We know some exact values of the trilogarithm $\operatorname{Li}_3$ function.

Known real analytic values for $\operatorname{Li}_3$:

  • $\operatorname{Li}_3(-1)=-\frac{3}{4} \zeta(3)$
  • $\operatorname{Li}_3(0)=0$
  • $\operatorname{Li}_3\left( \frac{1}{2} \right) = \frac{7}{8} \zeta(3) - \frac{1}{12} \pi^2 \ln 2 + \frac{1}{6} \ln^3 2$
  • $\operatorname{Li}_3(1) = \zeta(3),$ where $\zeta(3)$ is the Apéry's constant
  • $\operatorname{Li}_3\left(\phi^{-2}\right)=\frac{4}{5} \zeta(3) + \frac{2}{3} (\ln \phi)^3 - \frac{2}{15} \pi^2 \ln \phi,$ where $\phi$ is the golden ratio.

Using identities for the list above we could also get:

  • $\operatorname{Li}_3(2) = \frac{\pi^2}{4} \ln 2 + \frac{7}{8} \zeta(3) - \frac{\pi}{2} \ln^2(2) \cdot i,$ or we could write into this alternate form.
  • $\operatorname{Li}_3\left(\phi^2\right) = \frac{4}{5} \zeta(3) - \frac{2}{3} \ln^3 \phi + \frac{8\pi^2}{15}\ln \phi - 2\pi\ln^2(\phi) \cdot i,$ or there is an alternate form here.

We know even less about complex argumented values:

  • $\operatorname{Li}_3(i)=-\frac{3}{32}\zeta(3) +\frac{\pi^3}{32} i$
  • $\operatorname{Li}_3(-i)=-\frac{3}{32}\zeta(3) -\frac{\pi^3}{32} i.$

There are some partial result for complex cases:

  • $\Im\left[ \operatorname{Li}_3 \left( \frac{1 \pm i}{\sqrt{2}} \right) \right] = \pm \frac{7\pi^3}{256}$, and there is an expression for the real part in term of derivatives of digamma function,
  • other related specific values around the unit circle like this or this, etc.

Lucian said the following in this question:

  • $\Re\left[\text{Li}_3\left(\dfrac{1+i}2\right)\right]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3).$

While working on Lucian's problem I was able to specify this one

  • $\Re \left[\operatorname{Li}_3(1 \pm i)\right] = \frac{\pi^2}{32} \ln 2 + \frac{35}{64} \zeta(3).$

Than I got the idea to write a question, maybe someone could give us some more specific values of the trilogarithm function.

  • $\begingroup$ A somewhat related recent question. $\endgroup$ Commented Sep 17, 2014 at 20:11
  • $\begingroup$ You probably know that for all $x>1,\ \Im\big[\operatorname{Li}_3(x)\big]=-\frac{\pi}{2}\ln^2x$. $\endgroup$ Commented Sep 17, 2014 at 22:10
  • $\begingroup$ @VladimirReshetnikov could you give me some reference or a proof for this property? If we are here I'm also interested for a proof of this. $\endgroup$
    – user153012
    Commented Sep 18, 2014 at 8:05
  • $\begingroup$ It follows from the formula at Wolfram Functions. $\endgroup$ Commented Sep 18, 2014 at 16:23
  • $\begingroup$ @VladimirReshetnikov Sorry for that, but I don't exactly see how is it comes from the identity what you linked. Could you explain me by giving more details? By the ways using this identity then this more general formula is also true: $$\Im[\operatorname{Li}_3(z)]=-\frac14 \,\pi \, \left( \ln \left( {z}^{-1} \right) \right) ^{2} \left( \left| {\frac {-1+z}{z}} \right| z+z-1 \right) \left( -1+z \right) ^ {-1},$$ for all $1 \neq z>0$. Between $0<z<1$ it is zero. $\endgroup$
    – user153012
    Commented Sep 30, 2014 at 12:24

1 Answer 1


It appears that $$ \Im\left[\mathrm{Li}_3\left((-1)^{\frac{1}{2^n}}\right) \right]=\frac{(2^n-1)(2^{n+1}-1)}{3\cdot2^{3n+2}}\pi^3 $$ but that's by experimentation.

  • 1
    $\begingroup$ What is the relationship between $k$ and $n$? $\endgroup$
    – user153012
    Commented Oct 21, 2017 at 9:05
  • $\begingroup$ Thanks for pointing that out. The $k$ was supposed to be an $n$. $\endgroup$ Commented Oct 23, 2017 at 9:41
  • 1
    $\begingroup$ For example for $n=1$ the right-hand side equals zero, but the left-hand side is non-zero. Other values seem also incorrect. $\endgroup$
    – user153012
    Commented Oct 23, 2017 at 17:18
  • $\begingroup$ My apologies, two typos in one post. Have changed $2^{n-1}$ to $2^{n+1}$, which is what I originally meant. $\endgroup$ Commented Oct 24, 2017 at 9:03

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