Path integral identity I am reading the Background Field Methods in the EPFL Lectures on GR as an EFT. The authors use this identity on Page 23, Equation (174):
$$
\mathcal{N}^{-1}\int\mathcal{D}\phi\,\mathcal{D}\phi^*\exp\{-i\int\mathrm{d}^4x\,\phi(\square+v(x))\phi^*\} = \frac{\mathcal{N}^{-1}}{\det(\square+v(x))}\tag{174}
$$
which is just a standard Gaussian (path) integral. But they go on to use this:
$$
\frac{\mathcal{N}^{-1}}{\det(\square+v(x))} = \mathcal{N}^{-1}\exp\{-\int\mathrm{d}^4x\,\langle x|\text{Tr}\ln(\square+v(x))|x\rangle\}
\tag{174}$$
which I don't fully understand. Does this make use of the $\log \det \mathbf{A} = \text{Tr} \log \mathbf{A}$? If so, why is there an integral in the $\exp$ - shouldn't there be a corresponding delta function to 'cancel' it?
 A: *

*Yes OP is right: The lectures are using Gaussian path integration and the identity
$$ \det A~=~e^{{\rm tr}\ln A}\quad \Leftrightarrow \quad \frac{1}{\det A}~=~e^{-{\rm tr}\ln A}  \tag{i}$$
in eq. (174).


*Next the lectures are using the completeness relation $$\int\mathrm{d}^4x\,|x\rangle  \langle x|~=~{\bf 1} \tag{ii}$$
and
$$ {\rm tr}[A|x\rangle  \langle x|]~=~ \langle x|A|x\rangle\tag{iii} $$
in eq. (174).


*The surviving trace on the RHS of the last eq. (174) could e.g. be a flavor trace, and indicates that the trace in eq. (iii) is only a partial trace.
A: Yes, you are right -the determinant-logarithim identity has been used above. As for the exponential, the following (from the book Field Quantization by Greiner, Reinhardt, chapter 11, page no. 353) has been used - 
$ \int {d^D v \exp \big\{ {- \frac{1}{2} v^T A v} \big\}} = \big(2\pi\big)^{\frac{D}{2}} \exp \big\{ {- \frac {1}{2} Tr ln A} \big\} = \big(2\pi\big)^{\frac{D}{2}}\big(det A \big)^{-\frac{1}{2}} $
The proof of the above is also given therein. The $2\pi$'s are most probably already absorbed in the accompanying constant. The expression given in your question is (most probably) an action integral in Feynman (path integral) formalism and that is why the $-i$ is there - the crux of Feynman formalism is that action can be written as $ \exp \big\{\frac{-i S}{\hbar}\big\} $. Cheers!
