Simplify the difference of two dilogarithms--as in the logarithmic counterpart This question--pertaining to the quantum-information-theoretic topic of "bound entanglement"--stems from the question and answer to https://math.stackexchange.com/questions/3464105/if-possible-meaningfully-simplify-an-expression-involving-logs-polylogs-and-hy/3465129#3465129 
The answer (that is, the formula for the "bound entanglement probability") there contains the  expression
\begin{equation}
 \text{Li}_2\left(\frac{1}{18}
   \left(9+\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)-\text{Li}_2\left(\frac{1}{18}
   \left(9-\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right),
\end{equation}
where the polylogarithm (in particular, dilogarithm) is indicated. 
We have further observed as part of the simplification analysis in that answer--changing the subscript of Li from 2 to 1 (leading to the standard logarithmic framework)--that 
\begin{equation}
 \text{Li}_1\left(\frac{1}{18}
   \left(9+\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)-\text{Li}_1\left(\frac{1}{18}
   \left(9-\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right) = 
\end{equation}
\begin{equation}
\log \left(\frac{1}{18} \left(9+\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)-\log
   \left(\frac{1}{18} \left(9-\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)=
\end{equation}
\begin{equation}
2 \coth ^{-1}\left(\frac{9}{\sqrt{81-\frac{64}{\sqrt{3}}}}\right),
\end{equation}
So, is a "parallel" simplification possible in the original $\mbox{Li}_{2}$ case? (Of course, one might ask--more generally--about $\mbox{Li}_{n}$.) 
 A: $$\text{Li}_2\left(\tfrac{1}{2}+a\right)-\text{Li}_2\left(\tfrac{1}{2}-a\right)=-\int_{1/2-a}^{1/2+a}\frac{\log(1-t)}{t}\,dt,\;\;0<|{\rm Re}\,a|<1/2.$$
the integral of $\log(1-t)/t$ cannot be expressed in terms of elementary functions; it can be written in terms of special functions, but that brings us back to the polylog expression in the OP.

more generally, for any integer $n\in\mathbb{Z}$,
$$f_n(a)=\text{Li}_n\left(\tfrac{1}{2}+a\right)-\text{Li}_n\left(\tfrac{1}{2}-a\right)=\int_{1/2-a}^{1/2+a}\frac{{\rm Li}_{n-1}(t)}{t}\,dt,\;\;0<|{\rm Re}\,a|<1/2,$$
and since ${\rm Li}_{n-1}(t)$ is a rational function for $n\leq 1$ this gives a simplification of $f_n(a)$ in terms of elementary functions when $n=1,0,-1,-2,\ldots$.

For example,
$$f_1(a)=2 \tanh ^{-1}(2 a),\;\;f_0(a)=\frac{8 a}{1-4 a^2},$$
$$f_{-1}(a)=\frac{8 a \left(4 a^2+3\right)}{\left(1-4 a^2\right)^2},\;\;f_{-2}(a)=\frac{8 a \left(16 a^4+72 a^2+13\right)}{\left(1-4 a^2\right)^3}.$$

A: Mathematica's FullSimplify[] command can do nothing with the expression in question: 

So, it appears unlikely that this expression can be simplified. 
