Closed form of $\sum_{n=1}^\infty (-1)^n\frac{\sin(n\theta)}{n^3}$ for $\theta\in (-\pi,\pi)$ We have to find the closed form of the following series $$\sum_{n=1}^\infty (-1)^n\frac{\sin(n\theta)}{n^3}$$
for $\theta\in (-\pi,\pi)$. We tried to use the following form of the sine $$\sin(n\theta)=\frac{e^{in\theta}-e^{-in\theta}}{2i}$$
but after some calculations it didnt seem to work, because of the $n^3$ in the denominator. So we also tried filling in the power series of the sine, but this also didnt help us much. I think we are missing an important example of a power series. 
Thanks in advance

We have, assuming $\theta \neq 0$, $$\sum_{k\geq1}\left(-1\right)^{k}\frac{\sin\left(k\theta\right)}{k^{3}}=\theta^{3}\sum_{k\geq1}\left(-1\right)^{k}\frac{\sin\left(k\theta\right)}{k^{3}\theta^{3}}
 $$ and we can use the fact that the Mellin transform identity for harmonic sums with base function $g(x)$ is $$\mathfrak{M}\left(\underset{k\geq1}{\sum}\lambda_{k}g\left(\mu_{k}x\right),\, s\right)=\underset{k\geq1}{\sum}\frac{\lambda_{k}}{\mu_{k}^{s}}\, g\left(s\right)^{*}
 $$ where $g\left(s\right)^{*}$ is the Mellin transform of $g\left(x\right)$ . So in this case we have $$\lambda_{k}=\left(-1\right)^{k},\,\mu_{k}=k,\, g\left(x\right)=\frac{\sin\left(x\right)}{x^{3}}
 $$ and so its Mellin transform is $$g\left(s\right)^{*}=\Gamma\left(s-3\right)\sin\left(\frac{1}{2}\pi\left(s-3\right)\right)
 $$and observig that $$\underset{k\geq1}{\sum}\frac{\lambda_{k}}{\mu_{k}^{s}}=\left(2^{1-s}-1\right)\zeta\left(s\right)
 $$ we have $$\theta^{3}\sum_{k\geq1}\left(-1\right)^{k}\frac{\sin\left(k\theta\right)}{k^{3}\theta^{3}}=\frac{\theta^{3}}{2\pi i}\int_{\mathbb{C}}\left(2^{1-s}-1\right)\zeta\left(s\right)\Gamma\left(s-3\right)\sin\left(\frac{1}{2}\pi\left(s-3\right)\right)\theta^{-s}ds=
 $$ $$\frac{\theta^{3}}{2\pi i}\int_{\mathbb{C}}Q\left(s\right)\theta^{-s}ds.
 $$ Note that sine term cancels poles in at odd negative integers, zeta cancels poles at even negative integers and $\left(2^{1-s}-1\right)$ cancels the pole of zeta at $s=1
 $ . So we have poles only at $s=0,2$ and the compute is $$\underset{s=0}{\textrm{Res}}\left(Q\left(s\right)\theta^{-s}\right)=\frac{1}{12}
 $$ $$\underset{s=2}{\textrm{Res}}\left(Q\left(s\right)\theta^{-s}\right)=-\frac{\pi^{2}}{12\theta^{2}}
 $$ and so we have $$\sum_{k\geq1}\left(-1\right)^{k}\frac{\sin\left(k\theta\right)}{k^{3}}=\frac{\theta^{3}}{12}-\frac{\pi^{2}\theta}{12}.
 $$
