= Solution
A direct way to avoid the singular transform of $1/t^2$ is to invert a <triangular function>. Evenness gives
$$
\frac1{2\pi}\int_{-1}^1\pi(1-|\omega|)e^{i\omega t}\,d\omega=\int_0^1(1-\omega)\cos(\omega t)\,d\omega=\frac{1-\cos t}{t^2}.
$$
At $t=0$ the limit is $1/2$, so there is no singularity in the original signal. Thus
$$
\boxed{\widetilde s(\omega)=\pi(1-|\omega|)_+,\qquad (r)_+=\max(r,0).}
$$
Alternatively, the hint uses the <Hadamard finite-part reciprocal-power distribution>, whose <distributional Fourier transform> is $\mathcal F[\operatorname{fp}(1/t^2)]=-\pi|\omega|$; combining its three translates gives $\pi(|\omega-1|+|\omega+1|-2|\omega|)/2$, the same triangle. Treating $1/t^2$ as an ordinary integrable function would be invalid.
The message spectrum has height $\pi$ at zero, vanishing at $\omega=\pm1$. For $\Omega>1$, the modulated continuous spectrum consists of disjoint triangles of height $\pi/2$, centred at $\pm\Omega$, each with half-width one. In addition there is a <Dirac delta> carrier line of weight $\pi$ at each centre. These lines have specified area, not a finite height; arrows mark them schematically.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1-q4-modulation-spectra.png]
{title=Triangular message spectrum, separated sidebands, and overlapping sidebands with the prescribed carrier-rejecting filter}
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