TITLE: Conjecture $\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)=\frac{7\pi^2}{48}-\frac13\arctan^22-\frac16\arctan^23-\frac18\ln^2(\tfrac{18}5)$
QUESTION [20 upvotes]: 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 $p,q\in\mathbb Q$?

REPLY [4 votes]: 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 $$\Re\operatorname{Li}_2\left(\frac{1+ti}2\right)=\frac1{12}\pi^2-\frac12\arctan^2t-\frac18\ln^2\frac{1+t^2}4$$