# How to take the Fourier transform of the propagator of a vector field?

In the paper Wilson Loops in N=4 Supersymmetric Yang--Mills Theory, the authors give the following generalized Fourier transform for a propagator in $$d=2\omega$$ dimensions:

$$\int \frac{d^{2\omega}p}{(2\pi)^{2\omega}}\frac{e^{ip\cdot x}}{(p^2)^s} = \frac{\Gamma(\omega - s)}{4^s \pi^\omega \Gamma(s)} \frac{1}{(x^2)^{\omega-s}} \tag{1}$$

When the propagator has the form $$1/(p^2)^s$$ in Fourier space (i.e. for a scalar field), this is easy to use. Now what about the propagator of a vector field, which at one loop has the form

$$\frac{\delta_{\mu\nu} - p_\mu p_\nu/p^2}{(p^2)^s}? \tag{2}$$

The first term has the form given in $$(1)$$, but the second one has those annoying $$p_\mu$$ in the numerator. I would think that the integrand is symmetric, since complex exponential is a symmetric function (right?), and thus it is okay to use the replacement

$$p_\mu p_\nu \rightarrow \frac{1}{2\omega} p^2 \delta_{\mu\nu} \tag{3}$$

Then I can use $$(1)$$ in a straightforward way. I have never used that replacement outside of integrals for Feynman diagrams, so I thought I'd check here first. Is that correct?

• $(3)$ is valid when the integrand is spherically symmetric. The factor $e^{ip\cdot x}$ breaks this symmetry by introducing $x/|x|$ as a preferred direction. – MannyC Aug 26 '19 at 1:38

$$\int \frac{d^{2\omega} p}{(2\pi)^{2\omega}} \frac{p_\mu p_\nu}{(p^2)^{s+1}} e^{i p\cdot x} = -\partial_\mu\partial_\nu \int \frac{d^{2\omega} p}{(2\pi)^{2\omega}} \frac{e^{i p\cdot x}}{(p^2)^{s+1}} \tag{4}$$
and then use $$(1)$$ to integrate.