Obtaining Inverse Fourier Transforms by "FourierTransform" vs. "Integrate" Option

Question

I am trying to calculate the inverse Fourier transform of "1" using two approaches. One using the Integrate option and the other with InverseFourierTransform, so the function looks like the following, with "w" as the frequency.

Integrate[1*Exp[I w t], {w, -Infinity, Infinity}, Assumptions -> {Element[w, Reals], Element[t, Reals]}]

The integral does not converge in this range from -infinity to +infinity. If we use {{w, -1, 1}, then we get sinc function 2sin(t)/t.

And,

InverseFourierTransform [1, w, t]

Gives the desired delta function Sqrt[2 \[Pi]] DiracDelta[t]

Is there a way to obtain Dirac Delta as an answer with the Integrate option?

Answer 1

Here's a non-rigorous way to do it using the claim that $\frac1{2\varepsilon}\int_{-\varepsilon}^{\varepsilon}f(t-a)dt$ converges to $f(a)$. I haven't justified swapping the integrals, but we should have

$$\int_{-\varepsilon}^\varepsilon\int_{-\infty}^\infty e^{iw(t-a)}dwdt=\int_{-\infty}^\infty\int_{-\varepsilon}^\varepsilon e^{iw(t-a)}dtdw$$ Now the best I can do is to get it to work in cases

Integrate[Exp[I ω(t-a)],{ω,-∞,∞},{t,-ε,ε},Assumptions->{Element[ω|t|a,Reals]}]
Integrate[Exp[I ω t]   ,{ω,-∞,∞},{t,-ε,ε},Assumptions->{Element[ω|t  ,Reals]}]

Strangely, if you include the assumption that ε is real in the Integrate, it will produce horrible answers. Anyway, the results of those integrals are

ConditionalExpression[0, ε^2/a^2 < 1]
(2π ε)/Sqrt[ε^2]

so you can evaluate these limits right after the Integrates

Limit[%%/(2ε), ε->0, Direction->"FromAbove"]
Limit[% /(2ε), ε->0, Direction->"FromAbove"]

to find

ConditionalExpression[0, a != 0]
∞

Not as satisfying as the DiracDelta symbol, but something. I find it odd that the ConditionalExpression doesn't also capture the case when a=0, usually it works pretty slick like that.