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

  • 8
    $\begingroup$ 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 '18 at 0:44
  • $\begingroup$ @Artes Thank you for this information : ) $\endgroup$ – matheorem Jan 10 '18 at 1:03
  • $\begingroup$ Duplicate: mathematica.stackexchange.com/questions/110263/… $\endgroup$ – 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]
  • $\begingroup$ 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 '18 at 1:03
  • $\begingroup$ @matheorem See update. I also fixed my usage of FourierTransform. $\endgroup$ – Carl Woll Jan 10 '18 at 1:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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