# 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

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

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


\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.

• Thanks Felix. There is a more general formula $$\text{Li}_s(z)\Gamma(s)=\int^\infty_0 \frac{t^{s-1}}{e^t/z-1}\,dt$$ Jun 24, 2014 at 13:45
• @ZaidAlyafeai That's related to Bose-Einstein statistic. There is a nice appendix about that in this book. Jun 24, 2014 at 16:59

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.

• I still believe there is a closed form .It might involve $\zeta(4), \log^4,\log^3,...$. Aug 15, 2013 at 19:57
• @ZaidAlyafeai If anybody will be able to give such a closed form, I will gladly award a bounty to his answer. Aug 15, 2013 at 20:08
• By the way , you might be interested in integralsandseries.prophpbb.com/post968.html#p968 . Aug 15, 2013 at 20:13

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.

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