documentclass[12pt]{report}

usepackage{amsmath}
usepackage{physics}
usepackage{parskip}

newcommand{set}[1]{left{#1right}}

begin{document}

We wish to rewrite this using
begin{align*}
int d^N psi exp[-frac{1}{2}psi_i A_{ij} psi_j + J_i psi_i] &= (2pi)^{N/2} (det A)^{-1/2} exp[frac{1}{2} J_i A^{-1}_{ij} J_j]\
&=  (2pi)^{N/2}exp[-frac{1}{2}Tr(ln A)] exp[frac{1}{2} J_i A^{-1}_{ij} J_j]
end{align*}
or alternatively
$$int d^N psi exp[-frac{1}{2}psi_i A_{ij} psi_j + J_i psi_i] =  (2pi)^{N/2}exp[-frac{1}{2}Tr(ln A)] exp[frac{1}{2} J_i A^{-1}_{ij} J_j]$$
where we identify $J_i equiv sigma_i$ and $A_{ij}^{-1}/2 equiv K_{ij}^{-1}$ or $A_{ij} = K_{ij}/2$. This gives us
$$Z = (2pi)^{-N/2} exp[frac{1}{2}Tr(ln(K/2))] sum_{set{sigma_i=pm 1}}int d^Npsi exp[-frac{1}{4}psi_i K_{ij} psi_j + sigma_i psi_i] expBig[h_i sigma_iBig]$$
or
$$Z = (2pi)^{-N/2} exp[frac{1}{2}Tr(ln(K/2))] int d^Npsi exp[-frac{1}{4}sum_{ij} psi_i K_{ij} psi_j] sum_{set{sigma_i=pm 1}}  prod_{i} expBig[(h_i +psi_i)sigma_iBig]$$
performing the summation, we have
$$Z = (2pi)^{-N/2} exp[frac{1}{2}Tr(ln(K/2))] int d^Npsi exp[-frac{1}{4}sum_{ij} psi_i K_{ij} psi_j] prod_{i}2, coshBig[(h_i +psi_i)sigma_iBig]$$
After factoring out $N$ factors of $2$ from the product, the leading factor is simply an overall normalization which doesn't affect the physics, call it $mathcal{N}$, so we finally have the desired result
$$Z = mathcal{N} int d^Npsi ,expleft{-left[frac{1}{4}sum_{ij} psi_i K_{ij} psi_j - sum_{i} ln[cosh(h_i +psi_i)]right]right}$$
where
$$mathcal{N} = left(frac{2}{pi}right)^{N/2} exp[frac{1}{2}Tr(ln(K/2))]$$

end{document}

documentclass[12pt]{report}

usepackage{amsmath}
usepackage{physics}
usepackage{parskip}

newcommand{set}[1]{left{#1right}}

begin{document}

We wish to rewrite this using

begin{align*}
int d^N psi exp[-frac{1}{2}psi_i A_{ij} psi_j + J_i psi_i] &= (2pi)^{N/2} (det A)^{-1/2} exp[frac{1}{2} J_i A^{-1}_{ij} J_j]\
&=  (2pi)^{N/2}exp[-frac{1}{2}Tr(ln A)] exp[frac{1}{2} J_i A^{-1}_{ij} J_j]
end{align*}

or alternatively

$$int d^N psi exp[-frac{1}{2}psi_i A_{ij} psi_j + J_i psi_i] =  (2pi)^{N/2}exp[-frac{1}{2}Tr(ln A)] exp[frac{1}{2} J_i A^{-1}_{ij} J_j]$$

where we identify $J_i equiv sigma_i$ and $A_{ij}^{-1}/2 equiv K_{ij}^{-1}$ or $A_{ij} = K_{ij}/2$. This gives us

$$Z = (2pi)^{-N/2} exp[frac{1}{2}Tr(ln(K/2))] sum_{set{sigma_i=pm 1}}int d^Npsi exp[-frac{1}{4}psi_i K_{ij} psi_j + sigma_i psi_i] expBig[h_i sigma_iBig]$$

or

$$Z = (2pi)^{-N/2} exp[frac{1}{2}Tr(ln(K/2))] int d^Npsi exp[-frac{1}{4}sum_{ij} psi_i K_{ij} psi_j] sum_{set{sigma_i=pm 1}}  prod_{i} expBig[(h_i +psi_i)sigma_iBig]$$

performing the summation, we have

$$Z = (2pi)^{-N/2} exp[frac{1}{2}Tr(ln(K/2))] int d^Npsi exp[-frac{1}{4}sum_{ij} psi_i K_{ij} psi_j] prod_{i}2, coshBig[(h_i +psi_i)sigma_iBig]$$

After factoring out $N$ factors of $2$ from the product, the leading factor is simply an overall normalization which doesn't affect the physics, call it $mathcal{N}$, so we finally have the desired result

$$Z = mathcal{N} int d^Npsi ,expleft{-left[frac{1}{4}sum_{ij} psi_i K_{ij} psi_j - sum_{i} ln[cosh(h_i +psi_i)]right]right}$$

where

$$mathcal{N} = left(frac{2}{pi}right)^{N/2} exp[frac{1}{2}Tr(ln(K/2))]$$

end{document}