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 $~\Im~\text{Li}_3(1+i).$
 A: I'm seriously late for this party, but to complement Cleo's answer, if we consider a hypergeometric as a closed-form, then using a different hypergeometric, the family is,
$$\Im\left[\operatorname{Li}_\color{red}2\big(1+i\big)\right] =\sqrt2 \;{_3F_2}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)$$
$$\Im\left[\operatorname{Li}_\color{red}2\left(\frac{1+i}2\right)\right] =\frac{-1}{\sqrt2} \;{_3F_2}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)+\frac{3K}2$$
with Catalan's constant $K=\Im\left[\operatorname{Li}_\color{blue}2(i)\right]$ and,
$$\Im\left[\operatorname{Li}_\color{red}3\big(1+i\big)\right] =\sqrt2 \;{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)-\frac{\pi^3}{192}$$
$$\Im\left[\operatorname{Li}_\color{red}3\left(\frac{1+i}2\right)\right] =-\sqrt2 \;{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)+\frac{23\pi^3}{384}+\frac{3\pi\ln^2 2}{32}$$
where $\frac{\pi^3}{32}=\Im\left[\operatorname{Li}_\color{blue}3(i)\right]$. It is tempting to speculate what comes next.

Edit: July 2019. With insights from this answer, it turns out,
$$\Im\left[\operatorname{Li}_\color{red}4\big(1+i\big)\right] =\sqrt2 \;{_5F_4}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)-\frac{\pi^3\ln 2}{382}-\frac{\pi\zeta(3)}8+\frac{\Im\left[\operatorname{Li}_\color{blue}4(i)\right]}4$$
P.S. The first two hypergeometrics actually have a closed-form, but I didn't want to clutter my answer.
A: I'll sketch a simple solution to the fact that $$\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3),$$ as the OP stated.
First step: We easily prove this (possibly new?) result

$$\int_0^1 \frac{\log (x) \log (1-x)}{1-a x} \textrm{d}x$$
$$=\zeta (2)\frac{ \log (1-a)}{a}+\frac{\log ^3(1-a)}{6 a}+\frac{\text{Li}_3(a)}{a}-\frac{\text{Li}_3\left(\frac{a}{a-1}\right)}{a}. \tag1$$
The result above is easy to prove with the algebraic identities and the result $\displaystyle \int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$ from (Almost) Impossible Integrals, Sums, and Series, page $4$.

Second step: Plug in $(1)$ $a=-i$.
Third step: We need the integral 

$$\int_0^1 \frac{x\log(1- x)\log(x)}{1+x^2}\textrm{d}x=\frac{1}{16}\left(\frac{41}{4}\zeta(3)-9\log(2)\zeta(2)\right),$$
and this is calculated by real methods in the book (Almost) Impossible Integrals, Sums, and Series

Fourth step: We combine the results above and extract the desired value 
$$\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3).$$
In a similar style we derive other values mentioned by user153012 in his/her post.
More details may be found in the preprint,
A special way of extracting the real part of the Trilogarithm, Li_3((1±i)/2))

For those interested in various forms of $(1)$, with slightly different numerators, see these new entries in the mathematical literature, 

Let $a\le1$ be a real number. The following equalities hold: 
  \begin{equation*}
i) \ \int_0^1 \frac{\log (x)\operatorname{Li}_2(x) }{1-a x} \textrm{d}x=\frac{(\operatorname{Li}_2(a))^2}{2 a}+3\frac{\operatorname{Li}_4(a)}{a}-2\zeta(2)\frac{\operatorname{Li}_2(a)}{a};
\end{equation*}
\begin{equation*}
ii) \ \int_0^1 \frac{\log^2(x)\operatorname{Li}_3(x) }{1-a x} \textrm{d}x=20\frac{\operatorname{Li}_6(a)}{a}-12 \zeta(2)\frac{\operatorname{Li}_4(a)}{ a}+\frac{(\operatorname{Li}_3(a))^2}{a},
\end{equation*}

which appears in the preprint A simple idea to calculate a class of polylogarithmic integrals by using the Cauchy product of squared Polylogarithm function by Cornel I. Valean.
Just observe the result from $i)$ is the result in $(1)$ with $\log(1-x)$ replaced by $\operatorname{Li}_2(x)$.
Further Research: 
The new formulae might open the possibility of extracting the real parts of the more advanced versions, $\Re\bigg[\text{Li}_4\bigg(\dfrac{1\pm i}2\bigg)\bigg]$, which are known in the mathematical literature, or even approach the general case  $\Re\bigg[\text{Li}_n\bigg(\dfrac{1\pm i}2\bigg)\bigg]$.

