Conjecture $\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)=\frac{7\pi^2}{48}-\frac13\arctan^22-\frac16\arctan^23-\frac18\ln^2(\tfrac{18}5)$ I numerically discovered the following conjecture:
$$\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)\stackrel{\color{gray}?}=\frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac{\arctan^23}6-\frac18\ln^2\!\left(\frac{18}5\right).$$
It holds numerically with a precision of more than $30000$ decimal digits. 
Could you suggest any ideas how to prove it?
Can we find a closed form for $\Im\,\operatorname{Li}_2\left(\frac12+\frac i6\right)$?
Is there a general method to find closed forms of expressions of the form $\Re\,\operatorname{Li}_2(p+iq)$, $\Im\,\operatorname{Li}_2(p+iq)$ for $p,q\in\mathbb Q$?
 A: I don't know how you got your conjecture but I checked and on further simplifications I found it to be correct.
We know that :
$$
{Li}_{2}(\bar{z})=\bar{{Li}_{2}(z)}
$$
So :
$$
\Re{{Li}_{2}(z)}=\frac{\bar{{Li}_{2}(z)}+{Li}_{2}(z)}{2}
$$
So:
$$
\Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{{Li}_{2}(\frac{1}{2}+\frac{i}{6})+{Li}_{2}(\frac{1}{2}-\frac{i}{6})}{2}\\
=\frac{{Li}_{2}(\frac{1}{2}+\frac{i}{6})+{Li}_{2}(1-(\frac{1}{2}+\frac{i}{6}))}{2}
$$
Now let's use:
$$
{Li}_{2}(z)+{Li}_{2}(1-z)=\frac{{\pi}^{2}}{6}-\ln{z}\ln{(1-z)}
$$
So we get:
$$
\Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{\frac{{\pi}^{2}}{6}-\ln{(\frac{1}{2}+\frac{i}{6})}\ln{(1-(\frac{1}{2}+\frac{i}{6}))}}{2}\\
=\frac{\frac{{\pi}^{2}}{6}-\ln{(\frac{1}{2}+\frac{i}{6})}\ln{(\frac{1}{2}-\frac{i}{6})}}{2}
$$
Let's compute those logarithms:
$$
\ln(\frac{1}{2}\pm \frac{i}{6})=\ln{(\sqrt{\frac{1}{2^2}+\frac{1}{6^2}}{e}^{\pm i\arctan{\frac{1}{3}}})}\\
=\ln{(\sqrt{\frac{5}{18}}{e}^{\pm i\arctan{\frac{1}{3}}})}\\
=\frac{1}{2}\ln{(\frac{5}{18})}\pm i\arctan{\frac{1}{3}}
$$
Taking their product:
$$
\ln{(\frac{1}{2}+\frac{i}{6})}\ln{(\frac{1}{2}-\frac{i}{6})}=\frac{1}{4}{\ln{(\frac{5}{18}})}^{2}+{(\arctan{\frac{1}{3}})}^{2}
$$
Finally:
$$
\Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln{(\frac{18}{5}})}^{2}-\frac{1}{2}{(\arctan{1/3})}^{2}\\
=\frac{7(\pi)^2}{48}-\frac{{(\arctan{2})}^{2}}{3}-\frac{{(\arctan{3})}^{2}}{6}-\frac{1}{8} {\ln{\frac{18}{5}}}^{2}
$$
There's a general formula using the same method. But the imaginary part doesn't have a known closed form:
$$
\Re{{Li}_{2}(\frac{1}{2}+iq)}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln{(\frac{1+4q^2}{4})}}^{2}-\frac{{\arctan{(2q)}}^{2}}{2}
$$
A: This is an answer finding $\Re\operatorname{Li}_2\left(\frac{1+ti}2\right)$ via integration method. (Personally I don't like the reflection formula)
First, we restrict $t\in\mathbb R$. One can prove $\Re\operatorname{Li}_2\left(\frac{1+ti}2\right)$ is differentiable and the following changing the positions of signs is correct.
$$\begin{aligned}
\Re\operatorname{Li}_2\left(\frac{1+ti}2\right)&=-\frac12\int_0^1\ln\left(1-x+\frac{1+t^2}4x^2\right)\frac{d x}x\\
&=-\frac12\int\int_0^1\frac{\partial}{\partial t}\ln\left(1-x+\frac{1+t^2}4x^2\right)\frac{d x}xd t\\
&=-\int\int_0^1\frac{tx}{4-4x+(1+t^2)x^2}d xd t\\
&=-\int\left(\frac1{1+t^2}\arctan t+\frac12\cdot\frac t{1+t^2}\ln\frac{1+t^2}4\right)d t\\
&=-\frac12\arctan^2t-\frac18\ln^2\frac{1+t^2}4+C
\end{aligned}$$
Substitute $t=0$ into the equation, we get $\frac1{12}\pi^2-\frac12\ln^22=0-\frac18\ln^24+C$, or $C=\frac1{12}\pi^2$.
Hence $$\Re\operatorname{Li}_2\left(\frac{1+ti}2\right)=\frac1{12}\pi^2-\frac12\arctan^2t-\frac18\ln^2\frac{1+t^2}4$$
A: First of all we know that:

