It is not possible to diagonalize in the remaining variables, but we can “block-diagonalize” $\bar{k}^{TT}$ and $f^{TT}$ in the following way. Defining $$\hskip 1cm
 A=a\,\bar{k}^{TT}+b\,f^{TT} \qquad{\rm and}\qquad B=c\,\bar{k}^{TT}+f^{TT}$$ we replace the remainder of the TT path-integral expression by $$\hskip 1cm
 A(-\nabla^2+m_A)B+B(-\nabla^2+m_B)B \,.$$ This fixes all variables. Provided that $a$ and $b$ are not simultaneously zero the path-integral over $A$ yields a determinant depending on $m_A$ and a delta function for $B$. We find $m_A=-2/\ell^2+M^2$, where $M^2=M_-^2=M_+^2$ is the mass of the two propagating massive gravitons, thus $$\hskip 1cm
 Z_{\rm mlog}^{TT}= \big[\det(-\nabla^2-\frac{2}{\ell^2})_2^{TT}\big]^{-1/2}
                   \big[\det(-\nabla^2-\frac{2}{\ell^2}+M^2)_2^{TT}\big]^{-1} \,.$$ To obtain the result for the scalar sector we rescale $2\bar{K}=\bar{\sigma}\bar{K}'$, $\varepsilon=\alpha\varepsilon'$ with $\alpha$ as defined earlier, and shift $\bar{h}=h-\bar{K}'/2$, $\delta'=\delta+\ell^2/6\bar{F}$. Then we integrate over $\bar{h}$, $\varepsilon'$ and $\delta'$ to obtain a quadratic expression in $\bar{k}$ and $\bar{F}$. This we “block-diagonalize” again to find exactly (\[ZS\]), provided $\beta\neq-2b_2/\ell^2$. Setting $\beta=-2b_2/\ell^2$, together with equation (\[Mp=Mm\]), implies $$\hskip 1cm
 (b_2,\beta)=(0,0) \qquad{\rm or }\qquad (b_2,\beta)=(-2\sigma\ell^4,4\sigma\ell^2) \,.$$ The first solution is Einstein–Hilbert gravity while the second one is the tricritical point, so we already covered those.

The full one-loop gravity partition function for massive-log PET gravity reads $$\hskip 1cm
 Z_{\rm mlog~PET}^{\rm 1-loop}= Z_{\rm Einstein}\cdot (Z_\mathcal{M})^2 \,,$$ where we defined $\mathcal{M}^2=M^2+1/\ell^2$ and 