Identity involving PolyLog functions

Solution 1:

After some effort, I was able to find a relation of the sort I (almost) wanted. As anticipated by David H, the relation I really wanted probably does not exist. However, if you are willing to live with a relation that involves more than just $\mathrm{Li}_4(y)$, $\mathrm{Li}_4(1-y)$ and $\mathrm{Li}_4(\frac{y}{1-y})$, then the following is probably the most concise thing one can write down $$ \text{Li}_4(1-y)=\frac{1}{24} \pi ^2 \log ^2(1-y)-\frac{19 \pi ^4}{6480}+\frac{7}{48} \log ^4(1-y)-\frac{1}{6} \log ^3(1-y) \log (y)+\frac{1}{3} \text{Li}_4\left(-\frac{y}{(1-y)^2}\right)+\frac{1}{18} \text{Li}_4\left(-(1-y)^3\right)-\frac{3}{2} \text{Li}_4(-(1-y))-\frac{1}{3} \text{Li}_4\left(-\frac{y}{1-y}\right)+\frac{1}{6} \text{Li}_4\left(\frac{y}{1-y}\right)-\frac{1}{6} \text{Li}_4((1-y) y)+\frac{1}{6} \text{Li}_4\left(\frac{y^2}{(1-y)^2}\right)+\frac{1}{6} \text{Li}_4\left(-\frac{y^2}{1-y}\right)-\frac{1}{18} \text{Li}_4\left(\frac{y^3}{(1-y)^3}\right)\,. $$ One remarkable thing about the above expression is that it tells you immediately what is the singularity structure of $\text{Li}_4(x)$ for $x\sim1$. This identity can be found rather hidden (and with a typo) in Lewin L. Polylogarithms and associated functions (NH, 1981).