Inverse Square Law in $D$ dimensions (two cases)

Thank You, AccidentalForierTransform & Sean E. Lake!

(1) To obtain the correct answer for massless carrier one can use Schwinger parametrization and obtain the following expression: $$E(r)=-\frac{2^{D/2-1}}{r^{D-2}}\Gamma\left(\frac{D}{2}-1\right)\frac{1}{2(2\pi)^{D/2}}.$$ (2) Unfortunately, both cases (massive and massless carriers) have "bad behavior" for $D=2$. The gamma function has the pole at $z=0$. To deal with it, one can use dimensional regularization: replace $D\rightarrow D+2\epsilon$. Thus, the integration measure is to change: $$\frac{d^{D}k}{(2\pi)^D}\rightarrow \frac{d^{D+2\epsilon}k}{(2\pi)^{D+2\epsilon}},$$ but with this replacement, one should correct the dimensionality and regularization parameter $\mu$. Finally, the measure has the following form: $$\frac{d^{D+2\epsilon}k}{(2\pi)^{D+2\epsilon}}\mu^{-2\epsilon}.$$ This regularization provides the physically correct answer. The gamma function should be expanded into series: $$\Gamma(\epsilon)\approx\frac{1}{\epsilon}-\gamma.$$ And the fraction $(1/(\mu r))^{\epsilon}$ should be expanded too: $$\left({\mu r}\right)^{-\epsilon}\approx 1 - \ln (\mu r)\epsilon.$$

Considering all the above, the answer is $$E(r)=\frac{1}{2\pi}\ln(\mu r),$$ which has the correct dimensionality (in contrast to $-\ln r/(2\pi)$ which is "unphysical" due to the logarithm of length).

Comments:

1. dimensional regularization does not change the singularity character of the gamma function for $D=2$ because the expansion contains the pole at 0.
2. the Schwinger parametrization is very convinient way to calculate propagator-type because it allows to avoid the charade with hyperpsherical coordinates
3. of course, these tricks are easy for good physicists but I have not found any explanations and solutions for this problem

I'm also reading Zee's book. When attempting this question, I took a shortcut and considered the massless case $(m=0)$ from the get go. Then, I noticed that $E(r)$ is reduced to the Green's function for the $D$-dimensional Laplace equation. It is well known, or by Gauss' Law, one can find that $\nabla E(r) = \frac{1}{S_{D-1}} \propto 1/r^{D-1}$, where $S_{D-1}$ is the surface area of a $D$-dimensional sphere. Thus, $E(r)\propto 1/r^{D-2}$. In the case of $D=2$, $\nabla E(r) \propto 1/r \implies E(r) \propto \ln(r)$.