I want to calculate a functional determinant coming from a Gaussian path integral with operator Matrix $M$. The determinant is given by the product over the eigenvalues according to
$$\text{det}(M) = \prod_{\omega=-\infty}^{\infty} C $$ where $C$ is independent of $\omega$ and also $C>1$. Also note that $\omega$ takes on continous values from $-\infty$ to $\infty$. Although the product is divergent, I read that one can regularize such a product e.g. by using Zeta function regularization, but I am unsure if I did it the right way, especially because I am unsure how to handle the divergent integral which appears in my calculation: I know that for discrete sets, one defines the zeta-function as $\zeta(s) = \sum_n\frac{1}{n^s}$, so my first question is, if I can write the zeta function as
$$ \zeta(s) = \int_{-\infty}^{\infty} d\omega \, C^{-s}$$
If so, I can write $$\zeta(s) = C^{-s}\int_{-\infty}^{\infty} d\omega$$ and now I am unsure if I can regularize this divergent integral somehow. I noticed that it can be written as the fourier transformation of $\delta(0)$ and I was hoping that maybe terms $\sim\delta(0)$ have a well known regularization. If so, I am able to calculate $\zeta(s)$ and then write the determinant as $$\text{det}(M) = e^{-\zeta'(0)}$$ I extend and add to my original question to make it more clear what my problem is in the first place: I am working with a Keldysh path integral in a replica setup. But I think the problem I have can easily be demonstrated with the basic Keldysh path integral. In the free theory (no interactions) the partition sum has the form $$Z = \int D[\Phi^\dagger,\Phi]e^{iS_0[\Phi_c,\Phi_q]} $$
where the action is quadratic and in frequency space given as
$$ S_0 = \int d\omega \Phi^\dagger \underbrace{\begin{pmatrix} 0 & \omega-\omega_0 +i0^+\\ \omega-\omega_0-i0^+ & i0^+\end{pmatrix}}_{G_0^{-1}}\Phi$$ where $\Phi = \begin{pmatrix} \phi_c, \phi_q\end{pmatrix}^\intercal$ (Keldysh basis). Now, in the Keldysh formalism, per construction we have $Z=tr(\rho) = 1$, but I can also calculate the Gaussian integral and obtain $$ Z \sim \frac{1}{(\text{det}(G_0^{-1}))} = -\frac{1}{\prod_{\omega=-\infty}^\infty((\omega-\omega_0)^2+(i0^+)^2)}$$ which is some infinite continous product and therefore not (at least not obvious to me) 1. It gets even worse when I turn on interactions, then my matrix may become something like $$G_0^{-1} \rightarrow G^{-1} = \begin{pmatrix} 0 & \omega-\omega_0+iF\\ \omega-\omega_0-iF & iG\end{pmatrix}$$ with some functions (finite constants) $F$ and $G$. So how do these 2 things: $tr(\rho)=1$ on the one hand and a seemingly divergent determinant on the other hand match?Till now I never thought about how to handle these determinants, because when calculating correlators i can always just devide by Z=1 which cancels the determinant, divergent or not, similar to basic quantum field theory, where one normalizes by the free path integral: $$ \langle \hat{O}\rangle = \frac{\int D[\Phi^\dagger,\Phi]O(\Phi^\dagger,\Phi)e^{iS_0[\Phi^\dagger,\Phi]}}{\int D[\Phi^\dagger,\Phi]e^{iS_0[\Phi^\dagger,\Phi]}} $$
