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


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


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


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?

  • 10
    $\begingroup$ 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. $\endgroup$
    – Artes
    Jan 10, 2018 at 0:44
  • $\begingroup$ @Artes Thank you for this information : ) $\endgroup$
    – matheorem
    Jan 10, 2018 at 1:03
  • $\begingroup$ Duplicate: mathematica.stackexchange.com/questions/110263/… $\endgroup$
    – Michael E2
    Dec 27, 2018 at 16:23

1 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]
  • $\begingroup$ 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? $\endgroup$
    – matheorem
    Jan 10, 2018 at 1:03
  • $\begingroup$ @matheorem See update. I also fixed my usage of FourierTransform. $\endgroup$
    – Carl Woll
    Jan 10, 2018 at 1:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.