How to Fourier Transform $\frac{\sin(x)^2}{x^2}$ without Contour Integration. In our lecture we need to Fourier transform $\frac{\sin(x)^2}{x^2}$, i.e. compute the integral:
$$\int_{-\infty}^{\infty} \mathrm e^{-iy x}\frac{\sin(x)^2}{x^2} \mathrm dx$$ 
Since it's a lecture on partial differential equations and not complex analysis, I don't think contour integration can be the solution here. I already tried to rewrite $\sin x$ getting 
$$\int_{-\infty}^{\infty}\frac{-1}{4x^2} (\mathrm e^{ix(2-y)}-2\mathrm e^{-ixy}+\mathrm e^{-ix(y+2)})\mathrm dx$$
but now I'm still stuck with the computation of something like $$\int_{-\infty}^{\infty}\frac{\mathrm e^{-ixy}}{x^2}\mathrm dx$$
Any ideas or hints are greatly appreciated. Thanks in advance!

Due to parity it is enough to compute
$$\int_{-\infty}^{+\infty}\frac{\sin^2(x)\cos(xs)}{x^2}\,dx \stackrel{\text{def}}{=}\lim_{M\to +\infty}\int_{-M}^{M}\frac{\sin^2(x)\cos(xs)}{x^2}\,dx $$
and by integration by parts the RHS equals
$$ \lim_{M\to +\infty}\int_{-M}^{M}\frac{\frac{1}{2}\cos(xs)-\frac{1}{4}\cos(x(s+2))-\frac{1}{4}\cos(x(s-2))}{x^2}\,dx $$
or
$$ \lim_{M\to +\infty}\int_{-M}^{M}\frac{\frac{s+2}{4}\sin(x(s+2))+\frac{s-2}{4}\sin(x(s-2))+\frac{s}{2}\sin(xs)}{x}\,dx $$
where we may exploit the standard result
$$ \forall \alpha\in\mathbb{R},\qquad \int_{-\infty}^{+\infty}\frac{\sin(\alpha x)}{x}\,dx = \pi\,\text{Sign}(\alpha) $$
to get:
$$\int_{-\infty}^{+\infty}\frac{\sin^2(x)\cos(xs)}{x^2}\,dx =\frac{\pi}{4}\left[|s-2|+|s+2|-2|s|\right]. $$
The RHS is a piecewise-linear function, supported on $[-2,2]$, going from $0$ to $\pi$ on $[-2,0]$ and from $\pi$ to $0$ on $[0,2]$. Not by chance, it is a multiple of the convolution between $\mathbb{1}_{(-1,1)}$ and itself.
