Integrating $ \int\limits_{-\infty }^\infty {{\mathrm e^{\mathrm i\omega t}}\mathrm d\omega } = 2\pi\, \delta(t) $

Question

$\int\limits_{-\infty}^\infty {{e^{i\omega t}}d\omega} = 2\pi \delta \left( t \right)$ is generally accepted.

But

Integrate[E^(I w t), {w, -\[Infinity], \[Infinity]}, 
 Assumptions -> t \[Element] Reals]

gives

Integrate::idiv: Integral of E^(I t w) does not converge on {-[Infinity],[Infinity]}.

and

Integrate[E^(I w t), {w, -\[Infinity], \[Infinity]}, 
 PrincipalValue -> True, Assumptions -> t \[Element] Reals]

gives 0

So is it possible to get the correct delta function result?

Answer 1

(updated to use FourierTransform correctly)

You could use FourierTransform:

FourierTransform[1, ω, t, FourierParameters->{1,1}]

2 π DiracDelta[t]

To restrict the integration over the positive $t$ axis, include HeavisideTheta:

FourierTransform[HeavisideTheta[t], t, ω, FourierParameters->{1,1}]

I/ω + π DiracDelta[ω]