Ramanujan gave the following identities for the Dilogarithm function:

$$ \begin{align*} \operatorname{Li}_2\left(\frac{1}{3}\right)-\frac{1}{6}\operatorname{Li}_2\left(\frac{1}{9}\right) &=\frac{{\pi}^2}{18}-\frac{\log^23}{6} \\ \operatorname{Li}_2\left(-\frac{1}{3}\right)-\frac{1}{3}\operatorname{Li}_2\left(\frac{1}{9}\right) &=-\frac{{\pi}^2}{18}+\frac{1}{6}\log^23 \end{align*} $$ Now, I was wondering if there are similar identities for the trilogarithm? I found numerically that

$$\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)\stackrel?= -\frac{\log^3 3}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6} \tag{1}$$

  • I was not able to find equation $(1)$ anywhere in literature. Is it a new result?
  • How can we prove $(1)$? I believe that it must be true since it agrees to a lot of decimal places.
  • $\begingroup$ This is beautiful ! I computed the results with 10,000 decimal places; at this level, they are identical. How did you establish this relation ? $\endgroup$ – Claude Leibovici Feb 14 '14 at 8:41
  • $\begingroup$ @ClaudeLeibovici: I guessed that the answer should be of the form $X \log^3 3+Y \zeta(2)\log(3)+Z\zeta(3)$ for some rational $X,Y,Z$. Then with help of Mathematica I found that if we choose $X=-1/6,Y=1,Z=-13/6$ then the two expressions would be incredibly close. $\endgroup$ – Shobhit Bhatnagar Feb 14 '14 at 8:57
  • 1
    $\begingroup$ Incredibly close is just an understatement, I bet. Sure, I only compared 10,000 decimal places ! May be looking at Lerch function could help but I loop in circles ! Cheers and congratulations. $\endgroup$ – Claude Leibovici Feb 14 '14 at 9:12

Combining trilogarithm identities 1 and 2, one obtains the formula \begin{align} \operatorname{Li}_3\left(\frac{1-z}{1+z}\right)-\operatorname{Li}_3\left(-\frac{1-z}{1+z}\right)= 2\operatorname{Li}_3\left(1-z\right)+2\operatorname{Li}_3\left(\frac{1}{1+z}\right)- \frac12\operatorname{Li}_3\left(\frac{1}{1-z^2}\right)-\frac74\operatorname{Li}_3\left(1\right)\\ -\frac13\ln^3(z+1)+\frac{\pi^2}{6}\ln(z+1)+\frac{1}{12}\ln^3\left(z^2-1\right)+\frac{\pi^2}{12}\ln(z^2-1). \end{align}

Now it suffices to set $z=2$ and use that $\operatorname{Li}_3\left(1\right)=\zeta(3)$, $\operatorname{Li}_3\left(-1\right)=-\frac34\zeta(3)$.


By "anywhere in literature" did you include L. Lewin, Polylogarithms and Assoicated Functions ?

In the (fortcoming) solution for Monthly problem 11654, you will see that very equation. For the proof, our solver (Richard Stong) used these identities: $$ \mathrm{Li}_3\left(\frac{1-z}{1+z}\right) - \mathrm{Li}_3\left(\frac{z-1}{z+1}\right) = 2\;\mathrm{Li}_3(1-z) + 2\;\mathrm{Li}_3\left(\frac{1}{z+1}\right) - \frac{1}{2}\;\mathrm{Li}_3(1-z^2) - \frac{7}{4}\zeta(3)- \frac{1}{3}\big(\log(1+z)\big)^3 + \frac{\pi^2}{6}\log(1+z) $$ at $z=2$ and $$ \mathrm{Li}_3(z) = \mathrm{Li}_3\left(\frac{1}{z}\right) - \frac{1}{6}\big(\log(-z)\big)^3 - \frac{\pi^2}{6}\log(-z) $$ at $z=-3$. Stong cites the Lewin book.


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