Proof: The inverse Fourier transform of sinc is
This establishes that the algebraic area under sinc is 1 for every . Every delta function (impulse) must have this property. We now show that sinc also satisfies the sifting property in the limit as . This property fully establishes the limit as a valid impulse. That is, an impulse is any function having the property that
for every continuous function . In the present case, we need to show, specifically, that
Define sinc . Then by the power theorem (§B.9),
Then as , the limit converges to the algebraic area under , which is as desired:
We have thus established that
For related discussion, see [36, p. 127].