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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

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

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  • $\begingroup$ Have you got any solution for it? $\endgroup$ Aug 15, 2013 at 4:48

4 Answers 4

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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.

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  • $\begingroup$ Thanks Felix. There is a more general formula $$\text{Li}_s(z)\Gamma(s)=\int^\infty_0 \frac{t^{s-1}}{e^t/z-1}\,dt$$ $\endgroup$ Jun 24, 2014 at 13:45
  • $\begingroup$ @ZaidAlyafeai That's related to Bose-Einstein statistic. There is a nice appendix about that in this book. $\endgroup$ Jun 24, 2014 at 16:59
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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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  • $\begingroup$ I still believe there is a closed form .It might involve $\zeta(4), \log^4,\log^3,...$. $\endgroup$ Aug 15, 2013 at 19:57
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    $\begingroup$ @ZaidAlyafeai If anybody will be able to give such a closed form, I will gladly award a bounty to his answer. $\endgroup$ Aug 15, 2013 at 20:08
  • $\begingroup$ By the way , you might be interested in integralsandseries.prophpbb.com/post968.html#p968 . $\endgroup$ Aug 15, 2013 at 20:13
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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.

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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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    $\begingroup$ Where did the $14.56\ldots$ come from? $\endgroup$
    – Ron Gordon
    Aug 15, 2013 at 5:35
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    $\begingroup$ 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? $\endgroup$
    – Ron Gordon
    Aug 15, 2013 at 19:07
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    $\begingroup$ @RonGordon: Offcourse, it is a closed form and relates two polylogarithm functions. $\endgroup$ Aug 15, 2013 at 19:09
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    $\begingroup$ @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. $\endgroup$ Aug 15, 2013 at 19:13
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    $\begingroup$ @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 $\operatorname{Li}_4(2)$ $\endgroup$ Aug 17, 2013 at 0:23

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