The limit to show is the following: $$ \lim_{t\to \infty} \int_\mathbb{R}\left|\frac{-\sin x \sin tx}{x^2} \right|dx $$
- A direct splitting of the integral into $\int_{-\infty}^0+\int_0^{+\infty}$ in order to get rid of the absolute values does not seem to be possible, since the denominator is always positive and the sin products behave as shown below: (plotted in WolframAlpha)
- How can one proceed in such circumstances? Are there known identities that help simplify this limit computation?
Note that your integral $I(t)$ is such that (as the function under the integral sign is $\geq0$) $$I(t)\geq F(t)=\int_0^1\frac{|\sin(x)||\sin(tx)|}{x^2}dx$$
Now, if $x\in [0,1]$, we get (by the MVT) that $\sin(x)=x\cos(c)$ for some $c\in ]0,x[$, hence $\sin(x)\geq mx$ with $m=\cos(1)$.
Hence $$F(t)\geq m\int_0^1\frac{|\sin(tx)|}{x}dx$$ Change of variable $tx=u$ in the last integral gives:
$$F(t)\geq m\int_0^t\frac{|\sin(u)|}{u}du$$
As $\displaystyle \int_0^{+\infty}\frac{|\sin(u)|}{u}du=+\infty$, we get that $I(t)\to +\infty$ if $t \to +\infty$.