be simply $m^2$, so this is a trivial example. On the other hand, while renormalizing the two-point function for $\phi^3$ in $d=6$, we find that the Fourier transform of the two-point function is given by an expression of the form $$\label{eq:2_pt_usual_QFT}
 {\cal G}(k^2) = {\cal G}_{\rm Free}(k^2) + 
 {\cal G}_{\rm Free}(k^2) \left[ \Pi(k^2) \right]  {\cal G}_{\rm Free}(k^2) + \cdots \; ,$$ where ${\cal G}_{\rm Free}(k^2)$ is the Feynman propagator. $\Pi(k^2)$ is the contribution of the one-loop diagram (with external lines amputated) and is given by a quadratically divergent integral and the expression for dimensionally regulated $\Pi(k^2)$ is of the form $$\begin{aligned}
 \label{Pi_usual_QFT}
\Pi(k^2) = A_2\left( \frac{1}{\epsilon} \right) k^2 + A_0 \left( \frac{1}{\epsilon} \right) m^2 
+ {\cal F}(k^2,m,\mu) \; .\end{aligned}$$

The following points are important to notice

1.  $\Pi(k^2)$ has two kinds of contributions: (1) a polynomial in the external momentum $k^2$ (the $m^2$ term can be thought of as the term $(k^2)^0$), (2) another function ${\cal F}$, which depends on, among other things, the fake scale $\mu$.

2.  The coefficients of the polynomial in the external momentum $k^2$, the $A_i$, are divergent, they are functions of $1/\epsilon$, on the other hand, the function ${\cal F}$ is finite.

3.  The forms 