Computing position two-point function and Fourier transform of $p^4 \ln p$

Question

I am computing a two-point correlator in 4D Euclidean space and I am struggling with one particular term. I have found that in momentum space my correlator goes as $$\langle \mathcal{O}(p)\mathcal{O}(q) \rangle \sim \delta(p+q) p^4\ln p.$$ I now want to Fourier transform this to get the correlator in position space. In other words I need to compute the integral $$\langle \mathcal{O}(x)\mathcal{O}(y) \rangle\sim\int d^4 p \; p^4 \ln(p) e^{-ip\cdot(x-y)}.$$

Some hints I have gotten, but haven't brought me far are the following:

1. transform into polar coordinates, $$\int d p \, d\theta \, d\psi \, d\phi \; p^7 \sin \theta \sin^2 \psi \ln(p) e^{-i|p|(|x|-|y|)\cos \theta}.$$ I'm not sure if this is correct because we want to identify the $\theta$ angle as the one between the vectors $p$ and $x$.

2. use a regulator function of the form $e^{-\epsilon p}$ for the 'radial' integral. I am confused here because I have never worked explicitly with these regulators before. I don't know how to compute an integral of the type $$\int dp \; p^7 \ln p e^{-ip x}.$$

3. consider the integral representations of the Bessel functions. I think this means use $$J_0(x) = \frac{1}{2\pi} \int^\pi_{-\pi} e^{ix\sin \phi}d\phi.$$ I don't know how to proceed once we have the Bessel function.

Answer 1

The regulator here should play exactly the same role that it plays in the proof that $\frac{1}{|x - y|^{d - 2}}$ in momentum space is $\frac{1}{p^2}$. I.e. you get $\frac{1}{p^2 + m^2}$ for the Yukawa potential first and then take $m \to 0$. But there is a much faster way to solve this problem if you are not interested in the prefactor. Instead of introducing a scale for regularization, you can just exploit the fact that there is no scale to being with and use dimensional analysis.

Doing the same thing here, \begin{align} \left < \mathcal{O}(x) \mathcal{O}(y) \right > &\sim \int d^4p \; p^4 \log(p) e^{-ip \cdot (x - y)} \\ &= \partial_\alpha \int d^4 p \; p^{4 + \alpha} e^{-ip \cdot (x - y)} \bigl | _{\alpha = 0} \\ &\sim \partial_\alpha \frac{1}{|x - y|^{8 + \alpha}} \bigl |_{\alpha = 0} \\ &= -\frac{\log |x - y|}{|x - y|^8}. \end{align}