Can someone explain why it makes sense that the Fourier transform of the Dirac delta function is 1?

Question

I am learning about Fourier series and Fourier transforms. In class today we worked out the Fourier transform of the delta function and found it was equal to 1.

I don’t know why but I can’t really see why that makes sense. Can someone explain why this is a sensible answer?

Answer 1

The $\delta$ function is the identity for convolution: $$ (\delta * f)(x) = \int_{-\infty}^{+\infty} \delta(y) f(x-y)\;dy= f(x) \tag1$$ Fourier transform converts convolution $*$ to (pointwise)multiplication $\cdot$ like this: $$ \widehat{f * g} = \widehat{f}\cdot\widehat{g} \tag2$$ So for the Fourier transform $\widehat{\delta}$ (whatever it is) has the property $$ \widehat{\delta}\cdot\widehat{g} = \widehat{\delta * g} = \widehat{g}. $$ That is, $\widehat{\delta}$ is the identity for pointwise multiplication. So it must be the constant $1$.

Answer 2

The definition of $\delta$ is $$\int f(t)\,d\delta(t)=f(0).$$So $$\hat\delta(\xi)=\int e^{-it\xi}\,d\delta(t)=e^0=1.$$