8
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$\int\limits_{-\infty}^\infty {{e^{i\omega t}}d\omega} = 2\pi \delta \left( t \right)$ 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?

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3
  • 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, 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

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(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]
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2
  • $\begingroup$ 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

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