= Solution
Work first on the infinite chain, or with a periodized exponential kernel. Put $\rho=e^{-\kappa}$ and use lattice Fourier momentum $q\in[-\pi,\pi]$. Summing the geometric series gives the <exponentially coupled Ising inverse kernel>:
$$
J(q)=J\sum_{n\in\mathbb Z}\rho^{|n|}e^{iqn}
=J\frac{1-\rho^2}{1-2\rho\cos q+\rho^2}
=\frac{J\sinh\kappa}{\cosh\kappa-\cos q}>0.
$$
Thus
$$
J^{-1}(q)=\frac{\cosh\kappa-\cos q}{J\sinh\kappa}
=\frac{\tanh(\kappa/2)}J+\frac{1-\cos q}{J\sinh\kappa}.
$$
Although the original kernel is long-ranged, its inverse contains only a constant and a cosine and therefore is nearest-neighbor. Fourier orthogonality gives
$$
\sum_{ij}m_i[\mathsf J^{-1}]_{ij}m_j
=\frac1{2J\sinh\kappa}\sum_j(m_j-m_{j+1})^2
+\frac{\tanh(\kappa/2)}J\sum_jm_j^2.
$$
Substitution into part (a) yields
$$
\boxed{\mathcal Z=C\int\prod_jdm_j\,e^{-\sum_j[(m_j-m_{j+1})^2/(2J\sinh\kappa)+U(m_j)]},\quad
U(m)=\frac{\tanh(\kappa/2)}Jm^2-\log[2\cosh(2m+h)].}
$$
This exact bulk identity does not require replacing $\sinh\kappa$ by $\kappa$; small $\kappa$ is used later to motivate a smooth continuum field.
For a finite open chain the inverse has off-diagonal entries $-\rho/[J(1-\rho^2)]$, interior diagonal entries $(1+\rho^2)/[J(1-\rho^2)]$ and endpoint diagonal entries $1/[J(1-\rho^2)]$. Accordingly the displayed homogeneous potential acquires the endpoint correction $\rho(m_1^2+m_N^2)/[J(1+\rho)]$. These boundary terms do not change the bulk thermodynamic conclusion. Stating infinite-chain or periodized <boundary conditions> is necessary if the translation-invariant expression is asserted exactly.
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