Related and conjectured cases you may find here On the simplification of $\Re\operatorname{Li}_5(1+i)$ and its generalization, like $\Re\{\operatorname{Li}_4(1+i)\}$ and $\Re\{\operatorname{Li}_5(1+i)\}$.
A: 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.
A: If you consider a hypergeometric function to be a closed form, you can have the following result:
$$\Im\left[\operatorname{Li}_3\left(\frac{1+i}2\right)\right]=\frac{\pi^3}{128}+\frac\pi{32}\ln^22+\frac14\,{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,1,1\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,1\right).$$
And for the polylogarithm value that appears in another answer, you can have
$$\Im\Big[\operatorname{Li}_3\left(1+i\right)\Big]=\frac{3\pi^3}{64}+\frac\pi{16}\ln^22-\frac14\,{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,1,1\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,1\right).$$
A: Start with the trilog identity
$$\text{Li}_3(x)+\text{Li}_3(1-x)+\text{Li}_3\left(\frac{x-1}{x}\right)$$
$$=\zeta(3)+\frac16\ln^3(x)+\zeta(2)\ln(x)-\frac12\ln^2(x)\ln(1-x)\tag1$$
set $x=i$ and consider the real parts of the two sides we have
$$\Re\left\{\text{Li}_3(i)+\text{Li}_3(1-i)+\text{Li}_3\left(\frac{i-1}{i}\right)\right\}$$
$$=\zeta(3)+\Re\left\{\frac16\ln^3(i)+\zeta(2)\ln(i)-\frac12\ln^2(i)\ln(1-i)\right\}$$
note that $\frac{i-1}{i}=1+i$ and that $\Re \text{Li}_3(1+i)=\Re \text{Li}_3(1-i)$. This gives
$$\Re\left\{\text{Li}_3(i)+2\text{Li}_3(1+i)\right\}=\zeta(3)+\Re\left\{\frac16\ln^3(i)+\zeta(2)\ln(i)-\frac12\ln^2(i)\ln(1-i)\right\}$$
Substitute $\Re\text{Li}_3(i)=-\frac{3}{32}\zeta(3)$, $\ln(i)=\frac{\pi}{2}i$ and $\ln(1-i)=\frac12\ln(2)-\frac{\pi}{4}i$ we get
$$\Re\text{Li}_3(1+i)=\frac{3}{16}\ln(2)\zeta(2)+\frac{35}{64}\zeta(3)$$
now setting $x=1+i$ in $(1)$ and considering the real parts of both sides yields
$$\Re\left\{\text{Li}_3(1+i)+\text{Li}_3(-i)+\text{Li}_3\left(\frac{i}{1+i}\right)\right\}$$
$$=\zeta(3)+\Re\left\{\frac16\ln^3(1+i)+\zeta(2)\ln(1+i)-\frac12\ln^2(1+i)\ln(-i)\right\}$$
Note $\frac{i}{1+i}=\frac{1+i}{2}$ and substitute $\Re\text{Li}_3(1+i)=\frac{3}{16}\ln(2)\zeta(2)+\frac{35}{64}\zeta(3)$, $\Re\text{Li}_3(-i)=-\frac{3}{32}\zeta(3)$, $\ln(-i)=-\frac{\pi}{2}i$ and $\ln(1+i)=\frac12\ln(2)+\frac{\pi}{4}i$ we obtain
$$\Re\text{Li}_3\bigg(\dfrac{1+i}2\bigg)=\dfrac{\ln^3(2)}{48}-\dfrac{\ln(2)}{32}\zeta(2)+\dfrac{35}{64}\zeta(3)=\Re\text{Li}_3\bigg(\dfrac{1-i}2\bigg)$$
A: Form here we know that
$$\operatorname{Li}_3(z)-\operatorname{Li}_3\left(\frac{1}{z}\right)=-\frac{1}{6} \ln^3(z)-\frac{\pi\sqrt{-(z-1)^2}}{2(z-1)}\ln^2(z)+\frac{\pi^2}{3}\ln(z)\tag{1}.$$
If we put $z:=(1+i)/2$ into $(1)$ and get the imaginary part of it, we get
$$\Im\left[\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right] - \Im \left[\operatorname{Li}_3(1-i)\right] = \frac{7\pi^3}{128}+\frac{3\pi}{32}\ln^2 2.$$
@Tunk-Fey said that $\displaystyle\Im[\operatorname{Li}_3(1-i)]=-\Im[\operatorname{Li}_3(1+i)]$, so it is also true, that
$$\Im\left[\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right] + \Im \left[\operatorname{Li}_3(1+i)\right] = \frac{7\pi^3}{128}+\frac{3\pi}{32}\ln^2 2.$$
Sadly the only two exact complex value of $\operatorname{Li}_3$ function what I found is $\operatorname{Li}_3(\pm i).$ We could use it, but with known identities we coudn't break out of the prison of $\operatorname{Li}_3(1+i)$ or $\operatorname{Li}_3(1-i)$ with this approach.
I found nothing about it, but I think @Tunk-Fey's result is in general true, and I think that is true, that $\Im \operatorname{Li}_3(z)+\Im \operatorname{Li}_3(\overline{z}) = 0$ for all $z$ which have complex part. It isn't a good news, because using identities we get an unknown complex value of $\operatorname{Li}_3$ or a complex conjugate pair of the variable what we are looking for.
Of course there is relationship between polylogarithm function and generalized hypergeometric function. For $\operatorname{Li}_3$ we have
$$\operatorname{Li}_3(z) = z \;_{4}F_{3} (1,1,\dots,1; \,2,2,\dots,2; \,z).$$
So we could write the problem also into the form
$$\Im\left[\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right] = \Im\left[\frac{1+i}{2}{_4F_3}\left(\begin{array}c\ 1,1,1,1\\2,2,2\end{array}\middle|\,\frac{1+i}{2}\right) \right].$$
I don't know how @Cleo transformed it into the form that is given above, but it is really nice, how the imaginary part is eliminated.
If somebody could give some more complex valued exact solutions of $\operatorname{Li}_3$ function, then maybe I could get something more. But for now, I'm also waiting for a solution.

