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
\begin{equation}\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},
\end{equation}
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.

In order to avoid confusion about scaling factors, for definiteness we define the Fourier transform $\mathcal{F} : L^2(\RR) \to L^2(\RR)$ by
\[
\mathcal{F}(\xi) \DEF \int_{-\infty}^{\infty}{f(t) e^{-2\pi i t \xi}\,dt},
\]
we let $f, g : \RR \to \RR$ be the unnormalized and the normalized sinc functions, respectively,
\[
f(t) \DEF \frac{\sin{t}}{t}, \qquad g(t) \DEF f(\pi t) = \frac{\sin{\pi t}}{\pi t},
\]
and we note that these functions are both even. By tables of standard Fourier transforms, we have
\[
(\mathcal{F}f)(\xi) = \pi \RECT{(\pi \xi)},
\]
where the rectangular function appears in the right-hand side. From this it is clear that
\[
(\mathcal{F}f \cdot \mathcal{F}f)(\xi) = \pi^2 \RECT{\pi \xi}.
\]
Therefore, using the convolution theorem and another standard transform – namely, the transform dual to the one used above – we see that
\[
(f \ast f)(t) = \mathcal{F}^{-1}(\mathcal{F}f \cdot \mathcal{F}f)(t) = \pi^2 \frac{1}{\pi} g\left(-\frac{t}{\pi}\right) = \pi f(t)
\]
Returning to the original integral in \eqref{eq:I}, we see that it can be written as
\[
I(t_0) = \int_{-\infty}^{\infty}{f(\tau)f(2t_0 – \tau)\,d\tau} = (f\ast f)(2t_0) = \pi f(2t_0) = \pi \frac{\sin{2 t_0}}{2 t_0}
\]
which equals the result from the original forum contribution.

References

[1] Charles Link, An Integral Result from Parseval’s Theorem, published online in 2017.