How can one prove that $$ (\log\det\cal A=) \operatorname{Tr} \log \cal{A} = \int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr} e^{-s \mathcal{A}},$$ for a sufficiently well-behaved operator $\cal{A}?$ How mathematically rigorous is the expression?

I'm looking at the $d=2$ Euclidean case, as discussed for $\cal{A}=-\Delta + m^2$ in paragraph 32.2.1 of the book *Mirror Symmetry* by Vafa et al.