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Here are some suggestions, until somebody more knowledgeable appears.

1. Why does this definition make sense: Analytic continuation is unique, and hence if there is analytic continuation to neighbourhood of $0$, and the formula is valid in finite dimension, then why not take it. Unfortunately, we have in general that $$ \det (AB) \neq \det(A) \det(B).$$

2. Why does it possibly generalize the theory of the Riemann zeta function: If you take the circle with the Laplace operator $-\frac{\partial^2}{\partial^2 x}$ on $L^2[0,1]$ has eigenfunctions $\exp(2 \pi \textup{i} nx)$ for $n \in \mathbb{Z}$, you actually will get the Riemann zeta function as the spectral zeta function $\zeta_A(s) = 2 \zeta(2s)$.

3. Why does it not: Certain properties are not shared by the general spectral zeta function. There is in general no functional equation relating $\zeta_A(s)$ to $\zeta_A( D-s)$ for some $D$, and there is no Euler product.

But here is also a quote from page 20, of "Ten physical applications of spectral zeta functions" from Elizalde, pg.20

Actually, a universal definition for the determinant [...] is still missing.

Also Singer and Ray refer in http://www.sciencedirect.com/science/article/pii/0001870871900454 to Seeley http://www.ams.org/mathscinet/search/publications.html?pg1=IID&s1=157950 , who has proven the regularity at $0$. He has a nice formula there:

$$A^s=i/2π\int_{Γ} λ^s(A−λ)^{−1}dλ$$

Here are some suggestions, until somebody more knowledgeable appears.

1. Why does this definition make sense: Analytic continuation is unique, and hence if there is analytic continuation to neighbourhood of $0$, and the formula is valid in finite dimension, then why not take it. Unfortunately, we have in general that $$ \det (AB) \neq \det(A) \det(B).$$

2. Why does it possibly generalize the theory of the Riemann zeta function: If you take the circle with the Laplace operator $-\frac{\partial^2}{\partial^2 x}$ on $L^2[0,1]$ has eigenfunctions $\exp(2 \pi inx)$ for $n \in \mathbb{Z}$, you actually will get the Riemann zeta function as the spectral zeta function $\zeta_A(s) = 2 \zeta(2s)$.

3. Why does it not: Certain properties are not shared by the general spectral zeta function. There is in general no functional equation relating $\zeta_A(s)$ to $\zeta_A( D-s)$ for some $D$, and there is no Euler product.

But here is also a quote from page 20, of "Ten physical applications of spectral zeta functions" from Elizalde, pg.20

Actually, a universal definition for the determinant [...] is still missing.

Also Singer and Ray refer in http://www.sciencedirect.com/science/article/pii/0001870871900454 to Seeley http://www.ams.org/mathscinet/search/publications.html?pg1=IID&s1=157950 , who has proven the regularity at $0$. He has a nice formula there:

$$A^s=i/2π\int_{Γ} λ^s(A−λ)^{−1}dλ$$

Here are some suggestions, until somebody more knowledgeable appears.

1. Why does this definition make sense: Analytic continuation is unique, and hence if there is analytic continuation to neighbourhood of $0$, and the formula is valid in finite dimension, then why not take it. Unfortunately, we have in general that $$ \det (AB) \neq \det(A) \det(B).$$

2. Why does it possibly generalize the theory of the Riemann zeta function: If you take the circle with the Laplace operator $\frac{\partial^2}{\partial^2 x^2}$ on $L^2[0,2 \pi]$ has eigenfunctions $\exp(nx)$ for $n \in \mathbb{Z}$, you actually will get the Riemann zeta function as the spectral zeta function $\zeta_A(s) = 2 \zeta(2s)$.

3. Why does it not: Certain properties are not shared by the general spectral zeta function. There is in general no functional equation relating $\zeta_A(s)$ to $\zeta_A( D-s)$ for some $D$, and there is no Euler product.

