The next step consists of the trivial integration in Eq. (A.6) over Fourier components, which yields otherwise. In the last two expressions which give the zero-th order free energy F $_{0}$, we have defined the pseudo-phonon frequencies in the form and an additional parameter for λ$_{1}$ + 2 g = 0 and The calculation of the first order correction to free energy proceeds in a similar way. The usual trick consists in expanding the on-site and inter-site potential in Fourier basis, Then we evaluate the average potentials per particle 〈 V ( x$_{k}$ ) 〉 and 〈 V ( x$_{k}$ ) 〉 in (A.15) by integrating over Fourier components and finally invert the fourier transforms of Eq. (A.16). Consequently, the average value of the potential energies can be expressed in the form of smeared-out potentials, where we have defined two parameters ρ 2 and γ 2 which correspond to lattice displacement and two-point correlation function, that is V$_{ρ}$ ( η ) ≡ 〈 V ( x$_{k}$ ) 〉 = ∫ dy √ 2 πρ $_{2}$e − ( y$_{−}$η ) 2 2 ρ 2 V ( y ) , W$_{γ}$ ≡ 〈 V ( δx$_{k}$ ) 〉 = ∫ dy √ 2 πγ $_{2}$e − y 2 2 γ 2 W ( y ) , (A.17) ρ 2 ≡ 〈 x 2 k 〉 = ℏ 2 Nm ∑ p ω − 1 p coth ( β ℏ ω$_{p}$ 2 ) (A.18) V ( x$_{k}$ ) = ∫ dq 2 π ˜ V ( q ) e $^{iqx$_{k}$}$, W ( δx$_{k}$ ) = ∫ dq 2 π ˜ W ( q ) e $^{iqδx$_{k}$}$. (A.16) 〈 H − H$_{0}$ 〉 = 1 Z$_{0}$ ∫ D x e − S [ x ] { N ∑ k =1 [ V ( x$_{k}$ ) + W ( δx$_{k}$ )] − H$_{0}$ } (A.15) η = − λ$_{2}$ λ$_{1}$ + 2 g . (A.14) mω 2 p = λ$_{1}$ + 2 g − 4 g sin ( pπ N ) (A.13) Z$_{0}$ = N ∏ p =1 sin( pπ/N ) sinh( β ℏ ω$_{p}$/ 2) , (A.11) Z$_{0}$ = e Nβ λ 1 +2 g 2 η 2 N ∏ p =1 1 2 sinh( β ℏ ω$_{p}$/ 2) , (A.12) 16