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Write the finite-chain spin vector as $\sigma$ and the exchange matrix as $\mathsf{J}$. For $J>0$ and $\kappa >0$ this matrix is positive definite, as can be seen from the positive Fourier symbol in part (b), or from the explicit positive open-chain inverse. The linear-source multivariate Gaussian integral, obtained by completing the square, givesIndeed the exponent is $-(m-\mathsf{J}\sigma {)}^T{\mathsf{J}}^{-1}(m-\mathsf{J}\sigma )+{\sigma }^T\mathsf{J}\sigma$. Insert this Hubbard–Stratonovich transformation into the spin partition function. The field term remains $h{\sum }_i{\sigma }_i$, and each independent spin sum isConsequentlyThe interaction convention counts the full ordered pair sum, with no factor of one-half, so the auxiliary coupling is $2m_i{\sigma }_i$. Diagonal terms $J_{ii}{\sigma }_i^2$ only contribute a constant to the spin Hamiltonian; retaining them makes the given positive exchange matrix and normalization convenient. The auxiliary field is not itself the physical spin magnetization: the conditional spin expectation is $\tanh (2m_i+h)$, so $⟨{\sigma }_i⟩=⟨\tanh (2m_i+h)⟩$ in the auxiliary-field integral.

Because $A$ is a real symmetric positive-definite matrix, its inverse exists and all its eigenvalues are positive. Completing the square givesThe translation $v=u-A^{-1}b$ has unit Jacobian determinant. Choose an orthogonal matrix $O$ with $O^{T}AO=\mathrm{diag}(a_1,\dots ,a_n)$, where $a_i>0$. The change $v=Oz$ also has absolute Jacobian one. The remaining integral is the product of $n$ one-dimensional Gaussian integrals, each equal to $\sqrt{2\pi /a_i}$. Consequently the linear-source multivariate Gaussian integral isThe determinant square root is the positive one, as required by convergence of the original real integral.