# Euclidean propagator expression for massless particle

Let $$\Delta_F(\tilde{x})$$ denote the Feynman propagator in the Euclidean variable $$\tilde{x}$$, in $$D$$ dimensions, $$\Delta_F(\tilde{x}) = \int \frac{\text{d}^D\tilde{p}}{(2\pi)^D}\frac{e^{i\,\tilde{p}\cdot\tilde{x}}}{\tilde{p}^2+m^2}.\tag{1}$$

Since this expression is $$\mathrm{O}(D)$$ invariant, one can change variables to spherical coordinates and simplify the expression, yielding $$\Delta_F(\tilde{x}) = \frac{S_{D-2}}{(2\pi)^D}\int_0^{\infty}\int_0^{\pi} \text{d}\tilde{p}\text{d}\theta\,\frac{\tilde{p}^{D-1}}{\tilde{p}^2+m^2}e^{i\,\tilde{p}\cdot\tilde{x}}\, \left(\sin(\theta)\right)^{D-2}.\tag{2}$$

For $$m = 0$$, $$\Delta_F(\tilde{x}) = \frac{S_{D-2}}{(2\pi)^D}\int_0^{\infty}\int_0^{\pi} \text{d}\tilde{p}\text{d}\theta\,\tilde{p}^{D-3}\, \left(\sin(\theta)\right)^{D-2}e^{i\,|\tilde{p}||\tilde{x}|\cos(\theta)}.\tag{3}$$

However, I am supposed to get $$\Delta_F(\tilde{x}) = \frac{1}{(D-2)S_{D-1}}\frac{1}{r^{D-2}}.\tag{4}$$

Any ideas on how one can proceed further?

Edit 2: Using $$u = i\,|\tilde{p}||\tilde{x}|\text{cos}(\theta)$$ as suggested, $$\Delta_F(\tilde{x}) = \frac{S_{D-2}}{(2\pi)^D}\int_0^{\infty}\int_0^{\pi} \text{d}u \text{d}\theta\,(\tan(\theta))^{D-2}\,\frac{u^{D-3}e^{u}}{(ir)^{D-2}}.\tag{5}$$

Am I missing some identity that involves gamma functions?

• Doing the inverse Fourier transform is usually more familiar (from, e.g. the Born approximation for Coulomb scattering). So you might want to try the opposite order.
– Buzz
Nov 4, 2021 at 3:49

Here is one method:

1. Introduce a Schwinger parametrization of $$\frac{1}{\tilde{p}^2+m^2}$$.

2. Do the $$D$$-dimensional Gaussian integral over $$\tilde{p}$$.

3. Case $$m=0$$. Make a substitution in the remaining integral over the Schwinger parameter, so that it turns into a well-known integral representation for the $$\Gamma$$ function. (By the way, the $$r$$-dependence follows from dimensional analysis alone.)

4. Case $$m>0$$. Identify the remaining integral as an integral representation for the modified Bessel function of the second kind.

• This does, indeed, seem to solve the problem. Thank you kindly. Nov 4, 2021 at 14:23

$$e^{ip\cdot x}= e^{i|p|r \cos \theta}$$

• My bad, I forgot I had done that already. Let me edit the question. Nov 3, 2021 at 22:35
• It does provide the what the OP did wrong, as he said. Nov 3, 2021 at 23:49