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I would like to integate this equation:

Integrate[Exp[I*(t1-t2)*ω, {ω, -∞, ∞}]

According to a textbook, I know the answer is $2\pi\delta(t_1 - t_2)$, where $\delta$ is the DiracDelta function. But how can prove this directly in Mathematica?

enter image description here Thank you very much!

march
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user14634
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    Why do you think your target is true? Can you kindly base it? BTW, the so-called $\delta$-function is not a usual function, but a distribution. – user64494 Apr 17 '19 at 15:24
  • FourierTransform[Exp[I (x) w], w, t]. – march Apr 17 '19 at 15:32
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    @user64494 can you give a situation where it isn't true? I.e., can you give a function $f(x)$ that does not satisfy $\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}dy e^{i x y}f(x)= 2\pi f(0)$? Or maybe a restriction on the set to which $f$ must belong for this to be true? – Roman Apr 17 '19 at 16:44
  • Closely related: Teaching Mathematica more about DiracDelta and KroneckerDelta – Jens Apr 17 '19 at 17:28
  • @Roman: Putting $f(x):=1$ , you deal with a divergent double integral. In general, the notation $\int_{-\infty}^\infty f(x)\delta(x),dx$ makes no sense in traditional math. – user64494 Apr 17 '19 at 18:01
  • @user64494 I don't think $f(x)=1$ gives a divergent integral: in polar coordinates, $\int_{-\infty}^{\infty}e^{i x y}dx,dy=\int_0^{\infty}r,dr\int_0^{2\pi}d\phi \exp(\frac12i r^2 \sin(2\phi))=2\pi\int_0^{\infty}r J_0(r^2/2)dr = \lim_{s\to\infty} \pi s^2 \cdot {_1}F_2(\frac12;1,\frac32;-\frac{r^4}{16})=2\pi$ is quite regular. Sorry for the messed-up markup, I don't know how to do it in this forum. – Roman Apr 17 '19 at 18:27
  • @user64494， This eqation is very pupular in quantum physics. You can see the details of this equation in Wikipedia: https://en.wikipedia.org/wiki/Dirac_delta_function – user14634 Apr 18 '19 at 00:16
  • @Jens, Roman, march, user64494, Thank you for your kind help. – user14634 Apr 18 '19 at 00:32
  • @Roman: Upgrade your math. You calculate the iterated integral. However, the double integral does not exist. See https://books.google.by/books?id=D_XBAgAAQBAJ for more info. – user64494 Apr 18 '19 at 03:21
  • @Roman One needs to justify that the double integral is equal to the iterated integral in polar coordinates. If the integral does not diverge, then the order in which the iterated integral is done should not matter, which does not seem to be the case. If the integral is divergent (which I say it is), then various alternatives present themselves; however, Integrate does not avail itself of these alternatives, whereas, FourierTransform uses one, one that yields the desired output. – Michael E2 Apr 19 '19 at 03:39

3 Answers3

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It's a Fourier integral. With those, Mathematica can confidently venture into generalized function territory and yield things like DiracDelta (hazardous in general). However, it doesn't recognize this unless you formulate the integral as a Fourier transform.

InverseFourierTransform[1, \[Omega], t, FourierParameters -> {-1, 1}] /. t -> t1 - t2
(* 2 \[Pi] DiracDelta[t1 - t2] *)
John Doty
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    Thanks a lot for the answer. But this is not what I need. I hope to yield the DiracDelta from Intagration directly, without considering Fourier transformation. Is it possiple to realize this purpose by defining some rules in Intagration? – user14634 Apr 18 '19 at 00:36
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it Diverge look at examples here http://courses.washington.edu/ph227814/228/nb/Green.nb.pdf

Thus you must do it by FourierTransform 

k = t1 - t2;

FourierTransform[Exp[I k w], w, 0, FourierParameters -> {1, 1}]

enter image description here

nufaie
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    Thanks a lot for you kind help. But this is not what I need. I hope to yield the DiracDelta from Intagration directly, without considering Fourier transformation. Is it possiple to realize this purpose by defining some rules in Intagration? – user14634 Apr 18 '19 at 00:38
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Try this simple code which will do exactly what you wanted:

rule = Integrate[Exp[I*t_*x_], {x_, -∞, ∞}] :> 2 Pi DirectDelta[t];
Integrate[Exp[I*(t1-t2)*ω, {ω, -∞, ∞}] /. rule // Quiet

The purpose of Quiet is to ignore the error message from the Integrate which will return an unevaluted Integrate expression and then the replacement rule is to tell Mathematica we want to give the integral the value we want.

Somos
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