Sum of trigonometric infinite series I am trying to prove that for any $x\geq 1$ we have:
$$ \sum_{m=1}^{\infty} \frac{\sin\frac{(2m-1)\pi}{x}}{\left(\frac{(2m-1)\pi}{x}\right)^3} = \frac{x}{8}(x-1). $$
Could I have some help please? I am thinking that Fourier series could help, but I found nothing until now. Thank you very much!
 A: Yes.  Fourier series can help.
It is equivalent to finding the limiting function of the Fourier series 
$\sum\limits_{n = 1}^\infty  {\frac{{\sin \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^3}}}} $ .
Note that$\sum\limits_{n = 1}^\infty  {\frac{{\sin \left( {nt} \right)}}{n}} $ converges to $(\pi - t)/2$ for $0 < t < 2 \pi $ .
Observe that 
${\mathop{\rm Im}\nolimits} Log(1 - {e^{i2t}}) =  - 2\sum\limits_{n = 1}^\infty  {\frac{{\sin (2nt)}}{{2n}}} $ .

We can deduce this from
$Log(1 - {z^2}) =  - 2\sum\limits_{n = 1}^\infty  {\frac{{{z^{2n}}}}{{2n}}} $.
Hence $\sum\limits_{n = 1}^\infty  {\frac{{\sin (2nt)}}{{2n}}}  =  - \frac{1}{2}{\mathop{\rm Im}\nolimits} Log(1 - {e^{i2t}}) = \frac{1}{2}\frac{1}{2}(\pi  - 2t) = \frac{1}{4}(\pi  - 2t)$  for $ 0 < t < \pi $.
Therefore,
  $\sum\limits_{n = 1}^\infty  {\frac{{\sin ((2n - 1)t)}}{{2n - 1}}}  = \sum\limits_{n = 1}^\infty  {\frac{{\sin (nt)}}{n}}  - \sum\limits_{n = 1}^\infty  {\frac{{\sin (2nt)}}{{2n}}}  = \frac{1}{2}(\pi  - t) - \frac{1}{4}(\pi  - 2t) = \frac{\pi }{4}$  for $0 < t < \pi$ .
We may integrate the above Fourier series term by term to give the integral of the function on the right: 
$ - \sum\limits_{n = 1}^\infty  {\frac{{\cos \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^2}}}}  + \sum\limits_{n = 1}^\infty  {\frac{1}{{{{(2n - 1)}^2}}}}  = \frac{\pi }{4}t$ I.e.,  
$\sum\limits_{n = 1}^\infty  {\frac{{\cos \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^2}}}}  = \sum\limits_{n = 1}^\infty  {\frac{1}{{{{(2n - 1)}^2}}}}  - \frac{\pi }{4}t = \frac{{{\pi ^2}}}{8} - \frac{\pi }{4}t$ . 
Integrating again gives:
$\sum\limits_{n = 1}^\infty  {\frac{{\sin \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^3}}}}  = \frac{{{\pi ^2}}}{8}t - \frac{\pi }{8}{t^2}$
Now for $x \ge 1$ , $ t = \pi/x \le \pi $.
Substituting this value of $t$ in the above equation gives:  
$\sum\limits_{n = 1}^\infty  {\frac{{\sin \left( {(2n - 1){\textstyle{\pi  \over x}}} \right)}}{{{{(2n - 1)}^3}}}}  = \frac{{{\pi ^3}}}{8}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)$ , 
which is equivalent to your equation for $x \ge 1$.
A: Let $z=\frac{1}{x}$. We want to prove that over the interval $I=(0,1)$ we have:
$$ f(z)=\sum_{m\geq 1}\frac{\sin((2m-1)\pi z)}{(2m-1)^3} = \frac{\pi^3}{8} z(1-z).\tag{1} $$
That is not so difficult, since:
$$-\sum_{m=1}^{M}\frac{\sin((2m-1)\pi z)}{2m-1}\xrightarrow[L^2(0,1)]{}-\frac{\pi}{4},\tag{2} $$
so $f(z)$ is the Fourier series of a second-degree polynomial. 
Since $f(0)=f(1)=0$, that polynomial is a multiple of $x(1-x)$, then $(1)$ follows from:
$$ f\left(\frac{1}{2}\right)=\sum_{m\geq 1}\frac{(-1)^{m+1}}{(2m-1)^3}=\frac{\pi^3}{32}.\tag{3}$$
For a proof of the last identity, see this related question.

Notice that, if we know in advance what the RHS of $(1)$ should be, then it is enough to compute the Fourier series of $x(1-x)$ in $L^2(0,1)$ then check it matches the LHS, way easier.
