The $\operatorname{sinc}$ function is defined by

$\operatorname{sinc}(x) = \frac{\sin (x)}{x}.$

A standard contour integral tells us

$\int _{-\infty }^\infty \operatorname{sinc}(x) \; \mathrm{d}x = \pi .$

Alternatively, we can observe that

$\operatorname{sinc}(x) = \frac{1}{2} \int _{-1}^1 e^{i k t}\; \mathrm{d}k.$

So up to some factors $\operatorname{sinc}x$ is the Fourier transform of the indicator function of $[-1, 1]$. The preceding integral of $\operatorname{sinc}x$ can be thought of as the value of the Fourier transform at $0$. Applying the Fourier inversion formula and carefully keeping track of the coefficients gives us the previous calculation.

David and Jonathan Borwein 1 observed that we also have

\begin{aligned} \int _{-\infty }^\infty \operatorname{sinc}(x) \operatorname{sinc}\left(\frac{x}{3}\right) \; \mathrm{d}x & = \pi ,\\ \int _{-\infty }^\infty \operatorname{sinc}(x) \operatorname{sinc}\left(\frac{x}{3}\right)\operatorname{sinc}\left(\frac{x}{5}\right) \; \mathrm{d}x & = \pi ,\\ \int _{-\infty }^\infty \operatorname{sinc}(x) \operatorname{sinc}\left(\frac{x}{3}\right) \operatorname{sinc}\left(\frac{x}{5}\right) \operatorname{sinc}\left(\frac{x}{7}\right) \; \mathrm{d}x & = \pi . \end{aligned}

This pattern holds up until

$\int _{-\infty }^\infty \operatorname{sinc}(x) \operatorname{sinc}\left(\frac{x}{3}\right) \cdots \operatorname{sinc}\left(\frac{x}{13}\right) \; \mathrm{d}x = \pi .$

Afterwards, we have

$\int _{-\infty }^\infty \operatorname{sinc}(x) \operatorname{sinc}\left(\frac{x}{3}\right) \cdots \operatorname{sinc}\left(\frac{x}{15}\right) \; \mathrm{d}x = \frac{467807924713440738696537864469}{467807924720320453655260875000} \pi .$

As we keep going on, the value continues to decrease.