By the way, if your result about the real part is correct, then we can get a really beautiful closed formula for $\Re[\operatorname{Li}_3(1 \pm i)]$ using the partial results of your problem.
If we put again $z:=(1+i)/2$ into $(1)$ and now we get the real part of it, we have
$$\Re\left[\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right] - \Re \left[\operatorname{Li}_3(1-i)\right] = \frac{\ln^3 2}{48}-\frac{11\pi^2}{192}\ln 2.$$
Now if you're right, and
$$\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3),$$
then we get for $\Re \left[\operatorname{Li}_3(1 \pm i)\right]$ the following.
$$\Re \left[\operatorname{Li}_3(1 \pm i)\right] = \frac{\pi^2}{32} \ln 2 + \frac{35}{64} \zeta(3).$$
This seems to me numerically correct, and via your problem I could solve a really interesting related diamond.
A: This is equivalent to finding a closed form for the following series
$$I :=\mathrm{Im} \left[\mathrm{Li}_3\left(\frac{1+i}{2}\right)\right] = \sum_{k=1}^{\infty} \frac{\sin \pi k/4}{2^{k/2}k^3}.$$
I was able to find the following expression, which seems to be numerically correct

$$I=\frac{7\pi^3}{256}+\frac{3\pi}{64}\log^22+\frac{1}{2} \mathrm{Im}\left[\mathrm{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)-\mathrm{Li}_3(1+i)\right].$$

Update: Taking the advice of V.Rosssetto into account, we are reduced to

$$I=\frac{7\pi^3}{128}+\frac{3\pi}{32}\log^22-\mathrm{Im} \; \mathrm{Li}_3(1+i).$$

