Proving $\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)= -\frac{\log^33}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}$?

Question

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.

Answer 1

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 $\operatorname{Li}_3\left(1\right)=\zeta(3)$, $\operatorname{Li}_3\left(-1\right)=-\frac34\zeta(3)$.

Answer 2

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.

Answer 3

From \heef{http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog3/06/01/}{Wolfram Functions} (in functionals with six trilogarithms) we have the identity:

\begin{align*} & L{i_3}\left( \frac{1 - z}{1 + z} \right) - L{i_3}\left( -\frac{1 - z}{1 + z} \right) \\ &= -2L{i_3}\left( \frac{z}{z - 1} \right) - 2L{i_3}\left( \frac{z}{z + 1} \right) + \frac{1}{2}L{i_3}\left( \frac{z^2}{z^2 - 1} \right) + \\ &\quad \frac{7}{4}L{i_3}(1) + \frac{1}{4}\log\left( \frac{1 - z^2}{z^2} \right)\log^2\left( \frac{1 + z}{1 - z} \right) + \frac{1}{4}\pi^2\log\left( \frac{1 - z}{1 + z} \right) \end{align*}

And for $z = \frac{1}{2}$ arises:

\begin{align*} & L{i_3}\left( \frac{1}{3} \right) - L{i_3}\left( -\frac{1}{3} \right) \\ &= -2L{i_3}(-1) - 2L{i_3}\left( \frac{1}{3} \right) + \frac{1}{2}L{i_3}\left( -\frac{1}{3} \right) + \frac{7}{4}L{i_3}(1) + \frac{1}{4}\log^3(3) - \frac{1}{4}\pi^2\log(3) \end{align*}

Because $L{i_3}(1) = \zeta(3)$ and $L{i_3}(-1) = -\frac{3}{4}\zeta(3)$ , finally, we have:

$$3L{i_3}\left( \frac{1}{3} \right) - \frac{3}{2}L{i_3}\left( -\frac{1}{3} \right) = \frac{13}{4}\zeta(3) + \frac{1}{4}\log^3(3) - \frac{1}{4}\pi^2\log(3) \Rightarrow L{i_3}\left( -\frac{1}{3} \right) - 2L{i_3}\left( \frac{1}{3} \right) = -\frac{13}{6}\zeta(3) - \frac{1}{6}\log^3(3) + \frac{1}{6}\pi^2\log(3)$$

Answer 4

Here is a solution.

We are using the identity:

$${\rm Li}_3 \left ( \frac{1-z}{1+z} \right ) - {\rm Li}_3 \left ( - \frac{1-z}{1+z} \right )= -2 {\rm Li}_3 \left ( \frac{z}{z-1} \right ) - 2{\rm Li}_3 \left ( \frac{z}{z+1} \right )+ \frac{1}{2}{\rm Li}_3 \left ( \frac{z^2}{z^2-1} \right )+ \frac{7}{4}{\rm Li}_3 (1)+ \\ +\frac{1}{4}\log \left ( \frac{1-z^2}{z^2} \right )\log^2 \left ( \frac{1+z}{1-z} \right )+ \frac{1}{4}\pi^2 \log \left ( \frac{1-z}{1+z} \right )$$

along with some known results of the trilog , i.e ${\rm Li}_3(1)=\zeta(3), \; {\rm Li}_3(-1)=-\frac{3\zeta(3)}{4}$. Hence subbing $z=1/2$ we have the desired result.