Closed-form of $\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right)$ I've found the following identity while I was going through a quite difficult path.

$$
\Re\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right) = \frac{\pi^2}{24} -\frac{1}{2}\ln^2 2 - \frac{1}{4}\operatorname{Li}_2\left(\tfrac{1}{4}\right),$$

where $\operatorname{Li}_2$ is the dilogarithm function.
I think we could prove it directly from dilogarithm identites.

How could we prove the identity above?

Furthermore

Could we specify $\Im\operatorname{Li}_2\left(1 \pm i\sqrt{3}\right)$?

There is a similar question here.
 A: For first, we have:
$$I=\Re\,\text{Li}_2(1\pm i\sqrt{3})=\Re\int_{1}^{i\sqrt{3}}\frac{\log t}{1-t}\,dt=\frac{\pi^2}{6}+\Re\int_{0}^{i\sqrt{3}}\frac{\log t}{1-t}\,dt\tag{1}$$
hence:
$$ I = \frac{\pi^2}{6}-\int_{0}^{\sqrt{3}}\frac{\frac{\pi}{2}+t\log t}{1+t^2}\,dt\tag{2}=-\int_{0}^{\sqrt{3}}\frac{t\log t}{1+t^2}\,dt $$
but:
$$\begin{eqnarray*} \int\frac{t\log t}{1+t^2}\,dt &=& \frac{\log(t)\log(1+t^2)}{2}+\text{Li}_2(it)+\text{Li}_2(-it)\\&=&\frac{\log(t)\log(1+t^2)}{2}+\frac{1}{4}\,\text{Li}_2(-t^2)\tag{3}\end{eqnarray*}$$
follows from integration by parts and the dilogarithm reflection formula, hence I simply get:

$$ \Re\,\text{Li}_2(1\pm i\sqrt{3})=-\frac{\log(3)\log(4)+\text{Li}_2(-3)}{4}\tag{4}$$

and it should not be too difficult to convert $(4)$ into your expression through the dilogarithm functional identities. In a similar fashion:
$$ \Im\,\text{Li}_2(1+ i\sqrt{3})=\int_{0}^{\sqrt{3}}\frac{\frac{\pi t}{2}-\log t}{1+t^2}\,dt = \frac{\pi}{2}\log 2+\int_{\frac{1}{\sqrt{3}}}^{+\infty}\frac{\log t}{1+t^2}\,dt\tag{5}$$
hence:
$$ \Im\,\text{Li}_2(1+ i\sqrt{3})=\frac{\pi}{2}\log 2+\frac{\pi}{12}\log 3+\Im\,\text{Li}_2\left(\frac{i}{\sqrt{3}}\right)\tag{6}$$
or:

$$ \Im\,\text{Li}_2(1+ i\sqrt{3})=\frac{\pi}{12}\log(192)+\frac{1}{\sqrt{3}}\sum_{n\geq 0}\frac{(-1)^n}{3^n(2n+1)^2}.\tag{7}$$

giving a value of the inverse tangent integral. It depends on the value of the Clausen function $\text{Cl}_2\left(\frac{\pi}{3}\right)$, and that quantity depends on the values of the Barnes $G$-function at $\frac{1}{3}$ and $\frac{2}{3}$.
A: Here is a solution only using dilogarithm identities:

$$ \operatorname{Li}_2(z) + \operatorname{Li}_2(1-z) = \zeta(2) - \log z \log(1-z). \tag{1} $$
  $$ \operatorname{Li}_2(1-z) + \operatorname{Li}_2(1-\tfrac{1}{z}) = -\tfrac{1}{2}\log^2 z. \tag{2} $$
  $$ \operatorname{Li}_2(z) + \operatorname{Li}_2(-z) = \tfrac{1}{2}\operatorname{Li}_2(z^2). \tag{3} $$

We remark that each identity is true for all $z \in \Bbb{C}$ that avoids the principal branch cuts of the functions which are involved in it. Then


*

*Applying (1), we have
\begin{align*}
\operatorname{Li}_2(1\pm i\sqrt{3})
&= \zeta(2)-\log(1\pm i\sqrt{3})\log(\mp i\sqrt{3}) - \operatorname{Li}_2(\mp i\sqrt{3}) \\
&= \zeta(2)-\left( \log 2 \pm \tfrac{i\pi}{3} \right) \left( \tfrac{1}{2}\log 3 \mp \tfrac{i\pi}{2} \right) - \operatorname{Li}_2(\mp i\sqrt{3}).
\end{align*}
Taking real part of both sides, we get
$$ \Re \operatorname{Li}_2(1\pm i\sqrt{3}) = - \tfrac{1}{2}\log 2 \log 3 - \Re\operatorname{Li}_2(\mp i\sqrt{3}). $$

*Now we apply (3). Using the fact that $\overline{\operatorname{Li}_2(z)} = \operatorname{Li}_2(\bar{z})$, we have
$$ \Re\operatorname{Li}_2(\mp i\sqrt{3})
= \tfrac{1}{2}\left( \operatorname{Li}_2(i\sqrt{3}) + \operatorname{Li}_2(-i\sqrt{3}) \right)
= \tfrac{1}{4}\operatorname{Li}_2(-3). $$

*Finally, we use (2) with $z = 4$ and (1) with $z = 1/4$. Then
$$ \operatorname{Li}_2(-3) = \operatorname{Li}_2(\tfrac{1}{4}) - \zeta(2) + 2\log 2 \log(\tfrac{2}{3}). $$

*Combining these altogether, we have
\begin{align*}
\Re \operatorname{Li}_2(1\pm i\sqrt{3})
&= - \tfrac{1}{2}\log 2 \log 3 - \Re\operatorname{Li}_2(\mp i\sqrt{3}) \\
&= - \tfrac{1}{2}\log 2 \log 3 - \tfrac{1}{4}\operatorname{Li}_2(-3) \\
&= \tfrac{1}{4}\zeta(2) - \tfrac{1}{2}\log^2 2  - \tfrac{1}{4}\operatorname{Li}_2(\tfrac{1}{4}).
\end{align*}

P.s. I read Jack D'Aurizio's answer now and I see that all the essential idea is exactly the same. I should have checked his answer before I write it. ;(
A: To complete the excellent answer of @JackD'Aurizio:
\begin{align*}
\Im\operatorname{Li}_2\left(1+i\sqrt3\right)=\frac{\pi \ln 2}{2}+\frac{5\sqrt3}{72}\left[\psi_1\left(\frac13\right)-\psi_1\left(\frac23\right)\right],
\end{align*}
where $\psi_1(z)$ denotes the trigamma function.
