Real Part of the Dilogarithm It is well known that 
$$\frac{x-\pi}{2}=-\sum_{k\geq 1}\frac{\sin{kx}}{k}\forall x\in(0,\tau),$$
which gives
$$\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}=\sum_{k\geq 1}\frac{\cos(kx)}{k^2}.$$
Note that
$$\textrm{Li}_2(e^{ix})=\sum_{k\geq 1}\frac{\cos(kx)+i\sin(kx)}{k^2}$$
This means that 
$$\mathfrak{R}\textrm{Li}_2(e^{ix})=\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}$$
unless I'm wrong on one of the above statements. Now, with the previous restrictions on $x$, we have that this formula is valid for all complex numbers lying on the complex unit circle as inputs. My question is: is there a way to define something along these lines (a finite degree polynomial, preferably) for complex inputs of the dilogarithm that don't necessarily lie on the complex unit circle? More specifically,
$$\mathfrak{R}\textrm{Li}_2(re^{ix})=?$$
 A: To simplify, you can use the Inverse Tangent Integral $\operatorname{Ti_{2}}$ and the Clausen Functions $\operatorname{C_{n}}$ and $\operatorname{S_{n}}$.
$$
\begin{align*}
\operatorname{Li_{2}}\left( e^{x \cdot i} \right) &= \sum_{k = 1}^{\infty} \left( \frac{\cos\left( k \cdot x \right) + \sin\left( k \cdot x \right) \cdot i}{k^{2}} \right)\\
\operatorname{Li_{2}}\left( e^{x \cdot i} \right) &= \sum_{k = 1}^{\infty} \left( \frac{\cos\left( k \cdot x \right)}{k^{2}} \right) + \sum_{k = 1}^{\infty} \left( \frac{ \sin\left( k \cdot x \right)}{k^{2}} \right) \cdot i\\
\operatorname{Li_{2}}\left( e^{x \cdot i} \right) &= \operatorname{C_{2}}\left( x \right) + \operatorname{S_{2}}\left( x \right) \cdot i\\
\operatorname{Li_{2}}\left( e^{x \cdot i} \right) &= \frac{1}{6} \cdot \pi^{2} - \frac{1}{2} \cdot \pi \cdot x + \frac{1}{4} \cdot x^{2} + \operatorname{S_{2}}\left( x \right) \cdot i\\
\end{align*}
$$
$$\fbox{$\Re\left( \operatorname{Li_{2}}\left( e^{x \cdot i} \right) \right) = \frac{1}{6} \cdot \pi^{2} - \frac{1}{2} \cdot \pi \cdot x + \frac{1}{4} \cdot x^{2}$}$$
and
$$
\begin{align*}
\operatorname{Li_{2}}\left( x \cdot i \right) &= \sum_{k = 1}^{\infty} \left( \frac{\cos\left( k \cdot x \cdot i \right) + \sin\left( k \cdot x \cdot i \right) \cdot i}{k^{2}} \right)\\
\operatorname{Li_{2}}\left( x \cdot i  \right) &= \frac{1}{4} \cdot \operatorname{Li_{2}}\left( -x^{2} \right) + \operatorname{Ti_{2}}\left( x \right) \cdot i\\
\operatorname{Li_{2}}\left( x \cdot i  \right) &= \frac{1}{4} \cdot \operatorname{Li_{2}}\left( -x^{2} \right) + \frac{1}{4} \cdot \phi\left( -x^{2};\, 2,\, \frac{1}{2} \right) \cdot i\\
\end{align*}
$$
$$\fbox{$\Re\left( \operatorname{Li_{2}}\left( x \cdot i  \right) \right) = \frac{1}{4} \cdot \operatorname{Li_{2}}\left( -x^{2} \right)$}$$

You can write it even more generally:
To find the real part of the dilogarithm, you can use the Bloch-Wigner-Dilogarithm or Bloch–Wigner function $\operatorname{D_{2}}$, which is defined as $\operatorname{D_{2}}\left( z \right) \equiv \Im\left( \operatorname{Li_{2}}\left( z \right) \right) + \arg\left( 1 - z \right) \cdot \ln\left( \left| z \right| \right)$ for $z \in \mathbb{C} \backslash \left\{ 0,\, 1 \right\}$.
