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

Question

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

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

Answer 1

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.

Answer 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.

Answer 3

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.