Off-shell-ness of the Feynman propagator

I'm looking where exactly, during a construction of the Feynman propagator \(D_F(x_1,x_2)\), a particle goes off-shell.

It is on-shell all the way until before the last step:

\(D_F(x_1,x_2)=\frac i {(2 \pi)^4}\int d^3 k \int d \omega \, e^{-i \vec k (\vec x_1 - \vec x_2)} \frac 1 {\omega^2 - \omega_k^2 + i \epsilon} e^{i \omega (t_1-t_2)}\)

The particle is on-shell here because its 4-momentum is \((\omega_k, \vec k) \), where \(\omega_k^2 = \vec k^2+m^2\).

Integration variable of the first integral is 3-momentum \(\vec k\), where each one of the three component varies from \(- \infty\) to \(\infty\).

Integration variable of the second integral is energy \(\omega\), which also varies from \(- \infty\) to \(\infty\).

Now, we combine these integration variables into a new 4-vector \(k=(\omega,\vec k)\), where each component varies independently from \(- \infty\) to \(\infty\). This 4-vector is not a 4-momentum of anything and thus is off-shell for a simple reason that there is no shell for it to be on. It is just an integration variable:

\(D_F(x_1,x_2)=\frac i {(2 \pi)^4}\int d^4 k \, e^{-i \vec k (\vec x_1 - \vec x_2)} \frac 1 {\omega^2 - \omega_k^2 + i \epsilon} e^{i \omega (t_1-t_2)} = \frac i {(2 \pi)^4}\int d^4 k  \frac {e^{k (x_1 - x_2)}} {\omega^2 - \omega_k^2 + i \epsilon}\)

Being a generic 4-vector, \(k\) satisfies \(\omega^2=k^2+ \vec k^2\). 

Being an on-shell 4-momentum, \((\omega_k, \vec k) \) satisfies \(\omega_k^2 = \vec k^2+m^2\).

Substituting these above, we get the final form of the Feynman propagator:

\(D_F(x_1,x_2)= \frac i {(2 \pi)^4}\int d^4 k \frac {e^{k (x_1 - x_2)}} {k^2 - m^2 + i \epsilon}\)

The particle, which is on-shell, is not explicit in this form. Instead, we have a generic variable \(k\), which is a Lorentz invariant way to package the four integration variables, and which is not a 4-momentum of any particle. Evidently, there are no off-shell particles here.