intuition – What intuitive meaning “determinant” of a divergency (divergent integral or series) can have?

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 intuitively can indicate such determinant of a divirgent integral or series? Can it tell something about its properties?

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{xtoinfty}{operatorname{germ}} x;underset{xto0^+}{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{xtoinfty}{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{xtoinfty}{operatorname{germ}} (x+1/2) & e^{-gamma} \
2 pi delta (i) & e^{omega_+}-e^{omega_-}-1 & 0 & int_{-infty }^{infty } e^x , dx & underset{xtoinfty}{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{xtoinfty}{operatorname{germ}}frac{x^2}2;underset{xto0^+}{operatorname{germ}} frac1{x^2}\
& frac{tau ^2}{2}-frac{1}{24} & -frac1{12} & sum _{k=0}^{infty } k & underset{xtoinfty}{operatorname{germ}} left(frac{x^2}2-frac1{12}right) \
-pi delta”(0) &frac {tau^3}3 +fractau{12};frac{omega_+^4-omega_-^4}{12}& 0 & int_0^infty x^2dx;int_0^inftyfrac6{x^4}dx&underset{xtoinfty}{operatorname{germ}}frac{x^3}3;underset{xto0^+}{operatorname{germ}} frac2{x^3}\
pi^2delta(0)^2-pidelta(0)+1/4&omega_-^2&frac16&2 int_0^{infty } left(x-frac{1}{2}right) , dx+frac{1}{6}&underset{xtoinfty}{operatorname{germ}}B_2(x)&e^{-2gamma}\
pi^2delta(0)^2+pidelta(0)+1/4&omega_+^2&frac16&2 int_0^{infty } left(x+frac{1}{2}right) , dx+frac{1}{6}&underset{xtoinfty}{operatorname{germ}}B_2(x+1)&e^{-2gamma}\
pi^2delta(0)^2&tau^2&-frac1{12}&int_{-infty}^{infty } |x| , dx-frac{1}{12}&underset{xtoinfty}{operatorname{germ}}B_2(x+1/2)&frac{e^{-2gamma}}{16} \
&ln omega_++gamma&0&int_1^infty frac{dx}x;sum_{k=1}^infty frac1x -gamma&underset{xtoinfty}{operatorname{germ}}ln x\
-3pidelta”(0)-frac14 pidelta(0);pi^3delta(0)^3&tau^3&0&int_0^infty left(3x^2-frac1{4}right)dx&underset{xtoinfty}{operatorname{germ}}B_3(x+1/2)&frac{e^{-3gamma}}{64} \
frac{2pidelta(i)+1}{e-1}&e^{omega_-}&frac1{e-1}&frac1{e-1}+frac1{e-1}int_{-infty}^infty e^x dx&underset{xtoinfty}{operatorname{germ}} frac{e^x+1}{e-1}&frac1{sqrt{e}}\
frac{2pidelta(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{xtoinfty}{operatorname{germ}} frac{e^x+1}{1-e^{-1}}&sqrt{e}\