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!
On
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 $x \ge 1$.
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.