$$\fbox{$\begin{align*}
\Re\left( \operatorname{Li_{2}}\left( z \right) \right) &= \operatorname{Li_{2}}\left( z \right)- \left( \operatorname{D_{2}}\left( z \right) - \arg\left( 1 - z \right) \cdot \ln\left( \left| z \right| \right) \right) \cdot i\\
\end{align*}$}$$
Via using Ramakrishnan's equality we can get this relation too:
$$
\begin{align*}
\operatorname{D_{m}}\left( z \right) &\equiv \Re\left[ i^{m + 1} \cdot \left( \sum_{k = 1}^{m} \left( \frac{\left( -\ln\left( \left| z \right| \right) \right)^{m - k}}{\left( m - k \right)!} \cdot \operatorname{Li_{k}}\left( z \right) \right) - \frac{\left( -\ln\left( \left| z \right| \right) \right)^{m}}{2 \cdot m!} \right) \right]\\
\operatorname{D_{2}}\left( z \right) &= \Re\left[ i^{2 + 1} \cdot \left( \sum_{k = 1}^{2} \left( \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2 - k}}{\left( 2 - k \right)!} \cdot \operatorname{Li_{k}}\left( z \right) \right) - \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2}}{2 \cdot 2!} \right) \right]\\
\operatorname{D_{2}}\left( z \right) &= \Re\left[ -i \cdot \left( \sum_{k = 1}^{2} \left( \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2 - k}}{\left( 2 - k \right)!} \cdot \operatorname{Li_{k}}\left( z \right) \right) - \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2}}{2 \cdot 2!} \right) \right]\\
\operatorname{D_{2}}\left( z \right) &= \Re\left[ -i \cdot \left( \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2 - 1}}{\left( 2 - 1 \right)!} \cdot \operatorname{Li_{1}}\left( z \right) + \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2 - 2}}{\left( 2 - 2 \right)!} \cdot \operatorname{Li_{2}}\left( z \right) - \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2}}{2 \cdot 2!} \right) \right]\\
\operatorname{D_{2}}\left( z \right) &= \Re\left[ -i \cdot \left( \ln\left( \left| z \right| \right) \cdot \ln\left( 1 - z \right) + \operatorname{Li_{2}}\left( z \right) - \frac{\ln\left( \left| z \right| \right)}{4} \right) \right]\\
\\
\Im\left( \operatorname{Li_{2}}\left( z \right) \right) &= \Re\left[ -i \cdot \left( \ln\left( \left| z \right| \right) \cdot \ln\left( 1 - z \right) + \operatorname{Li_{2}}\left( z \right) - \frac{\ln\left( \left| z \right| \right)}{4} \right) \right] - \arg\left( 1 - z \right) \cdot \ln\left( \left| z \right| \right)\\
\end{align*}
$$
$$\fbox{$\begin{align*}
\Re\left( \operatorname{Li_{2}}\left( z \right) \right) = ~&\operatorname{Li_{2}}\left( z \right) - (\Re\left[ -i \cdot \left( \ln\left( \left| z \right| \right) \cdot \ln\left( 1 - z \right) + \operatorname{Li_{2}}\left( z \right) - \frac{\ln\left( \left| z \right| \right)}{4} \right) \right]\\
&- \arg\left( 1 - z \right) \cdot \ln\left( \left| z \right| \right) + \arg\left( 1 - z \right) \cdot \ln\left( \left| z \right| \right) ) \cdot i\\
\end{align*}$}$$
But there are no further elementary simplifications. especially not with polynomials of finite degree.
