The sinc function is defined by sinc(x) = sin(x)/x. Philip Woodward introduced the name of the function in 1952, saying it “occurs so often in Fourier analysis and its applications that it does seem to merit some notation of its own.”
Here’s an elegant equation involving the integrals of the sinc function:
When I ran across this recently I wondered two things: How hard is it to compute these two integrals? What are the corresponding results for the jinc function? The jinc function is analogous to sinc, but using a Bessel function in place of sine: jinc(x) = J1(x)/x.
The Fourier transform of the box function, the function box(x) that is 1 on the interval [-1/2, 1/2] and zero everywhere else, is sinc(π ω). (That’s one of the reasons sinc comes up so often in Fourier analysis, as Woodward observed.) So the Fourier transform of sinc(x) is π box(π x). The integral of a function is the value of its Fourier transform at zero, so sinc integrates to π. 
By Plancherel’s theorem, the integral of sinc2(x) is the integral of it’s Fourier transform squared, which equals π.
[There are several conventions for defining the Fourier transform. Here I’m using what I call the (-1, τ, 1) definition in these notes. See that page for other conventions and how to convert between them.]
Now for the jinc function. It also has a simple Fourier transform: f(ω) = 2 √(1 – (2πω)) for |x| < 1/2π and zero otherwise. As above, we can compute the integral of jinc over the real line by evaluating its Fourier transform at 0, which equals 2.
Also as above, the integral of jinc2 is the integral of its Fourier transform squared, which works out to 8/3π.
Update: See the next post for the analogous relations for sums.
More on sinc and jinc functions
 You may have a couple objections to this calculation. I found the Fourier transform of the box function was sinc, then concluded that the transform of sinc is the box function. But applying the Fourier transform twice doesn’t give you the original function back, right? When you transform f(x) twice you get f(-x), but the functions involved here are even, so f(-x) = f(x).
OK, but you may still have another objection: the sinc function does not have bounded L1 norm, so you can’t just take it’s Fourier transform. True, but you can justify the transform in terms of L2 theory or distribution theory.