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\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}{\sin\pars{y} \over
y\root{1 - y^{2}}}\,\dd y}
\\ = &\
\int_{0}^{1}{1 \over
\root{1 - y^{2}}}\
\overbrace{\pars{\int_{0}^{1}
\cos\pars{ky}\,\dd k}}^{\ds{\sin\pars{y} \over y}}\
\,\dd y
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}
{\cos\pars{ky} \over \root{1 - y^{2}}}\,\dd y\,\dd k =
{\pi \over 2}\int_{0}^{1}\on{J}_{0}\pars{k}\,\dd k
\end{align}
where I used a Bessel $\ds{\on{J}_{\nu}}$ Integral Representation.
$\ds{\on{\bf H}_{\nu}}$ is a Struve Function.
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}{\sin\pars{y} \over
y\root{1 - y^{2}}}\,\dd y}
\\[5mm] = &\
{\pi \over 2}\,\on{J}_{0}\pars{1} +
{\pi^{2} \over 4}\on{J}_{1}\pars{1}
\on{\bf H}_{0}\pars{1} -
{\pi^{2} \over 4}
\on{J}_{0}\pars{1}\on{\bf H}_{1}\pars{1}
\\[5mm] = &\
1.44470914981055930772056106554\ldots
\end{align}
The last result is given by Formula $\ds{{\bf 8}.}$,
page $660$, of Table of Integrals, Series and Products ( seventh edition ) by I. S. Gradshteyn and I. M. Ryzhik.
I "guess" those integrations use somehow the generating functions.
Here is a correct solution that you might want to compare with:
\begin{align*}
&\int_{0}^{\infty}\frac{\sin^2 x - x \sin x}{x^3} \, \mathrm{d}x \\
&= \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \int_{\varepsilon}^{R}\frac{\sin^2 x - x \sin x}{x^3} \, \mathrm{d}x \\
&= \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \biggl( \left[ -\frac{\sin^2 x}{2x^2} + \frac{\sin x}{x} \right]_{\varepsilon}^{R} + \int_{\varepsilon}^{R} \left( \frac{\sin(2x)}{2x^2} - \frac{\cos x}{x} \right) \, \mathrm{d}x \biggr) \\
&= -\frac{1}{2} + \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \biggl( \int_{2\varepsilon}^{2R} \frac{\sin x}{x^2} \, \mathrm{d}x - \int_{\varepsilon}^{R} \frac{\cos x}{x} \, \mathrm{d}x \biggr) \\
&= -\frac{1}{2} + \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \biggl( \left[ -\frac{\sin x}{x} \right]_{2\varepsilon}^{2R} + \int_{2\varepsilon}^{2R} \frac{\cos x}{x} \, \mathrm{d}x - \int_{\varepsilon}^{R} \frac{\cos x}{x} \, \mathrm{d}x \biggr) \\
&= \frac{1}{2} + \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \biggl( \int_{R}^{2R} \frac{\cos x}{x} \, \mathrm{d}x - \int_{\varepsilon}^{2\varepsilon} \frac{\cos x}{x} \, \mathrm{d}x \biggr).
\end{align*}
Now by noting that
\begin{align*}
\left|\int_{R}^{2R} \frac{\cos x}{x} \, \mathrm{d}x\right|
&= \left| \frac{\sin R}{R} - \frac{\sin (2R)}{2R} + \int_{R}^{2R} \frac{\sin x}{x^2} \, \mathrm{d}x \right| \\
&\leq \frac{1}{R} + \frac{1}{2R} + \int_{R}^{2R} \frac{1}{x^2} \, \mathrm{d}x \\
&= \frac{2}{R},
\end{align*}
it follows that
$$ \lim_{R\to\infty} \int_{R}^{2R} \frac{\cos x}{x} \, \mathrm{d}x = 0. $$
On the other hand, by substituting $x = \varepsilon u$,
$$ \int_{\varepsilon}^{2\varepsilon} \frac{\cos x}{x} \, \mathrm{d}x
= \int_{1}^{2} \frac{\cos(\varepsilon u)}{u} \, \mathrm{d}u
\xrightarrow{\varepsilon \to 0^+} \int_{1}^{2} \frac{1}{u} \, \mathrm{d}u = \log 2 $$
by the dominated convergence theorem. Therefore
$$ \int_{0}^{\infty}\frac{\sin^2 x - x \sin x}{x^3} \, \mathrm{d}x = \frac{1}{2} - \log 2. $$
Best Answer
Integrate by parts repeatedly to reduce the integral
\begin{align} \int_0^\infty \frac{t^2 - \sin^2 t}{t^4} dt =& \ -\frac13\int_0^\infty ({t^2 - \sin^2 t})\ d(\frac1{t^3})\\ \overset{ibp}=& \ -\frac16\int_0^\infty ({2t- \sin 2t})\ d(\frac1{t^2})\\ \overset{ibp} =& \ -\frac13 \int_0^\infty ({1- \cos 2t})\ d(\frac1{t})\\ \overset{ibp} =& \ \frac23\int_0^\infty \frac{ \sin 2t}t\ dt =\frac23 \cdot \frac \pi2=\frac\pi3 \end{align} where $\int_0^\infty \frac{ \sin 2t}tdt\overset{2t\to t}= \int_0^\infty \frac{ \sin t}tdt=\frac\pi2 $.