Radiation field of a scattering experiment moves faster than light
Consider an electron with 4-momentum $p$ which receives a sudden "kick" at $t=0$. After the collision, it has momentum $p^\prime$. Now I am interested in the radiation, that the particle is emitting. According to <PERSON> and <PERSON>, this field is: \begin{equation} A_\rho(x) = \text{Re} \int \frac{d^3k}{(2\pi)^3} \; e^{-ik\cdot x} \frac{-e}{\left|\mathbf{k} \right|} \left(\frac{p^\prime_\rho}{k \cdot p^\prime}- \frac{p_\rho}{k \cdot p} \right) \end{equation} Where $k^0 = \left|\mathbf{k}\right|$ is implicit. I was interested in the potential in real space, so I wanted to evaluate the integral. Using spherical coordinates I obtained: \begin{equation} A_\rho(x) = \int_0^\infty d\left|\mathbf{k}\right| \int_{-1}^1 d\cos \theta \int_0^{2\pi} d\phi \; \cos \left(-\left|\mathbf{k}\right| t +\left| \mathbf{k}\right| \left|\mathbf{x}\right| \cos \theta \right) \left[... \right] \end{equation} Where the expression in the brackets does not depend on $\left|\mathbf{k} \right|$. 402
Since the integrand is symmetric in $\left|\mathbf{k}\right|$, the corresponding integral can be evaluated from $-\infty$ to $\infty$ as well, if an additional factor $1/2$ is multiplied. Then the $\left|\mathbf{k}\right|$-integration should yield a delta distribution $\delta (-t + \left| \mathbf{x} \right| \cos \theta)$. But this implies, that the solution is only non-zero, if $\left|\mathbf{x}\right| > t$, i.e. the solution is outside the lightcone. Hence the field travelled faster than light or the field appears instantanous at $t=0$ in the entire space.
How can I make sense of this? Where did I make a mistake?
The integral can be analyzed further: Let $\mathbf{x} = \left|\mathbf{x}\right| e_z$ and $\mathbf{k} = \left| \mathbf{k} \right| (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$ and $\mathbf{p} = \left| \mathbf{p} \right| (\sin \theta_p \cos \phi_p, \sin \theta_p \sin \phi_p, \cos \theta_p)$. Then the integral becomes (essentially): \begin{equation} A_\rho (x) = \text{Re} \int_{-\infty}^\infty d \left|\mathbf{k}\right| \int_{-1}^1 d\cos \theta \int_0^{2\pi} d\phi \frac{e^{-i\left|\mathbf{k}\right| (x^0 - \left|\mathbf{x}\right| \cos \theta)}}{p^0 - \left| \mathbf{p} \right| (\sin \theta \sin \theta_p \cos\phi - \phi_p + \cos \theta \cos\theta_p)} \end{equation}
As mentioned, the $\left|\mathbf{k}\right|$ integration yields a delta distribution. Then $\cos \theta$ integration can be executed. 402