&= 2\delta_I^J\int\frac{d\nu d^{3+d}k}{(2\pi)^{4+d}}\frac{\omega-\nu-(p-k)_i\sigma^i}{k^2((\omega-\nu)^2+(p-k)^2)}~~,\label{eq:SE3bFlat}\\
\mathbf{SE3c} &= -2(\rho^a\rho^{a\dag})_I^J\int\frac{d\nu d^{3+d}k}{(2\pi)^{4+d}}\frac{\omega-\nu+(p-k)_i\sigma^i}{(\nu^2+k^2)((\omega-\nu)^2+(p-k)^2)}~~.\label{eq:SE3cFlat}\end{aligned}$$

For the $Z_1,Z_2$ counterterms we need to extract the coefficients of $p^2,\omega^2$ in the vector and scalar fields self-energies, while for the fermions those would be the coefficients of $\omega,p_i\sigma^i$. In order to obtain the cutoff scheme integrals we can just set $d\rightarrow 0$ and multiply the propagators by the relevant regulator functions. We will start with the dimensional regularization expressions.

### Dimensional regularization in flat space {#sssec:FlatSpaceDimReg}

The coefficients of external momenta can all be expressed by the following integral,

$$\begin{aligned}
\label{eq:Imn}
I_{m,n}(a) &\equiv \int\frac{d\nu d^{3+d}k}{(2\pi)^{4+d}}\frac{\nu^{2m}k^{2n-2m-2}}{(k^2+a^2)(\nu^2+k^2)^n}=\nonumber\\
&=\frac{1}{4\pi^3}\frac{\Gamma(n-m-\frac{1}{2})\Gamma(m+\frac{1}{2})}{\Gamma(n)}\left(\ln\left(\frac{\mu}{a}\right)+\left[\frac{1}{\epsilon}+\frac{1}{2}\ln(\pi)-\frac{\gamma}{2}+1\right]\right).\end{aligned}$$

Here, $\mu$ is the regularization scale and $\epsilon=-d$. The subtraction scheme we use is to set the square brackets term in $I(a)$ to zero. The relevant coefficients of $p,\omega$ in terms of $I_{m,n}$’s , for vector, scalar and fermion self energy diagrams, can be found in appendix \[ssapp:1LoopFlatDR\].

Note that $I_{m,n}(0)$ contains IR divergences, which of course do not exist on $S^3$. To deal with this we can write $I(0)=I(a)+[I(0)-I(a)]$. Since the term in square brackets contains no UV divergences it must match between the two schemes. So we are allowed to use only the IR regulated term $I(a)$ in order to compare between the 