Calculation of effective action

Question

For a massless Dirac particle by integrating fermion degree of freedom in path integral, effective action is resulted for gauge field

$$l(\psi,\bar\psi,A)=\bar\psi( \gamma^\mu (i \partial_\mu +A_\mu ) ) \psi $$

$$Z= \int D\psi D\bar\psi D A_\mu e^{(i \int d^3x l)}$$

$$S_{eff} =\int D\psi D\bar\psi e^{(i \int d^3x l)}$$

$$S_{eff} =-i ln (det ( \gamma^\mu (i \partial_\mu +A_\mu )))$$

I want to know:

How can I calculate the following equation?

$$S_{eff} =C_1 C_2 $$

where

$$C_1=- \frac{1}{12} \epsilon^{\mu\nu\rho} \int \frac{d^3p}{(2\pi)^3} tr[ [G(p)\partial_\mu G^{-1}(p)] [G(p)\partial_\nu G^{-1}(p)] [G(p)\partial_\rho G^{-1}(p)] ] $$

and

$$C_2= \int d^3x \epsilon^{\mu\nu\rho}A_\mu \partial_\nu A_\rho $$

$G(p)$ is fermion propagator and $G^{-1}(p)$ is its inverse.

Answer 1

$S_{eff} = -i\ln\det(\gamma^\mu(i\partial_\mu+A_\mu)) = -i\text{ tr}\ln(\gamma^\mu(i\partial_\mu+A_\mu)) = -i\text{ tr}\ln(i\gamma^\mu \partial_\mu(1+(i\gamma^\mu \partial_\mu)^{-1}\gamma^\mu A_\mu)) = -i\text{ tr}\ln(i\gamma^\mu \partial_\mu) -i\text{ tr}\ln(1+(i\gamma^\mu \partial_\mu)^{-1}\gamma^\mu A_\mu))$

The first term is $-i\ln\det(i\gamma^\mu (-i)p_\mu) = -i\ln\det(\not p)$ is infinite and so is ignored.

The second term can be expanded: $$\ln(1+(i\gamma^\mu \partial_\mu)^{-1}\gamma^\mu A_\mu))=\sum_{k=1}^\infty\dfrac{(-1)^k}k((i\gamma^\mu \partial_\mu)^{-1}\gamma^\mu A_\mu)^k=\sum_{k=1}^\infty\dfrac{(-i)^k}k\not\partial^{-k} \not A^k = \sum_{k=1}^\infty\dfrac{(-i)^k}kG^{k} \not A^k$$