The inhomogeneous wave equation $$\left(\frac{1}{c^2} \partial_t^2 - \Delta\right)G(\vec{r},t)=\delta^3(\vec{r})\delta(t) \tag{0}$$ for a point-source in three spatial dimensions can be tackled by means of the Fourier-space, that is writing $$G(\vec{r},t)=\frac{1}{(2\pi)^4} \int {\rm d}^3k \int {\rm d}\omega \, \tilde{G}(\vec{k},\omega) \, e^{i\vec{k}\vec{r}-i\omega t} \\ \delta(\vec{r})\delta(t)=\frac{1}{(2\pi)^4}\int {\rm d}^3k \int {\rm d}\omega \, e^{i\vec{k}\vec{r}-i\omega t} $$ and plugging these into (0), which gives $$\left(k^2-\frac{\omega^2}{c^2}\right)\tilde{G}=1$$ and so $$\tilde{G}=\frac{1}{k^2-\omega^2/c^2} + a_1(\vec{k}) \delta(k-\omega/c) + a_2(\vec{k})\delta(k+\omega/c)$$ with arbitrary functions $a_1$ and $a_2$. These terms actually only give the homogenous solution. The first term is responsible for the inhomogeneity. Then, $$G(\vec{r},t)=\frac{c^2}{(2\pi)^4} \int {\rm d}^3k \int {\rm d}\omega \, \frac{1}{k^2c^2-\omega^2} \, e^{i\vec{k}\vec{r}-i\omega t} \tag{1} \, .$$
For $t>0$, the $\omega$-integral, can be calculated by closing the contour in the lower-half complex plane. A physical causality argument then justifies pushing the poles at $\omega=\pm kc$ in the lower-half complex plane and the residue-theorem then gives $$G(\vec{r},t)=\frac{-ic}{16\pi^3} \int \frac{{\rm d}^3k}{k} \, e^{i\vec{k}\vec{r}} \left( e^{ikct} - e^{-ikct} \right) \, . \tag{2}$$ Evaluating this remaining integral is most easily done in 3d-spherical coordinates with the result $$G(\vec{r},t)=\frac{1}{4\pi r} \, \delta(t-r/c) \,. \tag{3}$$ The result has the interpretation, that an instantaneous perturbation at $t=0$ and $\vec{r}=0$, will only have an effect at $r$ after the retarded time $r/c$.
My question is now with regard to the version in 2+1D. The analysis above holds, except (2) becomes $$G(\vec{r},t)=\frac{-ic}{8\pi^2} \int \frac{{\rm d}^2k}{k} \, e^{i\vec{k}\vec{r}} \left( e^{ikct} - e^{-ikct} \right) \, . \tag{4}$$
With polar-coordinates (4) becomes $$G(\vec{r},t)=\frac{-ic}{8\pi^2} \int_0^{2\pi} {\rm d}\varphi \int_0^\infty {\rm d}k \, e^{ikr\cos\varphi} \left( e^{ikct} - e^{-ikct} \right) \, . $$ where wlog $\vec{r}=r \vec{e}_x$ was assumed. The $k$-integral does not converge, but can be continued analytically by introducing a dampening factor $e^{-ks}$ and taking the limit $s\to 0^+$ with the result $$G(\vec{r},t)=\frac{c}{4\pi^2} \int_0^{2\pi} {\rm d}\varphi \, \frac{ct}{c^2t^2-r^2\cos^2t}. $$ If $r>ct$, then the integral is only defined as a principal value and becomes $0$. It is only non-zero for $r<ct$ and can be done via the residue theorem $$G(\vec{r},t)=\frac{1}{2\pi} \frac{\Theta(ct-r)}{\sqrt{t^2-r^2/c^2}} \, . \tag{5}$$ The fact that $r<ct$ is causally plausible, since an instantaneous perturbation at $t=0$ and $\vec{r}=0$ can not have an effect at $r$ if $ct<r$ i.e. the wave hasn't had the time to travel the distance. But contrary to the 3+1D case, the Greens function in the 2+1D case does not appear to exhibit a sharp response i.e. the instantaneous perturbation at $t=0$ and $\vec{r}=0$ does only act at distance $r$ after the the time $r/c$, but the same position at distance $r$ will be affected at later times as well, after the pulse should have passed already.
Is there a mathematical error, or what is the interpration to this?
