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Marc Palm
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This is the picture, I have in mind.

Certainly analytic continuation gives a unique, well-defined value at $0$, if available, and generalizes Jacobi's formula: $$ \frac{\mathrm{d} \det(A)}{\mathrm{d} \alpha} = \det(A) \operatorname{tr}\left(A^{-1} \frac{\mathrm{d} A}{\mathrm{d} \alpha}\right)$$ from the finite dimensional situation.

If you take the circle with the Laplace operator, you actually will get the Riemann zeta function as the spectral zeta function $\zeta_A(s) = \zeta(2s)$.

But the crucial difference between the above spectral zeta functions and the number theoretic ones is that the general zeta function does not admit a functional equation, in fact will not analytic continue to a meromorphic function of all of $\mathbb{C}$. They do analytically continue around a neighborhood of zero, if your eigenvalues do not grow to fast.

They do not grow to fast for pseudo differential operators on compact manifolds by a generalization of Weyl's law.

Marc Palm
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