Expressing Feynman propagator with absolute value instead of Heaviside step function?

I recently saw an interesting paper: https://arxiv.org/abs/1912.13435, where the author wrote the Feynman propagator of free massless scalar field as
$$
D_F(x-x’) = -iintlimits_{-infty}^infty frac{domega}{2pi} e^{-iomega (t-t’)}frac{e^{i|omega||vec{x}-vec{x},’|}}{4pi |vec{x}-vec{x},’|}.
tag{1}
$$
(Please see their equation (2.18).)

While they wrote the Wightman propagator
$$
W_0(x-x’)=intlimits_0^infty frac{domega}{2pi} e^{-iomega (t-t’)}frac{textrm{sin}omega|vec{x}-vec{x},’|}{2pi |vec{x}-vec{x},’|}.tag{2}
$$
(This is their equation(2.19).)

Equation (2) is easy to derive, but when I tried to write the equation (1), I only got this
begin{align}
D_F(x-x’) &= theta(t-t’)W_0(x-x’) + theta(t’-t)W_0(x’-x)tag{3}
&=theta(t-t’)intlimits_0^infty frac{domega}{2pi} e^{-iomega (t-t’)}frac{textrm{sin}omega|vec{x}-vec{x},’|}{2pi |vec{x}-vec{x},’|} cr
&+ theta(t’-t)intlimits_0^infty frac{domega}{2pi} e^{-iomega (t’-t)}frac{textrm{sin}omega|vec{x}-vec{x},’|}{2pi |vec{x}-vec{x},’|}tag{4}
&=intlimits_0^infty frac{domega}{2pi} e^{-iomega |t-t’|}frac{textrm{sin}omega|vec{x}-vec{x},’|}{2pi |vec{x}-vec{x},’|}.tag{5}
end{align}
I have tried to separate the sin$omega |vec{x}-vec{x},’|$ into two Exponentials and switch the $omega to -omega$ in the integrand, but it didn’t look similary to (1) at last. I cannot get rid of the absolute value $|t-t’|$ and I also have problem to get the integral from $-infty$ to $infty$.

Honestly, I haven’t seen a Feynman propagator like equation (1) before, all the textbooks I met express it as a piecewise function or the addition of two theta functions as equation (3).

Do you have any idea of how to transform (5) to (1)? All suggestions would be welcome.