We all know that
$$\Re{ \left[{\rm Li}_{2}\left(\frac{1}{2}+iq\right) \right]}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln{\left(\frac{1+4q^2}{4}\right)}}^{2}-\frac{{\arctan^2{(2q)}}}{2}$$
where $q \in \mathbb{Q}$. How about the imaginary part of the equation? I have a feeling it involves beta Dirichlet function.
Addendum:
Sketch of proof:
Recall the fact that:
$${\rm Li}_2(\bar{z})=\overline{{\rm Li}_2(z)}$$
hence $$\Re{\rm Li}_2(z)=\frac{{\rm Li}_2(\bar{z}) + \overline{{\rm Li}_2(z)}}{2}$$
and then combine it with the very known functional equation
$${\rm Li}_2(z)+{\rm Li}_2(1-z)=\zeta(2)-\ln z \ln (1-z)$$
Thus the result. Maybe we can get the imaginary part by invoking the known relation:
$$\Im \left[ {\rm Li}_2 \right] =\frac{{\rm Li}_2(\bar{z}) - \overline{{\rm Li}_2(z)}}{2}$$