But here is also a quote from page 20, of "Ten physical applications of spectral zeta functions" from Elizalde, pg.20

Actually, a universal definition for the determinant [...] is still missing.

Also Singer and Ray refer in http://www.sciencedirect.com/science/article/pii/0001870871900454 to Seeley http://www.ams.org/mathscinet/search/publications.html?pg1=IID&s1=157950 , who has proven the regularity at $0$. He has a nice formula there:

$$A^s=i/2π\int_{Γ} λ^s(A−λ)^{−1}dλ$$

Here are some suggestions, until somebody more knowledgeable appears.

1. Why does this definition make sense: Analytic continuation is unique, and hence if there is analytic continuation to neighbourhood of $0$, and the formula is valid in finite dimension, then why not take it. Unfortunately, we have in general that $$ \det (AB) \neq \det(A) \det(B).$$

2. Why does it possibly generalize the theory of the Riemann zeta function: If you take the circle with the Laplace operator $-\frac{\partial^2}{\partial^2 x}$ on $L^2[0,1]$ has eigenfunctions $\exp(2 \pi \textup{i} nx)$ for $n \in \mathbb{Z}$, you actually will get the Riemann zeta function as the spectral zeta function $\zeta_A(s) = 2 \zeta(2s)$.

3. Why does it not: Certain properties are not shared by the general spectral zeta function. There is in general no functional equation relating $\zeta_A(s)$ to $\zeta_A( D-s)$ for some $D$, and there is no Euler product.

But here is also a quote from page 20, of "Ten physical applications of spectral zeta functions" from Elizalde, pg.20

Actually, a universal definition for the determinant [...] is still missing.

Also Singer and Ray refer in http://www.sciencedirect.com/science/article/pii/0001870871900454 to Seeley http://www.ams.org/mathscinet/search/publications.html?pg1=IID&s1=157950 , who has proven the regularity at $0$. He has a nice formula there:

$$A^s=i/2π\int_{Γ} λ^s(A−λ)^{−1}dλ$$

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 (I like the logarithmic derivative variant best): $$ \frac{ \det(A(s))'}{\det(A(s))} = \operatorname{tr}\left(A^{-1}(s) A(s)'\right)$$ from the finite dimensional situation.

Hint: Take $A(s) = \exp( A^s)$ above and imagine $A$ as diagonalized operator.

Here is another variant $$\det(\exp(tA)) = 1 + t\ Tr(A) + O(t^2)$$

from Geometric interpretation of trace,

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.

Here are some suggestions, until somebody more knowledgeable appears.

1. Why does this definition make sense: Analytic continuation is unique, and hence if there is analytic continuation to neighbourhood of $0$, and the formula is valid in finite dimension, then why not take it. Unfortunately, we have in general that $$ \det (AB) \neq \det(A) \det(B).$$

2. Why does it possibly generalize the theory of the Riemann zeta function: If you take the circle with the Laplace operator $\frac{\partial^2}{\partial^2 x^2}$ on $L^2[0,2 \pi]$ has eigenfunctions $\exp(nx)$ for $n \in \mathbb{Z}$, you actually will get the Riemann zeta function as the spectral zeta function $\zeta_A(s) = 2 \zeta(2s)$.

3. Why does it not: Certain properties are not shared by the general spectral zeta function. There is in general no functional equation relating $\zeta_A(s)$ to $\zeta_A( D-s)$ for some $D$, and there is no Euler product.

But here is also a quote from page 20, of "Ten physical applications of spectral zeta functions" from Elizalde, pg.20

Actually, a universal definition for the determinant [...] is still missing.

Also Singer and Ray refer in http://www.sciencedirect.com/science/article/pii/0001870871900454 to Seeley http://www.ams.org/mathscinet/search/publications.html?pg1=IID&s1=157950 , who has proven the regularity at $0$. He has a nice formula there:

$$A^s=i/2π\int_{Γ} λ^s(A−λ)^{−1}dλ$$