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
 A: 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 $$\sum_{k\geq1}\left(-1\right)^{k}\frac{\sin\left(k\theta\right)}{k^{3}}=\frac{\theta^{3}}{12}-\frac{\pi^{2}\theta}{12}.
 $$
A: 
it must be a lot simpler since it was in a test.

I guess that rules out Marco Cantarini's impressive use of the Mellin transform as the expected or intended solution - unless it was a test for a course dealing with the Mellin transform.
More likely, it was expected that one recognises a related Fourier series. Defining
$$f(\theta) := \sum_{n = 1}^\infty \frac{(-1)^n\sin (n\theta)}{n^3},$$
it is easy to see
$$f''(\theta) = \sum_{n = 1}^\infty \frac{(-1)^{n+1}\sin (n\theta)}{n},$$
and that is a Fourier series that one sees occasionally.
We have
$$\int_{-\pi}^\pi \theta \sin (n\theta)\,d\theta = \biggl[-\frac{\theta\cos (n\theta)}{n}\biggr]_{-\pi}^\pi + \frac{1}{n}\int_{-\pi}^\pi \cos (n\theta)\,d\theta = \frac{(-1)^{n+1}2\pi}{n},$$
and so
$$f''(\theta) = \frac{\theta}{2}.$$
Integrating that, it follows that
$$f'(\theta) = \frac{\theta^2}{4} + c,$$
and using the constraint
$$\int_{-\pi}^\pi f'(\theta)\,d\theta = 0$$
from the periodicity of $f$, we find $c = -\frac{\pi^2}{12}$. Then integrating once more, we obtain (this time, the integration constant is easily seen to be $0$)
$$f(\theta) = \frac{\theta^3}{12} - \frac{\pi^2\theta}{12}.$$
If one doesn't recognise the Fourier series of $f''$ [or maybe already that of $f'$], things are more difficult. However, if one has the idea to write
$$\sum_{n = 1}^\infty \frac{(-1)^{n+1}\sin (n\theta)}{n} = \operatorname{Im} \sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n} e^{in\theta},$$
which in the context of Fourier transforms doesn't seem a too far-fetched idea, one may recognise the Taylor series of
$$\log \bigl(1 + e^{i\theta}\bigr)$$
in that - for an attempt to find the value, it is not necessary to go into the details of convergence; if one has found the candidate, one can verify it in many ways. Now
$$1 + e^{i\theta} = \bigl(e^{i\theta/2} + e^{-i\theta/2}\bigr)e^{i\theta/2} = 2\cos \frac{\theta}{2}\cdot e^{i\theta/2},$$
so $\operatorname{Im} \log \bigl(1 + e^{i\theta}) = \frac{\theta}{2} + 2\pi k(\theta)$, and it's not hard to see that $k(\theta) \equiv 0$ is the correct choice.
If one feels at home with distributions, one can recognise the Fourier series of $f'''$ as that of $\frac{1}{2} - \pi \delta_\pi$, but that is probably also beyond what can be expected. 
A: Combining Marco's solution with the polylog solution ...
$$
\text{Im}\;\text{Li}_3(-e^{it})= \frac{t^3-\pi^2 t}{12},
\qquad -\pi < t < \pi
$$
But of course everyone knows that!
A: This looks like the Fourier series of an odd function, and it converges uniformly, since
$$
 \sum_{n=1}^\infty \Bigg|(-1)^n \frac{\sin(n\theta)}{n^3} \Bigg|\leq \sum_{n=1}^\infty \frac{1}{n^3}.
$$
You can research a bit through the literature and look for some text in which the exact value of the series 
$$\sum_{n=1}^\infty \frac{1}{n^6}
$$
is computed (if you know about Parseval's theorem you will understand why), and the function you are looking for will most probably appear in that text. 
Also, just as a matter of language, such series are not called power series, but trigonometric series (due to the sine). Power series have the form $\sum a_nz^n$.
