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: Had forgotten to add some extra steps.

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?

  • $\begingroup$ 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. $\endgroup$
    – Buzz
    Nov 4, 2021 at 3:49

2 Answers 2


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.

  • $\begingroup$ This does, indeed, seem to solve the problem. Thank you kindly. $\endgroup$
    – miniplanck
    Nov 4, 2021 at 14:23

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

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

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