# Failure of Fubini when integrating in the sense of distributions

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?

• I think that we should start with the expression that led you to this integral. Commented Jul 1 at 19:25

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 ?
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.
• Thanks for the answer. I am aware of these manipulations, but the question is why this is the "distinguished way" to do things. When you replaced the $d^3k$ by some spherical coordinates, a perfectly fine choice would have been to pick $\theta\in [0,\pi/2[$ and $k\in]-\infty,\infty[$ instead. Then you would get instead no principal value. (With a similar way to get the distribution, namely via $\epsilon |k|$.) Commented Jul 2 at 5:11