simplifying expressions – Want to realize this operation (multiplication of divergent integrals of polynomials) in Mathematica

I am currently researching divergent integrals.

  1. Definition. An extended number is an expression of the form $int_a^b f(x)dx$, where function $f(x)$ is defined almost everywhere at $(a,b)$. Generally (when Riemann or Lebesque sum converges or when the equivalence follows from the rules expressed below), an extended number can be equal to a real or complex number.

  2. There are four simple equivalence rules based on linearity:

$int_a^c f(x) dx=int_a^b f(x)dx+int_b^c f(x)dx$

$int_a^b (f(x)+g(x)) dx=int_a^b f(x)dx+int_a^b g(x)dx$

$int_a^b c f(x) dx =c int_a^b f(x) dx$

$int_{-infty}^{-a} f(x) dx=int_a^infty f(-x) dx$

  1. There is one complicated Laplace-transform based rule:

$int_0^infty f(x)dx=int_0^inftymathcal{L}_t(t f(t))left(xright)dx=int_0^inftyfrac1xmathcal{L}^{-1}_t( f(t))left(xright)dx$

  1. There is a rule that allows to represent divergent integrals of polynomials via the most basic divergent integral $tau=int_0^infty dx$:

$int_0^infty x^n dx=frac{left(tau +frac{1}{2}right)^{n+2}-left(tau -frac{1}{2}right)^{n+2}}{(n+1)(n+2)}$

  1. Following Laplace transform, there is a similar rule (for $n>1$):

$int_0^infty frac1{x^n} dx=frac1{(n-1)!}int_0^infty x^{n-2} dx=frac{left(tau +frac{1}{2}right)^{n}-left(tau -frac{1}{2}right)^{n}}{(n-1)n!}$

  1. There is the opposite rule, converting in the opposite direction:

$tau^n=B_n(1/2)+nint_0^infty B_{n-1}(x+1/2)dx$


That said, one can use these rules to multiply divergent integrals of polynomials.

Example.

$int_0^infty left(2x^3-3x^2+x-4right) dx cdot int_0^infty left(2x^2-3x+1right) dx=left(frac{2 tau ^3}{3}-frac{3 tau ^2}{2}+frac{7 tau }{6}-frac{1}{8}right)
left(frac{tau ^4}{2}-tau ^3+frac{3 tau ^2}{4}-frac{17 tau
}{4}+frac{23}{480}right)=frac{tau ^7}{3}-frac{17 tau ^6}{12}+frac{31 tau ^5}{12}-frac{83 tau ^4}{16}+frac{5333
tau ^3}{720}-frac{4919 tau ^2}{960}+frac{1691 tau }{2880}-frac{23}{3840}=int_0^{infty } left(frac{7 x^6}{3}-frac{17 x^5}{2}+10 x^4-frac{41 x^3}{3}+frac{1007
x^2}{60}-frac{63 x}{10}-frac{113}{120}right) dx+frac{127}{420}$


I did the previous example by hand. Is it possible to optimally realize it in Mathematica?

I mean, 1. Enter the coefficients of two polynomials under the integrals 2. Obtain the coefficients of the resulting polynomial under integral plus the free term.

I suspect, this may be some kind of convolution.