Arovas and Auerbach, in their paper titled "Functional integral theories of low-dimensional quantum Heisenberg models" try to compute the free energy of $SU(N)$ models with the large $N$ limits.

In section 3/4 etc. they integrate out quadratic bosonic fields to obtain a free energy:

$Z_{F}=\int \mathscr{D}[b, \bar{b} ; Q, \bar{Q} ; \lambda] \exp \left(-{\mathcal{J}}_{F}[b, \bar{b} ; Q, \bar{Q} ; \lambda]\right)$ ${\mathcal{J}}_{F}=\int_{0}^{\beta} d \tau\left[\frac{1}{2} \sum_{i, \alpha}\left(\bar{b}_{\alpha i} \dot{b}_{\alpha i}-\dot{\bar{b}}_{\alpha i} b_{\alpha i}\right)+N \sum_{\langle i, j\rangle} \bar{Q}_{i j} Q_{i j}+\underset{\langle i, j\rangle}{\sum}\left(\bar{Q}_{i j} \bar{b}_{\alpha i} b_{\alpha j}+Q_{i j} b_{\alpha i} \bar{b}_{\alpha j}\right)+\sum_{i, \alpha} \lambda_{i}\left(\bar{b}_{\alpha i} b_{\alpha i}-S\right)\right)$

where $b$ is a bosonic creation operator, and $Q$s are just HS transformation parameters introduced to convert a quartic bosonic term into a quadratic term, while $\lambda$ is a lagrange constraint parameter.

And since $\mathcal{Z} = \exp(-\mathcal{N}\beta F)$, we find after treating $Q_{i,j} = Q, \lambda_i = \lambda$ and integrating out the quadratic b terms,

$F^{\mathrm{MF}}=\frac{1}{2} z Q^{2}-S \lambda+\frac{1}{\beta} \int \frac{d^{d} k}{(2 \pi)^{d}} \ln \left(1-e^{-\beta \omega_{\mathrm{k}}}\right)$

Now the first two terms are self evident, but I'm not sure how to carry out the integral to obtain the last term (due to the inability to integrate out the quadratic boson terms). I wanted to know how to do this.


Physics aside, quadratic action is just a Gaussian integral: $$ \int e^{-\frac{1}{2}\sum_{i,j=1}^nA_{ij}x_i x_j + \sum_{i=1}^nB_ix_i}d^nx = \int e^{-\frac{1}{2}\mathbf{x}^T\mathbf{A}\mathbf{x} + \mathbf{B}^T\mathbf{x}}d^nx \sqrt{\frac{(2\pi)^n}{\det \mathbf{A}}}e^{\frac{1}{2}\mathbf{B}^T\mathbf{A}^{-1}\mathbf{B}} $$


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