I'm working through a paper that involves a problem concerning the calculation of the Imaginary part of the derivative of the Hurwitz-Zeta function $\zeta_H(z,a)$ with respect to $z$, evaluated at a certain $-n$ such that $n \in \mathbb{N}_0$. More specifically, at a certain point I am faced with the expression \begin{equation} \Im Z = -\frac{1}{2} \Im\left[\zeta'_H\left(-1, \frac{1}{2} + i(a+b) \right)\right] -\frac{1}{2} \Im \left[\zeta'_H\left(-1, \frac{1}{2} + i(a-b) \right)\right] \end{equation} where $b > a > 0$.
It can be shown that \begin{equation} \Im\left[\zeta'_H \left(-1, \frac{1}{2} + \frac{ix}{2} \right) \right] = \frac{1}{8 \pi} \left[ \text{Li}_2 \left(-e^{\pi x} \right) - \text{Li}_2 \left(-e^{-\pi x} \right) \right] \end{equation} And if I apply this in $\Im Z$, I end up with
\begin{equation} \Im Z = -\frac{1}{16\pi} \left[ \text{Li}_2 \left(-e^{2 \pi (a+b)} \right) - \text{Li}_2 \left(-e^{-2\pi(a+b)} \right) + \text{Li}_2 \left(-e^{2\pi(a-b)} \right) - \text{Li}_2 \left(-e^{-2\pi(a-b)} \right)\right] \end{equation}
However, the result they obtained in the paper is
\begin{equation} \Im Z = \frac{1}{8 \pi} \left[ \text{Li}_2 \left(-e^{-2 \pi (a+b)} \right) - \text{Li}_2 \left(-e^{-2\pi(b-a)} \right)\right] \end{equation}
So I'm wondering if there is any relation between dilogarithms that allows the transformation from what I obtained into the paper's result.