Closed-form of a special value dilogarithm identity

Question

Let $c$ be the following.

$$c = \frac{1+i\sqrt 3}{3}\operatorname{Li}_2\left(1-\frac{i\sqrt 3}{3}\right)+\operatorname{Li}_2\left(\frac 34 + \frac{i\sqrt 3}{4}\right) + \frac{1-i\sqrt{3}}{3}\operatorname{Li}_2\left(\frac{1}{2}+\frac{i\sqrt 3}{6}\right),$$

where $\operatorname{Li}_2$ is the dilogarithm function.

This $c$ appeared while combining answers to this question. Results from there we know that

$$\Re(c) = -\frac{\pi^2}{54}-\frac{1}{2}\ln^2 2+\frac{\pi\,\sqrt{3}}{6} \ln 2 - \frac{\pi\,\sqrt{3}}{9} \ln 3 + \frac{1}{6} \ln 2 \ln 3 + \frac{1}{4}\psi_1\left(\frac{1}{3}\right)+\frac{1}{6}\operatorname{Li}_2\left(-\frac{1}{3}\right),$$

where $\psi_1$ is the trigamma function.

My questions.

• $1^\text{st}$ Question. Could we evaluate $\Re(c)$ also via some dilogarithm identity?

• $2^\text{nd}$ Question. Could we specify a closed-form of $\Im(c)$ too?

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Edit. From the analysis by @David H we have more results. Let

$$\begin{align} c_1 & = \frac{1+i\sqrt 3}{3}\operatorname{Li}_2\left(1-\frac{i\sqrt 3}{3}\right)\\ c_2 & = \operatorname{Li}_2\left(\frac 34 + \frac{i\sqrt 3}{4}\right)\\ c_3 & = \frac{1-i\sqrt{3}}{3}\operatorname{Li}_2\left(\frac{1}{2}+\frac{i\sqrt 3}{6}\right). \end{align}$$

Therefore of course $c=c_1+c_2+c_3$. David H has shown that

$$\Re(c_2) = \frac{7\pi^2}{72}+\frac{\ln^2{(3)}}{8}-\frac{\ln^2{\left(2\right)}}{2}+\frac14\operatorname{Li}_{2}{\left(-\frac13\right)}.$$

From here it's easy to see, that

$$\Re(c_1+c_3) = -\frac{25\,\pi^2}{216}-\frac{1}{8}\ln^2 3 + \frac{\pi\,\sqrt{3}}{6}\ln 2 - \frac{\pi\,\sqrt{3}}{9} \ln 3 + \frac{1}{6} \ln 2 \ln 3 + \frac{1}{4} \psi_1\left(\frac{1}{3}\right) - \frac{1}{12} \operatorname{Li}_2 \left( - \frac{1}{3} \right).$$

Answer 1

The following is a partial answer to the first question.

For any real number $a\in\mathbb{R}$ and any two complex numers $z_1,z_2\in\mathbb{C}$, the operations $\Re{(\cdot)}$ and $\Im{(\cdot)}$ obey the following algebraic properties:

$$\begin{cases} \Re{(z_1)}:=\frac{z_1+\bar{z_1}}{2}\\ \Im{(z_1)}:=\frac{z_1-\bar{z_1}}{2i}\\ \Re{(z_1+z_2)}=\Re{(z_1)}+\Re{(z_2)}\\ \Im{(z_1+z_2)}=\Im{(z_1)}+\Im{(z_2)}\\ \Re{(az_1)}=a\,\Re{(z_1)}\\ \Im{(az_1)}=a\,\Im{(z_1)}\\ \Re{(iz_1)}=-\Im{(z_1)}\\ \Im{(iz_1)}=\Re{(z_1)}.\\ \end{cases}$$

Using the above properties, we may simplify $\Re{(c)}$ as follows:

$$\begin{align} \Re{(c)} &=\Re{\left[\operatorname{Li}_{2}{\left(\frac34+i\frac{\sqrt{3}}{4}\right)}+\frac{1+i\sqrt{3}}{3}\operatorname{Li}_{2}{\left(1-i\frac{\sqrt{3}}{3}\right)}+\frac{1-i\sqrt{3}}{3}\operatorname{Li}_{2}{\left(\frac12+i\frac{\sqrt{3}}{6}\right)}\right]}\\ &=\Re{\left[\operatorname{Li}_{2}{\left(\frac34+i\frac{\sqrt{3}}{4}\right)}\right]}+\Re{\left[\frac{1+i\sqrt{3}}{3}\operatorname{Li}_{2}{\left(1-i\frac{\sqrt{3}}{3}\right)}\right]}\\ &~~~~~ +\Re{\left[\frac{1-i\sqrt{3}}{3}\operatorname{Li}_{2}{\left(\frac12+i\frac{\sqrt{3}}{6}\right)}\right]}\\ &=\Re{\left[\operatorname{Li}_{2}{\left(\frac34+i\frac{\sqrt{3}}{4}\right)}\right]}+\frac13\Re{\left[\operatorname{Li}_{2}{\left(1-i\frac{\sqrt{3}}{3}\right)}\right]}-\frac{1}{\sqrt{3}}\Im{\left[\operatorname{Li}_{2}{\left(1-i\frac{\sqrt{3}}{3}\right)}\right]}\\ &~~~~~ +\frac13\Re{\left[\operatorname{Li}_{2}{\left(\frac12+i\frac{\sqrt{3}}{6}\right)}\right]}+\frac{1}{\sqrt{3}}\Im{\left[\operatorname{Li}_{2}{\left(\frac12+i\frac{\sqrt{3}}{6}\right)}\right]}.\\ \end{align}$$

