$$\operatorname{Tr}(A) =\int_{\mathbb{R}^4} dx \ A(x,x)$$ together with the matrix trace including the Clifford algebra. By Laplace transformation one has $$S(\mathcal{D}_A) =\int_0^\infty dt \operatorname{Tr}(e^{-t\mathcal{D}^2_A})\tilde{\chi}(t)  \label{Action1}$$ where $\tilde{\chi}$ is the inverse Laplace transform of $\chi(s)$, $$\chi(s) = \int_0^\infty dt e^{-st}\tilde{\chi}(t).$$ The trace in is given by: $$\operatorname{Tr}(e^{-t\mathcal{D}^2_A}) =\int_{\mathbb{R}^4} dx \ \textrm{tr}(e^{-t\mathcal{D}^2_A})(x,x)$$ Assuming the trace of the heat kernel $e^{-t\mathcal{D}^2_A}$ has an asymptotic expansion $$\operatorname{Tr}(e^{-t\mathcal{D}^2_A})= \sum^\infty_{n=-\delta} a_n(\mathcal{D}^2_A)t^{n} \ , \ \delta \in \mathbb{N}  \label{heat}$$ we obtain replacing the previous expansion into $$S(\mathcal{D}_A) =\sum_{n=-\delta}^\infty  a_n(\mathcal{D}^2_A)\int_0^\infty dt \ t^{n} \tilde{\chi}(t) \label{heatd}$$ To compute the integrals we have to consider separately the cases $ n\notin\mathbb{N}$ and $ n\in\mathbb{N}$: $$\chi_n =\Bigg\{
\begin{array}{ll}
\frac{1}{\Gamma(-n)}\int_0^\infty ds \ s^{-n-1} \chi(s) & \textrm{for} \ n\notin\mathbb{N}  \\
(-1)^{n-\delta} \chi^{(n)}(0) & \textrm{for} \ n\in\mathbb{N}  
\end{array}\label{chi}$$ Due to the nature of the $\chi(t)$ function (usually one chose a characteristic function), we can assume $\chi(0)$ much bigger then the derivatives $\chi^{(m)}(0)$ for any $m>0$ appearing in . Consequently in the expansion we will take in account only the finite or singular part for $t\to 0$

Our strategy to compute the action is to use the relation , therefore after explicitly expressed $A_0$ and $A_1$ we proceed to the calculus of the traces and in the 