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!

This is trivial from the Inversion Theorem for the Fourier transform.
Note that below I leave out all the $\pi$'s and $2\pi$'s; to actually get the right answer you'll need to look up some definitions and insert a few constants.
Note first that $\frac{\sin(x)}x=\hat f(x)$, where $f=\frac12\chi_{[-1,1]}$.
So $\frac{\sin^2(x)}{x^2}=\widehat{f*f}$. Since all our functions are even the inversion theorem shows that the Fourier transform of $\frac{\sin^2(x)}{x^2}$ is $f*f$, which you can easily calculate. (In fact $f*f$ should be a triangle function supported on $[-2,2]$, which agrees with the other answer.)
