Imaginary part of dilogarithm I have evaluated a certain real-valued, finite integral with no general elementary solution, but which I have been able to prove equals the imaginary part of some dilogarithms and can write in the form
$\Im{\left(\operatorname{Li_2}\left(r\cdot e^{i\theta}\right)\right)}=\tfrac{1}{2i}\left({\operatorname{Li_2}\left(r\cdot e^{i\theta}\right)}-{\operatorname{Li_2}\left(r\cdot e^{-i\theta}\right)}\right)$
similar to the integral in this question and where we can assume $r≥0$ and $0<\theta<\pi$.
Since the input variables are real-valued and the answer is real-valued, I'd prefer a solution that avoids intermediate complex notation while at the same time keeping the underlying integral/series implicit.  I'm aware of some "trivial" solutions:
$\lim\limits_{\theta \to 0^+}\Im{\left(\operatorname{Li_2}\left(r\cdot e^{i\theta}\right)\right)}=\begin{cases}0 & \text{if } r≤1\\ \pi\log{(r)} & \text{if } r>1\end{cases}$
$\Im{\left(\operatorname{Li_2}\left(r\cdot e^{i\pi/2}\right)\right)}=\operatorname{Ti_2}\left(r\right)$
$\Im{\left(\operatorname{Li_2}\left(1\cdot e^{i\theta}\right)\right)}=\operatorname{Cl_2}\left(\theta\right)$
Is it possible to separate $r$ and $\theta$ more generically?  That is to say, I prefer to write the solution in terms like the inverse tangent integral $\operatorname{Ti_2}\left(r\right)$ or the Clausen function $\operatorname{Cl_2}\left(\theta\right)$ where the arguments stay real-valued.  It is also acceptable to use other related functions such as the trigamma function, zeta functions, or even the original dilogarithm with real argument, as long as you don't get more abstract like hypergeometric functions.
If not generically, are there particular values of $\theta$ where this is possible such as at angles that are rational numbers times $\pi$?  $\theta=\pi/3$ or $2\pi/5$ or $\pi/4$ for instance?  I suppose this is my main goal, to get a function of $r$ for certain fixed $\theta$ other than $0$ and $\pi/2$.
If it helps, you can restrict your answer to $0≤r≤1$ or smaller and $0<\theta<\pi/2$.  Can you extend your answer to $\left|\Re{\left(r\cdot e^{i\theta}\right)}\right|≤1$?  Can you extend it all the way to $0≤r≤2$ or more?
See also a related question about the real part.
 A: To find the imaginary 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 $z \in \mathbb{C} \backslash \left\{ 0,\, 1 \right\}$.
$$
\fbox{$\begin{align*}
\Im\left( \operatorname{Li_{2}}\left( z \right) \right) &= \operatorname{D_{2}}\left( z \right) - \arg\left( 1 - z \right) \cdot \ln\left( \left| z \right| \right)\\
\end{align*}$}
$$
The complex-plot of $\Im\left( \operatorname{Li_{2}}\left( z \right) \right)$ from $\Re(z) = -10$ to $\Re(z) = 10$ and $\Im(z) = -10$ to $\Im(z) = 10$:
$Im(Li(z))$" />
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*}
$$
$$
\fbox{$\begin{align*}
\Im\left( \operatorname{Li_{2}}\left( z \right) \right) &= \Re\left[ -i \cdot \left( \ln\left( \left| z \right| \right) \cdot \ln\left( 1 - z \right) + \operatorname{Li_{2}}\left( z \right) - \frac{\ln\left( \left| z \right| \right)}{4} \right) \right] - \arg\left( 1 - z \right) \cdot \ln\left( \left| z \right| \right)\\
\end{align*}$}
$$
A: Kevin's answer wasn't exactly what I wanted but helpfully pointed toward the formula I do want.  We have the following identity in equation 2 of the earlier linked notes
$\operatorname{D}(z)=\frac{1}{2}\left(\operatorname{D}\left(\frac{z}{\bar{z}}\right)+\operatorname{D}\left(\frac{1-1/z}{1-1/\bar{z}}\right)+\operatorname{D}\left(\frac{1/(1-z)}{1/(1-\bar{z})}\right)\right)$
This may not look like an improvement at first, but the real magic is that the function arguments all lie on the unit circle so that we can collapse everything into a sum of three Clausen functions.  With $z=r\cdot e^{i\theta}$ as above, this is
$\begin{align}\Im{\operatorname{Li_2}}\left(r\cdot e^{i\theta}\right)+\arg{\left(1-r\cdot e^{i\theta}\right)}\log{(r)}&=\frac{1}{2}\left(\Im{\operatorname{Li_2}}\left(e^{i2\theta}\right)+\Im{\operatorname{Li_2}}\left(e^{i\arg{\left(\tfrac{e^{-i\theta}-r}{e^{i\theta}-r}\right)}}\right)+\Im{\operatorname{Li_2}}\left(e^{i\arg{\left(\tfrac{1-r\cdot e^{-i\theta}}{1-r\cdot e^{-i\theta}}\right)}}\right)\right) \\ &= \frac{1}{2}\left(\operatorname{Cl_2}\left(2\theta\right)+\operatorname{Cl_2}\left(\arg{\left(\tfrac{e^{-i\theta}-r}{e^{i\theta}-r}\right)}\right)+\operatorname{Cl_2}\left(\arg{\left(\tfrac{1-r\cdot e^{-i\theta}}{1-r\cdot e^{-i\theta}}\right)}\right)\right)\end{align}$
$\begin{align}\Im{\operatorname{Li_2}}\left(r\cdot e^{i\theta}\right) &= \frac{1}{2}\left(\operatorname{Cl_2}\left(2\theta\right)+\operatorname{Cl_2}\left(2\arctan{\left(\frac{\sin(\theta)}{r-\cos{(\theta)}}\right)}\right)+\operatorname{Cl_2}\left(2\arctan{\left(\frac{r\sin(\theta)}{1-r\cos{(\theta)}}\right)}\right)\right) \\ &\qquad +2\arctan{\left(\frac{-1+r\cos{(\theta)}+\sqrt{1+r^2-2r\cos{(\theta)}}}{r\sin{(\theta)}}\right)}\log{(r)} \end{align}$
The singularities are removable.
Let's check the three "trivial" examples.
For the limit $\theta\to 0^+$, all Clausen terms vanish.  The remaining part is more easily written with $z=x+iy$ to get $2\arctan{\left(\frac{-1+x+\sqrt{(-1+x)^2-y^2}}{y}\right)}\log{(r)}$, which has the correct limiting behavior as $y\to 0^+$.
When $\theta=\frac{\pi}{2}$, the first Clausen term vanishes, and what remains is $\frac{1}{2}\left(\operatorname{Cl_2}\left(2\arctan{\left(\frac{1}{r}\right)}\right)+\operatorname{Cl_2}\left(2\arctan{(r)}\right)\right) +2\arctan{\left(\frac{-1+\sqrt{1+r^2}}{r}\right)}\log{(r)}$
Making the substitution $r=\tan{(u)}$ gives a known identity between $\operatorname{Ti_2}$ and $\operatorname{Cl_2}$.
Lastly, when $r=1$, we get $\operatorname{Cl_2}(\theta)=\frac{1}{2}\left(\operatorname{Cl_2}(2\theta)+2\operatorname{Cl_2}(\pi-\theta)\right)$, which is the duplication formula.