$$
\operatorname{Li}_2(z) = -\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac{1}{2}\ln^2(1-z), \quad z \notin (1,\infty).\tag{$\diamondsuit$}
$$

Furthermore, we have the following relationship between the dilogarithm and the Clausen functions:

$$\operatorname{Li}_2\left(e^{i\theta}\right) = \operatorname{Sl}_2(\theta)+i\operatorname{Cl}_2(\theta), \quad \theta \in [0,2\pi).\tag{$\heartsuit$}$$

where $\operatorname{Cl}_2$ and $\operatorname{Sl}_2$ are the standard Clausen functions, defined as:
$$\begin{align}
\operatorname{Cl}_2(\theta) &= \sum_{k=1}^{\infty}\frac{\sin(k\theta)}{k^2}, \\
\operatorname{Sl}_2(\theta) &= \sum_{k=1}^{\infty}\frac{\cos(k\theta)}{k^2}.
\end{align}$$
Using the relationship between SL-type Clausen functions and Bernoulli polynomials, we have that

$$
\operatorname{Sl}_2(\theta) = \frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}, \quad \theta \in [0,2\pi).\tag{$\spadesuit$}
$$


Now let $z:=\tfrac{1}{2}+\tfrac{i}{6}$. Because $\left|\tfrac{z}{z-1}\right| = 1$, the equation
$$
e^{i\theta} = \frac{z}{z-1} = -\frac{4}{5}-\frac{3}{5}i,
$$
has the only solution $\theta = \arctan\left(\tfrac{3}{4}\right) + \pi$ in $[0,2\pi)$.
Because of $(\diamondsuit)$ and $(\heartsuit)$ we have
$$
\operatorname{Li}_2(z) = -\color{red}{\operatorname{Sl}_2(\theta)} - i \color{green}{\operatorname{Cl}_2(\theta)} - \color{blue}{\frac{1}{2}\ln^2(1-z)},
$$
for $z=\tfrac{1}{2}+\tfrac{i}{6}$ and $\theta = \arctan\left(\tfrac{3}{4}\right) + \pi$.
For the logarithm term we get
$$
\Re{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{1}{8}\left(\ln^2\left(\frac{18}{5}\right)-(\pi-2\arctan 3)^2\right)
$$
and
$$
\Im{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} =  \frac{1}{4}\ln\left(\frac{18}{5}\right)(\pi-2\arctan 3).
$$
We know that $\color{red}{\operatorname{Sl}_2(\theta)}$ and $\color{green}{\operatorname{Cl}_2(\theta)}$ are real quantities. By using $(\spadesuit)$ for the SL-type Clausen term we get
$$
\color{red}{\operatorname{Sl}_2(\theta)} = \frac{\pi^2}{12}-\frac{1}{4}\arctan^2\left(\frac{3}{4}\right).
$$
Now we could obtain your conjectured closed-form:
$$
\Re\left[\operatorname{Li}_2(z)\right] = -\color{red}{\operatorname{Sl}_2(\theta)} - \Re{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac{\arctan^23}6-\frac18\ln^2\!\left(\frac{18}5\right).
$$
For the imaginary part we have
$$\begin{align}
\Im\left[\operatorname{Li}_2(z)\right] &= -\color{green}{\operatorname{Cl}_2(\theta)} - \Im{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} \\ &= -\operatorname{Cl}_2\left(\arctan\left(\frac{3}{4}\right)+\pi\right)-\frac{1}{4}\ln\left(\frac{18}{5}\right)(\pi-2\arctan 3).
\end{align}$$
By using $(\diamondsuit), (\heartsuit)$ and $(\spadesuit)$, you could generalize this process for all $z \in \mathbb{C}$ such that $\left|\frac{z}{z-1}\right|=1$.
