Evaluate $\int_0^\infty \frac{t^2 - \sin^2 t}{t^4} dt$ 
Evaluate $\int_0^\infty \frac{t^2 - \sin^2 t}{t^4} dt$.

The integrand is clearly positive as for all $t > 0, t > |\sin t|.$ I'm not sure if finding the Fourier series of some functions might be useful. Or maybe a Taylor expansion or telescoping sum might be useful? $\sin^2 t$ has period $\pi$, so we can split the integral into the sum $\sum_{k=0}^{\infty} \int_{k\pi}^{(k+1)\pi} \frac{t^2 - \sin^2 t}{t^4} dt.$ This sum, even if valid, doesn't seem very useful.

In hindsight, it seems the question was fairly straightforward, provided one can accept the fact that $\int_0^\infty (\sin x)/xdx = \pi/2.$ Many proofs are provided here for instance.


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
$.
