Using the decomposition $$\frac{e^t\sin x}{1–2e^t\cos x+e^{2t}} =\frac{1}{2i}\left(\frac{1}{1-e^{t+ix}}-\frac{1}{1-e^{t-ix}}\right),$$ we obtain for $\Re(z)>1$ that \begin{align*} \int_0^\infty \frac{t^{z-1}e^t\sin x}{1–2e^t\cos x+e^{2t}}\,dt &=\frac{\Gamma(z)}{2i}\left(\mathrm{Li}_z(e^{ix})-\mathrm{Li}_z(e^{-ix})\right)\\ &=\frac{\Gamma(z)}{2i}\sum_{k=1}^\infty\frac{e^{ikx}-e^{-ikx}}{k^z}\\ &=\Gamma(z)\sum_{k=1}^\infty\frac{\sin(kx)}{k^z}.\end{align*}

Using the decomposition $$\frac{e^t\sin x}{1–2e^t\cos x+e^{2t}} =\frac{1}{2i}\left(\frac{1}{1-e^{t+ix}}-\frac{1}{1-e^{t-ix}}\right),$$ we obtain for $\Re(z)>1$ (using the Bose-Einstein integral) that \begin{align*} \frac{1}{\Gamma(z)}\int_0^\infty \frac{t^{z-1}e^t\sin x}{1–2e^t\cos x+e^{2t}}\,dt &=\frac{1}{2i}\left(\mathrm{Li}_z(e^{ix})-\mathrm{Li}_z(e^{-ix})\right)\\ &=\frac{1}{2i}\sum_{k=1}^\infty\frac{e^{ikx}-e^{-ikx}}{k^z}\\ &=\sum_{k=1}^\infty\frac{\sin(kx)}{k^z}.\end{align*}