 Solution

Specify the Fourier transform before performing the Gaussian integral:At $u=0$ the functional is diagonal in both component and momentum, with inverse kernels $q^2+t+r$, $q^2+t-r$, and $q^2+t$. The condition $0<r<t$ makes all three positive. The covariance of a Gaussian distribution with statistical weight $e^{-F/(k_{B}T)}$ is $k_{B}T$ times the inverse kernel, henceIf $F$ is already the dimensionless free energy divided by $k_{B}T$, the numerators are one instead. Translational invariance forces opposite momenta to pair. There is no complex conjugate on the second field, so the delta function is at $\mathbf{p}=-\mathbf{q}$, not at $\mathbf{p}=\mathbf{q}$. A correlation with $m_j(\mathbf{p}{)}^{\ast }$ would have the latter convention. Rotational symmetry in real space leaves dependence on $q^2$ alone once the momentum-conservation delta function is factored out. The correlation lengths are ${\xi }_1=(t+r{)}^{-1/2}$ and ${\xi }_2=(t-r{)}^{-1/2}$; the second component becomes soft as $r\uparrow t$. The skipped letter in the source is preserved: this solution is part (e), with no inferred part (d).