<s>[INST] Closed Form for the Imaginary Part of $\text{Li}_3\Big(\frac{1+i}2\Big)$ 
$\qquad\qquad$ Is there any closed form expression for the imaginary part of $~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)$ ?


Motivation: We already know that $~\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3)$, 
so I was wondering whether a similar closed form expression might also exist for its
imaginary part as well. Thank you !

Apparently, $~\Im~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)~+~\Im~\text{Li}_3(1+i)~=~\dfrac7{128}\cdot\pi^3~+~\dfrac3{32}\cdot\pi\cdot\ln^22,~$ so the question
is equivalent to asking for the closed form of the imaginary part of $~\Im~\text{Li}_3(1+i).$
 [/INST] Using the relation between 3 trilogarithms given here:
\begin{equation}
 \operatorname{Li}_3(z)=-\operatorname{Li}_3(\frac{z}{z-1})-\operatorname{Li}_3(1-z)+\frac{1}{6}\log^3(1-z)-\frac{1}{2}\log(z)\log^2(1-z)+\frac{\pi^2}{6}\log(1-z)+\zeta(3)
\end{equation} 
with $z=(1+i)/2$, we have $z/(z-1)=-i$ and $1-z=(1-i)/2$. With the known specific value (see here)
\begin{align}
 \operatorname{Li}_3(-i)&=-\frac{3}{32}\zeta(3)-i\frac{\pi^3}{32} 
\end{align}and remarking that $\operatorname{Li}_n(z)=\overline{\operatorname{Li}_n(\overline{z})}$,
\begin{equation}
 \operatorname{Li}_3(z)+\operatorname{Li}_3(1-z)=2\Re\operatorname{Li}_3(\frac{1+i}{2})
\end{equation} 
and
\begin{align}
 \log\left( z \right)&=-\frac{\log 2}{2}+i\frac{\pi}{4}\\
 \log\left(1- z \right)&=-\frac{\log 2}{2}-i\frac{\pi}{4}
\end{align}
we deduce
\begin{align}
 \Re\operatorname{Li}_3(\frac{1+i}{2})&=\frac{1}{2}\left[
 \frac{3}{32}\zeta(3)+i\frac{\pi^3}{32}
+ \frac{1}{6}\left( -\frac{\log 2}{2}-i\frac{\pi}{4} \right)^3-\right.\\
&\left.\frac{1}{2}\left( -\frac{\log 2}{2}+i\frac{\pi}{4} \right))\left( -\frac{\log 2}{2}-i\frac{\pi}{4} \right)^2+\frac{\pi^2}{6}\left( -\frac{\log 2}{2}-i\frac{\pi}{4} \right)+\zeta(3)
 \right]\\
 &=\frac{35}{64}\zeta(3)+\frac{\log^32}{48}-\frac{5}{192}\pi^2\ln 2
\end{align}
as expected.
</s>