that the correlator $\langle\zeta_{\vec k}(\eta)\zeta_{\vec k'}(\eta)\rangle$ shall be of the form $$\langle\zeta_{\vec k}(\eta)\zeta_{\vec k'}(\eta)\rangle = 
 \frac{\delta^3(\vec k + \vec k')}{k^3} \times \left({\rm rest} \right) \; ,$$ where the rest terms have to be dimensionless. In the one loop calculation, the rest terms shall contain UV divergent momentum integral. Thus, apart from a few numerical factors and a factor of $(c_3^2 M^8/\epsilon^4 M_{\rm Pl}^8)$, the rest terms shall be of the form $$\label{eq:loop_SZ_gen}
 \int d^3{\vec k}_1 d^3{\vec k}_2 \delta^3( {\vec k} + {\vec k}_1 + {\vec k}_2 ) 
 f(\eta, H, {\vec k}, {\vec k}_1, {\vec k}_2) \; .$$ Power counting makes it clear that the mass dimension of $f$ is $-3$. Since this integral is dimensionless, naively, we’d expect that on dimensional regularization it would give $$\left( \frac{k}{\mu} \right)^\delta F(\delta) \; ,$$ where $F(\delta)$ is a dimensionless function (which contains poles of $\delta = D-3$). This result shall not leave $\tilde{\cal G}$ invariant under the transformation $x \rightarrow \Lambda x$, $k \rightarrow k/\Lambda$ and $a \rightarrow a/\Lambda$. Thus, it is not correct and a detailed calculation (by Senatore and Zaldarriaga [@SZ2010]) shows that in fact the UV divergent momentum integral gives $$\left( \frac{k}{\mu} \right)^\delta F(\delta) G(\delta,k,-\eta, H) \; ,$$ and where 