Calculating an auto-convolution integral by Fourier transforms

Recently I came across a forum contribution [1] where the author describes how to use the unitarity of the Fourier transform on $L^2(\RR)$ to compute the definite integral
\label{eq:I}
I(t_0) \DEF \int_{-\infty}^{\infty}{\frac{\sin(\tau – t_0)}{(\tau – t_0)}\frac{\sin(\tau + t_0)}{(\tau + t_0)}\,d\tau},

where $t_0 \in \RR$ is a constant. In the comments it is also argued that one may alternatively use contour integration. I liked the article, but wondered whether the symmetry of the problem would perhaps admit a simpler approach. Here I give a third method that uses Fourier transforms and convolutions.

Continuity of polynomial roots

It was recently brought up how to show that the $n$ roots of a real or complex polynomial depend continuously on the polynomial’s coefficients. Although I have used this proposition numerous times, implicitly and explicitly, I realized that I never saw a proof of it.

Perhaps the most obvious approach would try to apply the Implicit Function Theorem but, as you may know or can easily check, such an attempt would only work for roots that are simple. Indeed, the very failure of the Implicit Function Theorem in case of non-simple roots is one of the subjects studied in local bifurcation theory. For an example, see this discussion of the Bogdanov-Takens bifurcation.

Returning to the original proposition, here is an elementary proof using only the Fundamental Theorem of Algebra and some simple estimates.