Generalizing the alternating series of the cubes of the reciprocals of odd numbers The series
$$
S_{1/2} := \sum_{k\in\mathbb Z} \frac{(-1)^k}{(2k+1)^3}
$$
can be summed using a rather standard trick, namely by defining a suitable meromorphic function
$$
f(z) := \frac{\csc(\pi z)}{(2z + 1)^3}
$$
and observing that its contour integral on an infinite-radius circle around the origin evaluates to both 0 (by asymptotic analysis) and $ 2\pi \mathrm{i}\left(S_{1/2}/\pi - \pi^2/16\right) $ (by the residue theorem), which entails $ S_{1/2} = \pi^3/16 $.
I am trying to use the same approach to compute a generalized version of that series,
$$
S_\alpha := \sum_{k\in\mathbb Z} \frac{\sin\left[(2k + 1)\pi\alpha\right]}{(2k+1)^3},
$$
which, I have reasons to believe, should be equal to $ S_\alpha = \pi^3\alpha (1-\alpha)/4 $.
I can now proceed in a similar way as before and note that
$$
f_\alpha(z) := \frac{\sin\left[(2z + 1)\pi\alpha\right] \cot(\pi z)}{(2z + 1)^3}
$$
has the property that
$$
\lim_{n\to\infty} \frac{1}{2\pi\mathrm{i}} \oint_{C_n} f_\alpha(z)\,\mathrm{d}z = \frac{S_\alpha}{\pi} + \frac{\pi^2\alpha}{8},
$$
where $ {\{C_n\}}_n $ is a sequence of positive-oriented circles with radius going to infinity without ever intercepting any poles.
The problem is that the left hand side is not 0 anymore (because the integrand is now unbounded in the complex plane), so this does not give any useful information unless I am able to also compute the residue at infinity of $ f_\alpha $.
However, calculating the residue of $ f_\alpha(1/z)/z^2 $ is hard, because the function develops an essential singularity at $ 1/z = 0 $, and the residue itself ends up being expressed as a nasty series—much nastier-looking, in fact, than the one I was trying to compute in the first place. 
I would appreciate any help, whether a simple hint or a full solution, to crack this series. Thank you very much in advance.
 A: We may use the following 

Lemma. $$ g_1(x)=\sum_{n\geq 0}\frac{\sin((2n+1)x)}{2n+1} $$ is the
  Fourier series of a rectangle wave that equals $\frac{\pi}{4}$ over
  $(0,\pi)$ and $-\frac{\pi}{4}$ over $(\pi,2\pi)$.

By integrating twice $g_1(x)$, we get that
$$ g_3(x) = \sum_{n\geq 0}\frac{\sin((2n+1)x)}{(2n+1)^3}$$
is the Fourier series of a piecewise-polynomial function, that is odd and equals $\frac{\pi x(\pi -x)}{8}$ over the interval $(0,\pi)$. That proves your conjecture. I believe you my find interesting to also know that
$$\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{E_{2m}}{2\cdot(2m)!}\left(\frac{\pi}{2}\right)^{2m+1}$$
that is proved here.
A: Using the Fourier series for the Bernoulli polynomials we have 
$$\sum\limits_{k=1}^\infty \frac{\sin(2\pi x k)}{k^3}=\frac{(2\pi)^3 B_3(x)}{2\cdot 3!}$$ 
for $\enspace 0\leq x<1\enspace $ and with the Bernoulli polynomial $\enspace \displaystyle B_3(x)=x^3-\frac{3}{2}x^2+\frac{1}{2}x$ . 
It follows (as assumed): 
$\hspace{1cm}\displaystyle \sum\limits_{k=-\infty}^\infty \frac{\sin((2k+1)\pi x)}{(2k+1)^3}=2\sum\limits_{k=0}^\infty \frac{\sin((2k+1)\pi x)}{(2k+1)^3}= $
$\hspace{5.3cm}\displaystyle =2\left(\frac{(2\pi)^3 B_3(\frac{x}{2})}{2\cdot 3!}-\frac{1}{2^3}\frac{(2\pi)^3 B_3(x)}{2\cdot 3!}\right)=\frac{\pi^3}{4}x(1-x)$

Additional informations:
The general case of the Fourier series for the Bernoulli polynomials is
$$\sum\limits_{k=1}^\infty \frac{\operatorname{Im}(i^{m-1}e^{i2\pi xk})}{k^m}=\frac{(-2\pi)^m B_m(x)}{2\cdot m!}$$
for $m\in\mathbb{N}$, $0\leq x <1$ and $x\ne 0$ for $m=1$ .
In the literature are often used the Euler numbers $E_{2m}$ instead of $-\frac{4^{2m+1}}{2m+1}B_{2m+1}(\frac{1}{4})$ 
which are defined by $\frac{1}{\cosh x}=\sum\limits_{k=0}^\infty\frac{x^k}{k!}E_k$ . 
