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 Peskin and Schroeder, 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 $\left|\mathbf{k} \right|$. 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. (From now on $ \cos \theta = \frac{x^0}{\left|\mathbf{x}\right|}$ is implicit)
What remains is:
\begin{equation}
\frac{1}{\left|\mathbf{x}\right|} \Theta(|\mathbf{x}| - x^0) \int_0^{2\pi} d\phi \; \frac{2\pi}{a - b \cos \theta}
\end{equation}
Where $a= p^0 - \left|\mathbf{p}\right| \cos\theta \cos \theta_p$ and $b = \left|\mathbf{p}\right| \sin\theta \sin\theta_p$. One can easily show that $a>b$. This integral can then be solved using the substitution $z = e^{i\phi}$. Then the integral is a contour integral with $\left|z \right| = 1$ and the poles are at $\frac{a}{b} \pm \sqrt{\left(\frac{a}{b}\right]^2 - 1}$ (here the restriction $a>b$ is important). So only one of the poles lies within the interior of the contour and we obtain the result:
\begin{equation}
= \frac{1}{\left|\mathbf{x}\right|} \Theta (\left|\mathbf{x} \right| - x^0 ) \frac{(2\pi)^2}{\sqrt{a^2 - b^2}}
\end{equation}
So now the entire solution for $t>0$ is:
\begin{equation}
A^\mu = \frac{ep^{\prime \mu}}{4\pi m |\mathbf{x} - \frac{\mathbf{p}^\prime}{m}t|} + \frac{e}{4\pi |\mathbf{x}|} \left( \frac{p^\mu}{\sqrt{a^2 -b^2}} - \frac{p^{\prime \mu}}{\sqrt{a^{\prime 2} - b^{\prime 2}}} \right) \Theta(|\mathbf{x}| - x^0)
\end{equation}
And for $t<0$ the solution is:
\begin{equation}
A^\mu = \frac{ep^{ \mu}}{4\pi m |\mathbf{x}- \frac{\mathbf{p}}{m}t|}
\end{equation}
At time $t = 0$ these solutions should be equal. For this it would be required that $a^2 - b^2 = m^2$. However we imidietly see that:
\begin{equation}
a^2 - b^2 = (p^{0})^2 - \left| \mathbf{p} \right|^2 \sin^2 \theta_p
\end{equation}
So it almost works but the $\sin \theta_p$ destroys it.
I also plotted the solution to get a better overview. [1]: https://i.stack.imgur.com/RmkEI.png
We clearly see that from the second to the third picture the distribution is much slimmer.