Evaluating FourierTransform like integral manually

I am trying to evaluate

Integrate[x^2*Exp[I k (x - 1)], {k, -∞, ∞}, {x, -∞, ∞}]

Since $$\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ik(x-1)} d{k}$$ is $$\delta(x-1)$$, answer should be $$\int_{-\infty}^{\infty} x^2[2\pi \delta(x-1)] dx =2\pi$$.

However, Mathematica says that the integral does not converge.

• Am I missing something? – Archisman Panigrahi Jun 26 at 10:53
• The integral diverges. Probably better to use FourierTransform in such cases. – Daniel Lichtblau Jun 26 at 15:25
• Integrate doesn't venture into generalized function territory for divergent integrals, so it won't yield an expression with DiracDelta in it. – John Doty Jun 26 at 19:39
• The improper integral under consideration diverges. – user64494 Jun 27 at 5:31

It looks like OP is interested in seeing how the integral can be done without using a black box like FourierTransform function. As mentioned in the comments, when we do not think of this integral as a distribution, it behaves badly and seems to diverge. Therefore, it is useful to regularize the integral by introducing a small parameter ϵ which makes it convergent. For example

integral[k_] = Integrate[Exp[-ϵ k^2 + I k f], k] where Erfi is the imaginary error function. The above result is an anti-derivative, which we can trivially verify to be correct by taking the derivative:

D[integral[k], k] // ExpandAll We also can use that anti-derivative to evaluate the integral over -∞ < k < ∞:

Assuming[Element[{ϵ, f, k}, Reals] && ϵ > 0,
result = Series[integral[k] - integral[-k], {k, ∞, 0}] // Normal
] The Series expansion was not perfect since the exponential function is non-perturbative in powers of k; but it was good enough, since we can now see that all the terms still dependent on k go to zero for k -> ∞. So that actually

result = (E^(-(f^2/(4 ϵ))) Sqrt[π])/Sqrt[ϵ];

Of course, in Mathematica we can arrive at this directly without making reference to any anti-derivative

result = Integrate[Exp[-ϵ k^2 + I k f], {k, -∞, ∞}] You might notice that this result is essentially a limit representation of the DiracDelta function: However, it is not necessary for us to recognize this and substitute anything by hand. We can simply proceed with the second integration by specifying f = x-1 and multiplying by x^2:

Integrate[
x^2 (E^(-((x - 1)^2/(4 ϵ))) (Sqrt[π]))/ Sqrt[ϵ]
, {x, -∞, ∞}] Sure enough, you get the expected result, and can now drop the ϵ regulator.

While ϵ>0 is a requirement for convergence, we can pick it to be as small as we like. Asymptotically, in the limit ϵ->+0 the result becomes indistinguishable from the one where ϵ is actually zero. This is what is implied when we say we "drop" the regulator. Additionally, recall that ϵ was not there initially and was introduced as a tool to be able to calculate. So at the end we certainly are looking for ways to minimize/remove the effect of ϵ to have an equality between initial and final expression where ϵ does not appear, in a consistent manner. That is what we do by "dropping" ϵ as described above.

• Can you kindly base your "can now drop the ϵ regulator"? Pay your attention at Re[[Epsilon]>0]. – user64494 Jun 27 at 5:28
• @user64494 I added some discussion to the answer. – Kagaratsch Jun 27 at 13:01
• (1) Dropping the regulator is actually how one defines the generalized integral (the "usual" integral is divergent). You get an expression that has a limiting value, and define that limit as the generalized integral. (2) I liked (and upvoted) this answer. Just wanted to point out what it is that is being defined (you knew but I'm not sure it was clear). – Daniel Lichtblau Jun 28 at 16:16

After a variable substitition $$x\to x+1$$,

Sqrt[2π]*FourierTransform[(x + 1)^2, x, k]

Sqrt[2π] (Sqrt[2π] DiracDelta[k] - 2 I Sqrt[2π] DiracDelta'[k] - Sqrt[2π] DiracDelta''[k])

Integrate[%, {k, -∞, ∞}]

• The integral Integrate[%, {k, -∞, ∞}] in the above makes no sense (e.g. see en.wikipedia.org/wiki/Dirac_delta_function ). – user64494 Jun 27 at 5:30
• @user64494 Can you please give more details about why you think it makes no sense to integrate the Dirac $\delta$-distribution and its derivatives? From what I know, $\int_{-\infty}^{\infty}\delta(k)dk=1$ is a central property of this distribution, and $\int_{-\infty}^{\infty}\delta'(k)dk=\int_{-\infty}^{\infty}\delta''(k)dk=0$ as well. – Roman Jun 27 at 5:49