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.
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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: $$ \color{#f00}{{2 \over \pi}\sum_{n = 1}^{\infty} \int_{0}^{\infty}{\sin\pars{\bracks{n + 1}x}\sin\pars{nx} \over \pars{xn^{2}}^{2}}\,\dd x} = {2 \over \pi}\sum_{n = 1}^{\infty}{1 \over n^{4}}\pars{{\pi \over 2}\,n} = \sum_{n = 1}^{\infty}{1 \over n^{3}} = \color{#f00}{\zeta\pars{3}} $$