Failure of Fubini when integrating in the sense of distributions

Question

Similar questions have been asked, but I find it hard to apply it to this case. The following is a physics motivated problem and should illustrate how Fubini fails.

The main question is then: How do I interpret the following integral, since Fubini fails:

$$\int \frac{d^3k}{k^2} e^{i(k-p \cos \theta)|t|},$$

where $\cos\theta$ is the angle of $k$ and $p$. Please read the following section for the illustrated example

We can now try to integrate this using two different spherical coordinates (as I would do any other spherical integral). First, I integrate over $\theta\in[0,\pi]$ and $k\in[0,\infty[$. And the other way is from $\theta\in[0,\pi/2]$ and $k\in]-\infty,\infty[$.

We get in the first case: $$\int_0^\infty dk \frac{2e^{ik|t|}\sin(p|t|)}{p|t|}$$

and in the second: $$\int_{-\infty}^\infty dk e^{ik|t|}\frac{1-e^{-ip|t|}}{ip|t|}$$

Ok, clearly, the second case can be evauated to $\delta(t)$, which then becomes a real quantity since the $ip|t|$ cancels when $t$ goes to 0. But the first case gives something totally different. Beside the $\delta(t)$, I also get (something ill defined, but we can pretend there is some kind of cut-off since the context is physics for now) $\mathcal{P}(1/|t|)$. There are ways to make this expression good, but no matter what, it seems that the two results are vastly different.

So again: How do I interpret the following integral, since Fubini fails: $$\int \frac{d^3k}{k^2} e^{i(k-p \cos \theta)|t|},$$

EDIT: From suggestion, here is the original expression: $$\int \frac{d^3k}{E_kE_{p-k}} e^{i(E_k+E_{p-k})|t|}$$ where $$E_k=\sqrt{k^2+m^2}$$ and $$E_{p-k}=\sqrt{(\vec{k}-\vec{p})^2+m^2}$$

I then extracted the leading order behaviour in $1/k$ which should be interpreted as a distribution. This lead to the expression in the question.

EDIT2: I realized that the expression in the question does not come from the one in the previous edit, since this one has a $|k|$ in the exponent rather than $k$. Then in that case it actually works out to be the same. But maybe take the expression in the question for granted instead?

Answer 1

Let me replace $|t|$ by $t$ alone for the sake of simplicity. We may indeed use spherical coordinates, so that $$ \begin{align} \int_{\Bbb{R}^3}\frac{\mathrm{d}^3\vec{k}}{k^2}\, e^{it(k-p\cos\theta)} &= \int_0^\infty\mathrm{d}k \int_0^\pi\mathrm{d}\theta \int_0^{2\pi}\mathrm{d}\phi\; e^{it(k-p\cos\theta)}\sin\theta \\ &= 2\pi \int_0^\infty\mathrm{d}k\; \left.\frac{e^{it(k-p\cos\theta)}}{ipt}\right|_{\theta=0}^{\theta=\pi} \\ &= 4\pi\operatorname{sinc}(pt) \int_0^\infty e^{itk} \,\mathrm{d}k \end{align} $$ which is what you found $-$ except for the extra $2\pi$-prefactor coming from the integration with respect to the second angular variable. Now, what to do with the remaining radial integral ?

As you noticed, the result is somewhat awkward to determine without invoking ill-defined divergent expressions. Actually, it corresponds to the Fourier transform of the Heaviside function and it is known to be delicate to manipulate due to its inherent distributional nature.

This problem is usually circumvented by considering the Laplace transform instead, since the Fourier transform can be viewed as a limit case of the Laplace transform. The same argument is invoked when the Fourier transform of the Coulomb potential is computed as the limit of a screened analog.

Concretely, a small negative real part is introduced as follows in order to ensure convergence : $$ \int_0^\infty e^{ik(t+i\varepsilon)} \,\mathrm{d}k = \left.\frac{e^{ik(t+i\varepsilon)}}{i(t+i\varepsilon)}\right|_{k=0}^{k=\infty} = \frac{i}{t+i\varepsilon}, $$ with $\varepsilon$ positive $-$ it is replaced by the symbol $0^+$ sometimes. However, even if this formal "move" suggests to conclude by taking the limit $\varepsilon \to 0$ naively (because of the singularity at $t = 0$), it is not possible as such and has to be understood throught Sokhotski-Plemelj theorem / Kramers-Kronig relations, stating $$ \frac{i}{t+i\varepsilon} = \pi\delta(t) + \operatorname{p.v.} \frac{i}{t}, $$ where $\operatorname{p.v.}$ denotes the Cauchy principal value.