Be $A_{ij}$  an antisymmetric matrix. Then I can easily write
$$
\int \exp\left(-\frac{1}{2}\sum_{i,j}x_i A_{ij} x_j+\sum_{i} B_i x_i\right)\;  d^nx=
\sqrt{(2\pi)^n}\exp\left\{-\frac{1}{2}\mathrm{Tr}\log A\right\}\exp\left\{\frac{1}{2}\vec{B}^{T}A^{-1}\vec{B}\right\}.
$$
and this formula holds both in the finite and inifinite dimensional case.

If I consider a functional integration, the result is usually expressed as
$$
\int \exp\left( - \frac 1 2 \varphi  \hat A  \varphi +J  \varphi \right) D\varphi \; \propto \;
\exp \left( {1\over 2} \int d^4x \; d^4y J\left ( x \right ) D\left ( x - y \right )  J\left( y \right )  \right)
$$
or in some cases as
$$
\int \exp\left( - \frac 1 2 \varphi  \hat A  \varphi +J  \varphi \right) D\varphi \; =\\\mathcal{N}\,\exp\left\{-\frac{1}{2}\mathrm{Tr}\log \hat A \right\}\,
\exp \left( {1\over 2} \int dx \; dx^\prime J\left ( x \right ) A^{-1}\left ( x - x^\prime \right )  J\left( x^\prime \right )  \right)
$$
where $\varphi  \hat A \varphi = \int dx dx^\prime \varphi(x) A(x,x^\prime) \varphi(x^\prime) $ and $J\varphi  = \int dx \varphi(x) J(x) $.

A more complicated situation I have to face now would be
$$
\int D\psi D\varphi 
\exp\left\{ - \frac 1 2 \left[ \varphi  \hat A  \varphi + \psi  \hat B  \psi 
+ \varphi \psi \hat C \psi \varphi \right] \right\} 
$$
my idea in this case would be rewriting it as
$$
\int D\psi 
\exp\left\{ - \frac 1 2 \psi  \hat B  \psi 
\right\} 
\int
D\varphi 
\exp\left\{ - \frac 1 2 \varphi \left[   \hat A   
+ \psi \hat C \psi \right]\varphi \right\} 
$$
in such a way to obtain something similar to 
$$
\int D\psi 
\exp\left\{ - \frac 1 2 \psi  \hat B  \psi  \right\}
\exp \left\{ - \frac{1}{2} \mathrm{Tr} \log \left[   \hat A   
+ \psi \hat C \psi \right]\right\} 
$$
but I don't know how to treat the Trace in this case. Would this procedure be acceptable, and how should I deal with the trace operator?