Feynman propagator as a distribution There is something probably simple which is confusing me. In his Quantum Field Theory, Folland defines (or derives) the Feynman propagator as
$$\Delta_F(t, \mathbf{x}) = i \int_{\mathbb{R}^3} \frac{e^{-i \omega_{\mathbf{p}} |t| + i \mathbf{p} \cdot \mathbf{x}}}{2 \omega_{\mathbf{p}}} \frac{d^3 \mathbf{p}}{(2\pi)^3} $$
where as usual $\omega_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2 + m^2}$ for an $m \geq 0$. As he says, this is supposed to be interpreted as a tempered distribution on $\mathbb{R}^4$. My question is: how do we interpret it as such? If I merely take the above formula and integrate against some Schwartz function, as such:
$$\Delta_F(f) = i \int_{\mathbb{R}^4} f(x) \left[ \int_{\mathbb{R}^3} \frac{e^{-i \omega_{\mathbf{p}} |t| + i \mathbf{p} \cdot \mathbf{x}}}{2 \omega_{\mathbf{p}}} \frac{d^3 \mathbf{p}}{(2\pi)^3}\right] d^4 x$$
I don't seem to get something convergent, because for any fixed $x = (t, \mathbf{x})$ the integral in $\mathbf{p}$ diverges. What am I missing here?
 A: I think I get it now. If, for a fixed $t$ and $\mathbf{p}$ we integrate with respect to $d^3 \mathbf{x}$ in
$$\Delta_F(f) = i \int_{\mathbb{R}^4} \int_{\mathbb{R}^3} \frac{e^{-i \omega_{\mathbf{p}} |t|} e^{i \mathbf{p} \cdot \mathbf{x}}}{2 \omega_{\mathbf{p}}} f(x) \frac{d^3 \mathbf{p}}{(2\pi)^3} d^4 x$$
we get
$$\Delta_F(f) = i \int_{-\infty}^{+\infty} \int_{\mathbb{R}^3} \frac{e^{-i \omega_{\mathbf{p}} |t|}}{2 \omega_{\mathbf{p}}} \check{f}(t, \mathbf{p}) d^3 \mathbf{p} dt $$
where $\check{f}(t, \mathbf{p}) = \int_{\mathbb{R}^3} e^{i \mathbf{p} \cdot \mathbf{x}} f(t, \mathbf{x}) d^3 \mathbf{x}/(2\pi)^3$ is the usual Euclidean inverse Fourier transform. This is now an honestly convergent integral because $\check{f}$ is a Schwartz function (since $f$ is). Furthermore note that
$$ \Delta_F(\hat{f}) = i \int_{-\infty}^{+\infty} \int_{\mathbb{R}^3} \frac{e^{-i \omega_{\mathbf{p}} |t|}}{2 \omega_{\mathbf{p}}} \check{\hat{f}}(t, \mathbf{p}) d^3 \mathbf{p} dt = i \int_{\mathbb{R}^4} \frac{e^{-i \omega_{\mathbf{p}} |t|}}{2 \omega_{\mathbf{p}}} f(t, \mathbf{p}) d^3 \mathbf{p} dt$$
due to the Fourier inversion theorem that $\check{\hat{f}} = f$ (here $\hat{f}$ means the Fourier transform with respect only to the spatial variables, similar to $\check{f}$ above). Thus, as a distribution, the (spatial) Fourier transform of $\Delta_F$ is given by integrating against
$$(\mathcal{F}_{\mathbf{x}} \Delta_F)(t, \mathbf{p}) = i\frac{e^{-i \omega_{\mathbf{p}} |t|}}{2 \omega_{\mathbf{p}}}$$
as Folland claims. Thanks to J.G. for pointing me in the right direction.
