What physical meaning “determinant” of a divergency (divergent integral or series) can have? Is there a parallel with functional determinant?

I am working on the algebra of "divergencies", that is, infinite integrals, series and germs. So, I decided to construct something similar to determinant of a matrix of these entities.

$$\det w=\exp(\operatorname{reg }\ln w)$$ which is analogous to how determinant of a matrix can be expressed, except we take finite part (regularize) instead of taking trace.

Like determinant of a matrix, determinant of a divergency can be negative. It does not follow the requirements for a norm (Pythagorean theorem), and is not even continuous. Still, it has some usual properties, like $$1/\det w=\det 1/w$$.

Below is a table of some divergencies with their finite parts and determinants.

An interesting property is that the constant $$e^{-\gamma}$$ often appears in the expressions for these determinants.

I wonder, what can indicate such determinant of a divergent integral or series? Can it tell something about its properties?

Given that the regular part (which is analog of trace) is defined as $$\operatorname{reg}(\omega_-+x)^a= B_a(x)=-a\zeta(1-a,x)$$ I see some similarity with functional determinant that also includes a zeta function.

$$\begin{array}{cccccc} \text{Delta form} & \text{In terms of } \tau, \omega_+,\omega_- & \text{Finite part} & \text{Integral or series form} & \text{Germ form} &\text{Determinant}\\ \pi \delta (0) & \tau & 0 & \int_0^{\infty } \, dx;\int_0^{\infty } \frac{1}{x^2} \, dx & \underset{x\to\infty}{\operatorname{germ}} x;\underset{x\to0^+}{\operatorname{germ}}\frac1x&\frac{e^{-\gamma}}4 \\ \pi \delta (0)-\frac{1}{2} & \omega _-;\tau-\frac{1}{2} & -\frac{1}{2} & \sum _{k=1}^{\infty } 1 & \underset{x\to\infty}{\operatorname{germ}} (x-1/2) &e^{-\gamma} \\ \pi \delta (0)+\frac{1}{2} & \omega _+;\tau+\frac{1}{2} & \frac{1}{2} &\sum _{k=0}^{\infty } 1 & \underset{x\to\infty}{\operatorname{germ}} (x+1/2) & e^{-\gamma} \\ 2 \pi \delta (i) & e^{\omega_+}-e^{\omega_-}-1 & 0 & \int_{-\infty }^{\infty } e^x \, dx & \underset{x\to\infty}{\operatorname{germ}} e^x \\ & \frac{\tau ^2}{2}+\frac{1}{24};\frac{\omega_+^3-\omega_-^3}6 & 0 & \int_0^{\infty} x \, dx;\int_0^\infty \frac2{x^3}dx & \underset{x\to\infty}{\operatorname{germ}}\frac{x^2}2;\underset{x\to0^+}{\operatorname{germ}} \frac1{x^2}\\ & \frac{\tau ^2}{2}-\frac{1}{24} & -\frac1{12} & \sum _{k=0}^{\infty } k & \underset{x\to\infty}{\operatorname{germ}} \left(\frac{x^2}2-\frac1{12}\right) \\ -\pi \delta''(0) &\frac {\tau^3}3 +\frac\tau{12};\frac{\omega_+^4-\omega_-^4}{12}& 0 & \int_0^\infty x^2dx;\int_0^\infty\frac6{x^4}dx&\underset{x\to\infty}{\operatorname{germ}}\frac{x^3}3;\underset{x\to0^+}{\operatorname{germ}} \frac2{x^3}\\ \pi^2\delta(0)^2-\pi\delta(0)+1/4&\omega_-^2&\frac16&2 \int_0^{\infty } \left(x-\frac{1}{2}\right) \, dx+\frac{1}{6}&\underset{x\to\infty}{\operatorname{germ}}B_2(x)&e^{-2\gamma}\\ \pi^2\delta(0)^2+\pi\delta(0)+1/4&\omega_+^2&\frac16&2 \int_0^{\infty } \left(x+\frac{1}{2}\right) \, dx+\frac{1}{6}&\underset{x\to\infty}{\operatorname{germ}}B_2(x+1)&e^{-2\gamma}\\ \pi^2\delta(0)^2&\tau^2&-\frac1{12}&\int_{-\infty}^{\infty } |x| \, dx-\frac{1}{12}&\underset{x\to\infty}{\operatorname{germ}}B_2(x+1/2)&\frac{e^{-2\gamma}}{16} \\ &\ln \omega_++\gamma&0&\int_1^\infty \frac{dx}x;\sum_{k=1}^\infty \frac1x -\gamma&\underset{x\to\infty}{\operatorname{germ}}\ln x\\ -3\pi\delta''(0)-\frac14 \pi\delta(0);\pi^3\delta(0)^3&\tau^3&0&\int_0^\infty \left(3x^2-\frac1{4}\right)dx&\underset{x\to\infty}{\operatorname{germ}}B_3(x+1/2)&\frac{e^{-3\gamma}}{64} \\ \frac{2\pi\delta(i)+1}{e-1}&e^{\omega_-}&\frac1{e-1}&\frac1{e-1}+\frac1{e-1}\int_{-\infty}^\infty e^x dx&\underset{x\to\infty}{\operatorname{germ}} \frac{e^x+1}{e-1}&\frac1{\sqrt{e}}\\ \frac{2\pi\delta(i)+1}{1-e^{-1}}&e^{\omega_+}&\frac1{1-e^{-1}}&\frac1{1-e^{-1}}+\frac1{1-e^{-1}}\int_{-\infty}^\infty e^x dx&\underset{x\to\infty}{\operatorname{germ}} \frac{e^x+1}{1-e^{-1}}&\sqrt{e}\\ &(-1)^\tau&\frac\pi{2}&&&1\\ \end{array}$$