On calculations from $\zeta(3)=\frac{2}{\pi}\sum_{n=1}^\infty\int_0^\infty \frac{\sin ((n+1)x)\sin (nx)}{(xn^2)^2}dx$ I was inspired in a formula that I've found in Internet, page 5 of Jameson's notes about Frullani integrals, to ask to Wolfram Alpha this integral $$\int_0^\infty\frac{\sin ax\sin bx}{x^2}dx=\frac{\pi}{4} \left(  \left| a+b \right|- \left| a+b \right|\right), $$
that I presume well known. Then I did the specialisation $a=n+1$, $b=n$ for a fixed integer $n\geq 1$ and after I've multiplied by $\frac{1}{n^4}$ one has if there are no mistakes $$\zeta(3)=\frac{2}{\pi}\sum_{n=1}^\infty\int_0^\infty \frac{\sin ((n+1)x)\sin (nx)}{(xn^2)^2}dx.$$
My goal is learn more mathematics to encourage myself to study more.

Question. It's possible do more interesting calculations with nice mathematical content to deduce some identity from previous approach/identity? Thanks in advance.

My attempt was that I know that it is possible to ask to a CAS by the series $\sum_{n=1}^\infty\frac{\sin ((n+1)x)\sin (nx)}{n^4}$, but I don't know cvery well what's means the result. Also I know that I can do the change $y=xn^2$ of variable in previous integral to get it as $$\int_0^\infty \frac{\sin (\frac{n+1}{n^2}y)\sin (\frac{y}{n})}{(yn)^2}dy.$$
But neither I don't know if such change of variables will be useful.
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
&\color{#f00}{{2 \over \pi}\sum_{n = 1}^{\infty}
\int_{0}^{\infty}{\sin\pars{\bracks{n + 1}x}\sin\pars{nx} \over \pars{xn^{2}}^{2}}\,\dd x}
\\[5mm] = &\
{2 \over \pi}\sum_{n = 1}^{\infty}{1 \over n^{4}}\int_{0}^{\infty}
{\sin\pars{\bracks{n + 1}x} \over x}\,{\sin\pars{nx} \over x}\,\dd x
\label{1}\tag{1}
\end{align}
Albeit the integration is an elementary one, it's useful to know that
David Borwein and Jonathan Borwein set the following identity:
$$
\int_{0}^{\infty}\prod_{k = 0}^{n}{\sin\pars{a_{k}x} \over x}\,\dd x =
{\pi \over 2}\prod_{k = 1}^{n}a_{k}\,,\qquad
a_{k} \in \mathbb{R}\,,k = 0,1,\ldots,n\,,\quad a_{0} \geq \sum_{k = 1}^{n}\verts{a_{k}}
$$
With this identity, \eqref{1} becomes:
$$
\color{#f00}{{2 \over \pi}\sum_{n = 1}^{\infty}
\int_{0}^{\infty}{\sin\pars{\bracks{n + 1}x}\sin\pars{nx} \over \pars{xn^{2}}^{2}}\,\dd x} =
{2 \over \pi}\sum_{n = 1}^{\infty}{1 \over n^{4}}\pars{{\pi \over 2}\,n} =
\sum_{n = 1}^{\infty}{1 \over n^{3}} =
\color{#f00}{\zeta\pars{3}}
$$

By the way; unfortunately, Jonathan Borwein died this month ( $02$-aug-$2016$ ).

