= Solution
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, gives
$$
\int_{\mathbb R^N}d^Nm\,\exp[-m^T\mathsf J^{-1}m+2m^T\sigma]
=\pi^{N/2}(\det\mathsf J)^{1/2}\exp[\sigma^T\mathsf J\sigma].
$$
Indeed 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 is
$$
\sum_{\sigma_i=\pm1}e^{(2m_i+h)\sigma_i}=2\cosh(2m_i+h).
$$
Consequently
$$
\boxed{Z=C\int_{\mathbb R^N}d^Nm\,
\exp\left[-m^T\mathsf J^{-1}m+\sum_i\log\{2\cosh(2m_i+h)\}\right],\quad
C=\frac1{\pi^{N/2}\sqrt{\det\mathsf J}}.}
$$
The 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 $\langle\sigma_i\rangle=\langle\tanh(2m_i+h)\rangle$ in the auxiliary-field integral.
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