Deriving 2D Coulomb potential from Born approximation

It is known that one could recover electrostatic Coulomb potential from QED as a nonrelativistic Born approximation. The 3D case could be found in the standard textbook P&S Intro to QFT. This method indeed could (or should can) be generalized to arbitrary spatial dimension $d$. Theories in different dimension share free propagator in common form
$$
G_{00}(q)=frac{i}{q^2+iepsilon}
$$
due to the fundamental quadratic principle in physics. In the nonrelativistic limit,
$$
G_{00}(q)simfrac{i}{-|mathbf q|^2+iepsilon}+mathcal O(|mathbf q|^4).
$$
The Coulomb potential is its spaitial Fourier transformation
$$
intmathrm d^dmathbf q frac{ie^{imathbf qcdotmathbf r}}{-|mathbf q|^2+iepsilon}=intmathrm dqmathrm dOmega frac{iq^{d-1}e^{iqrcostheta}}{-q^2+iepsilon}=frac{1}{r^{d-2}}intmathrm dumathrm dOmega frac{iu^{d-1}e^{iucostheta}}{-u^2+iepsilon’}
$$
Assuming the integral has been properly regularized, it is a constant independent of $r$ so that Coulomb potential $$V(r)=frac C{r^{d-2}}.$$

The problem arises in the case $d=2$. In this case $V(r)$ is a constant, which violates the result from applying Gauss’s law in 2D $$V(r)=Cln(frac rL).$$

So what is wrong here?

—-(update)—-

Inspired by the post mentioned in @Chiral Anormal’s comment, I found the integral could be expressed with help of Bessel functions
$$
intmathrm dqmathrm dthetafrac{qe^{iqrcostheta}}{-q^2+iepsilon}=2piintmathrm d q frac{qJ_0(qr)}{-q^2+iepsilon}=-2pi K_0(-i^{3/2}rsqrt{epsilon})
$$

As is commonly done in QFT the divergence was arranged by $epsilon$ and asymptotically
$$
-2pi K_0(-i^{3/2}rsqrt{epsilon})sim 2pi ln(rsqrt{epsilon})+mathcal O(1),quad epsilonrightarrow 0
$$
as expected. The "characteristic length" $L$ in logarithmic plays a role as the regulator was argued by @Luboš Motl in Coulomb potential in 2D. It partially solves my question. However, still, I am troubled about the consistency of the QED propagator in 1+2d spacetime and the gap in my argument, i.e., how the last integral depends on $r$.