 Solution

Expand the Grassmann fields in fermionic Matsubara frequencies ${\omega }_l=(2l+1)\pi /\beta$. For $U=0$, the Grassmann Gaussian integral factorizes into thermal fermionic one-level thermal determinants:Each factor counts the empty and occupied states of one spin orbital. Equivalently, the regularized Matsubara sum gives the logarithm of that factor; the raw infinite determinant requires thermal normalization of a fermion determinant. ThereforeThe printed $F$ is the grand potential $\Omega$, because the trace contains $H-\mu \hat{N}$. The Helmholtz free energy at fixed mean particle number is $F_{\mathrm{H}}=\Omega +\mu ⟨\hat{N}⟩$.