Recently I came across a forum contribution, where the author Charles Link 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 \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.