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

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

• Documentation pages of DiracDelta say: "Integrate never gives DiracDelta as an integral of smooth functions:... FourierTransform can give DiracDelta", there are appropriate examples, as well. – Artes Jan 10 '18 at 0:44
• @Artes Thank you for this information : ) – matheorem Jan 10 '18 at 1:03
• – Michael E2 Dec 27 '18 at 16:23

(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]

• Thank you so much. +1 What about $\int_0^\infty {{e^{i\omega t}}dt} = i{\cal P}\frac{1}{\omega }{\rm{ + }}\pi \delta \left( \omega \right)$? Is it possible? – matheorem Jan 10 '18 at 1:03
• @matheorem See update. I also fixed my usage of FourierTransform. – Carl Woll Jan 10 '18 at 1:21