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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

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

Zaid Alyafeai
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4 Answers4

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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.

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Start wearing purple
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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.

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

\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.

Felix Marin
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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.

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Mhenni Benghorbal
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    Where did the $14.56\ldots$ come from? – Ron Gordon Aug 15 '13 at 05:35
  • @RonGordon: I posted the closed form. – Mhenni Benghorbal Aug 15 '13 at 18:58
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    Mhenni, how is this a closed form? The OP asked for a "closed form" of $\text{Li}_4(1/2)$, and you return $\text{Li}_4(2)$. No matter what definition we may attribute to the term "closed form," this cannot possibly be it, the relation between the quantities notwithstanding. Also, what's up with the imaginary piece? – Ron Gordon Aug 15 '13 at 19:07
  • @MhenniBenghorbal I would add to Ron Gordon's comment that the expression $\mathrm{Li}_4(2)$ is meaningless, as the standardly defined polylogarithm has a branch cut running from $1$ to $\infty$. – Start wearing purple Aug 15 '13 at 19:08
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    @RonGordon: Offcourse, it is a closed form and relates two polylogarithm functions. – Mhenni Benghorbal Aug 15 '13 at 19:09
  • @O.L.: Now, you are talking about some thing else which is which branch you need to consider. So, this is an evaluation issue. – Mhenni Benghorbal Aug 15 '13 at 19:11
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    @MhenniBenghorbal Of course. I only want to say that for the expression to make sense, one has to specify the branch. Even so, your statement remains a polylogarithm identity, not a closed form evaluation. – Start wearing purple Aug 15 '13 at 19:13
  • Let's assume for the moment that @O.L.'s comment is not important and that $\text{Li}_4(2)$ is well-defined. Even in that case, what you are saying makes no sense. How is the relationship you allege of any use here? It is still a series, or an integral, that has to be computed somehow, with the same level of difficulty. – Ron Gordon Aug 15 '13 at 19:14
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    @RonGordan , I think this is a good thing that Mhenni posted that , It is well known to extend the polylogarithm for the value $2$ because we can have nice closed forms . We can get closed forms for $\operatorname{Li}_2(2),\operatorname{Li}_3(2)$ the problem seems to continue with evaluating $\operatorname{Li}_4(2)$ – Zaid Alyafeai Aug 17 '13 at 00:23
  • @ZaidAlyafeai: I am glad you like Mhenni's contribution here. I disagree that this is a useful solution, though, because this is simply not a closed form for anything. If the problem were to relate the two values of the polylog, then kudos to Mhenni (although he also never gave a hint as to where such a relationship comes from). But, that is not the stated problem; rather, it was to ask whether there was a closed form of $\text{Li}_4(1/2)$. – Ron Gordon Aug 17 '13 at 02:32
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    As O.L. points out, this is extremely unlikely. Mhenni's solution is not a closed form, as it expresses the desired quantity in terms of another that has exactly the same problems of not knowing whether a simple closed form exists. Actually, it is worse, because now we must define a particular branch cut to make the evaluation even meaningful. This solution replaces one brick wall with another; it provides no insight as to how this quantity is computed practically. – Ron Gordon Aug 17 '13 at 02:35