= Solution
Write the interaction <matrix> as $\mathsf J$ to distinguish it from its scalar strength. The printed pair sum contains both orientations and has no factor $1/2$. For a <positive-definite matrix> $\mathsf J$, completion of the square gives the <Hubbard–Stratonovich transformation>
$$
\int_{\mathbb R^N}d^Nm\,e^{-m^T\mathsf J^{-1}m+2m^T\sigma}
=\pi^{N/2}\sqrt{\det\mathsf J}\,e^{\sigma^T\mathsf J\sigma},
$$
since $m^T\mathsf J^{-1}m-2m^T\sigma=(m-\mathsf J\sigma)^T\mathsf J^{-1}(m-\mathsf J\sigma)-\sigma^T\mathsf J\sigma$. For the exponential kernel with $J>0$ and $\kappa>0$, positivity follows from the positive Fourier symbol derived in part (b), or from its finite open-chain <covariance> <matrix>.
Insert this identity into the <Ising model> partition sum and interchange the finite spin sum with the convergent <Gaussian integral>. The spin variables now factor independently:
$$
\sum_{\sigma_i=\pm1}e^{(2m_i+h)\sigma_i}=2\cosh(2m_i+h).
$$
It follows that
$$
\boxed{\mathcal Z=C\int_{\mathbb R^N}\prod_i dm_i\,
\exp\left[-\sum_{ij}m_i[\mathsf J^{-1}]_{ij}m_j+\sum_i\log(2\cosh(2m_i+h))\right],}
$$
with $C=(\pi^N\det\mathsf J)^{-1/2}$ for the measure used here. The factor two in the spin-field coupling is required by the double pair sum; replacing it by one would decouple a <statistical Hamiltonian> with a different normalization. The diagonal $J_{ii}\sigma_i^2=J$ is a spin-independent energy. Including it as printed makes the positive Gaussian identity directly applicable; if diagonal self-couplings are removed they should first be restored as an irrelevant constant before this real auxiliary-field decoupling.
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