the closed of $\operatorname{Li}_1 \left( \frac{1}{2}\right),\operatorname{Li}_2 \left( \frac{1}{2}\right),\operatorname{Li}_3 \left( \frac{1}{2}\right)$

To put it in an integral representation, the problem asks to solve

$$\int^1_0 \frac{\log(x)^3}{2-x}\, dx$$

#### Solutions Collecting From Web of "Quadirlogarithm value $\operatorname{Li}_4 \left( \frac{1}{2}\right)$"

$\ds{\int_{0}^{1}{\ln^{3}\pars{x} \over 2 – x}\,\dd x:\ {\large ?}}$

\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.

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.

Using Borwein paper (1996), the quadrilogarithm value can be expressed by:

$Li_{4} (\frac{1}{2}) = \frac{\pi^4}{360} – \frac{(\log 2)^4}{24} + \frac{\pi^2 (\log 2)^2}{24} – \frac{1}{2} \zeta(\overline 3 , \overline 1)$

Where we introduced the alternate multiple zeta function as:

$\zeta(\overline a , \overline b) = \sum_{m>n>0} \frac{(-1)^{m+n}}{m^a n^b}$

Higher values can be evaluated by multiple zeta functions.

Related techniques. You can have the following new identity

$$\frac{1}{6}\int^1_0 \frac{\log(x)^3}{x-2} dx= \operatorname{Li}_4 \left( \frac{1}{2}\right) = 2\zeta(4) – \operatorname{Li}_4(2)-i\frac{\pi\ln^3(2)}{6}+\frac{{\pi }^{2} \ln^2\left( 2 \right)}{6}-\frac{\ln^4\left( 2\right)}{24}$$

Note that, the above gives a relation between $\operatorname{Li}_4\left( \frac{1}{2}\right)$ and $\operatorname{Li}_4\left( {2}\right)$ which is nice.