Now, Landen's dilogarithm identity states:

$$\operatorname{Li}_{2}{\left(z\right)}=-\operatorname{Li}_{2}{\left(\frac{z}{z-1}\right)}-\frac12\ln^2{\left(1-z\right)};~z\notin[1,\infty).$$

Then, letting $z=\frac34+i\frac{\sqrt{3}}{4}$ we have $\frac{z}{z-1}=-i\,\sqrt{3}$, and thus:

$$\begin{align} \operatorname{Li}_{2}{\left(\frac34+i\frac{\sqrt{3}}{4}\right)} &=-\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}-\frac12\ln^2{\left(\frac14-i\frac{\sqrt{3}}{4}\right)}\\ &=-\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}-\frac12\ln^2{\left(\frac12e^{-\frac{i\pi}{3}}\right)}\\ &=-\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}-\frac12\left(\ln{\left(\frac12\right)}-\frac{i\pi}{3}\right)^2\\ &=-\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}-\frac12\left(-\ln{\left(2\right)}-\frac{i\pi}{3}\right)^2\\ &=-\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}-\frac12\left(-\frac{\pi^2}{9}+\ln^2{\left(2\right)}+\frac{2i\pi\ln{(2)}}{3}\right)\\ &=-\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}+\frac{\pi^2}{18}-\frac{\ln^2{\left(2\right)}}{2}-\frac{i\pi\ln{(2)}}{3}\\ &=\frac{\pi^2}{18}-\frac{\ln^2{\left(2\right)}}{2}-\frac{i\pi\ln{(2)}}{3}-\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}.\\ \end{align}$$

Taking the real component of this dilogarithmic term yields:

$$\begin{align} \Re{\left[\operatorname{Li}_{2}{\left(\frac34+i\frac{\sqrt{3}}{4}\right)}\right]} &=\Re{\left[\frac{\pi^2}{18}-\frac{\ln^2{\left(2\right)}}{2}-\frac{i\pi\ln{(2)}}{3}-\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}\right]}\\ &=\frac{\pi^2}{18}-\frac{\ln^2{\left(2\right)}}{2}-\Re{\left[\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}\right]}\\ &=\frac{\pi^2}{18}-\frac{\ln^2{\left(2\right)}}{2}-\frac{\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}+\overline{\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}}}{2}\\ &=\frac{\pi^2}{18}-\frac{\ln^2{\left(2\right)}}{2}-\frac{\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}+\operatorname{Li}_{2}{\left(\overline{-i\,\sqrt{3}}\right)}}{2}\\ &=\frac{\pi^2}{18}-\frac{\ln^2{\left(2\right)}}{2}-\frac{\operatorname{Li}_{2}{\left(-i\,\sqrt{3}\right)}+\operatorname{Li}_{2}{\left(i\,\sqrt{3}\right)}}{2}\\ &=\frac{\pi^2}{18}-\frac{\ln^2{\left(2\right)}}{2}-\frac{\operatorname{Li}_{2}{\left(-3\right)}}{4},\\ \end{align}$$

where in going from the third to the fourth line we've used the mirror symmetry of the dilogarithm, and to obtain the last line we've used the dilogarithmic identity,

$$\operatorname{Li}_{2}{\left(z\right)}+\operatorname{Li}_{2}{\left(-z\right)}=\frac12\operatorname{Li}_{2}{\left(z^2\right)}.$$

One more dilogarithmic identity we shall need relates the dilogarithms of reciprocal arguments:

$$\operatorname{Li}_{2}{\left(z\right)}=-\operatorname{Li}_{2}{\left(\frac{1}{z}\right)}-\frac12\ln^2{(-z)}-\frac{\pi^2}{6};~z\notin[0,1].$$

Hence,

$$\begin{align} \Re{\left[\operatorname{Li}_{2}{\left(\frac34+i\frac{\sqrt{3}}{4}\right)}\right]} &=\frac{\pi^2}{18}-\frac{\ln^2{\left(2\right)}}{2}-\frac{\operatorname{Li}_{2}{\left(-3\right)}}{4}\\ &=\frac{\pi^2}{18}-\frac{\ln^2{\left(2\right)}}{2}-\frac{-\operatorname{Li}_{2}{\left(-\frac13\right)}-\frac12\ln^2{(3)}-\frac{\pi^2}{6}}{4}\\ &=\frac{7\pi^2}{72}+\frac{\ln^2{(3)}}{8}-\frac{\ln^2{\left(2\right)}}{2}+\frac14\operatorname{Li}_{2}{\left(-\frac13\right)}.\\ \end{align}$$