I am currently researching divergent integrals.

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.

There are four simple equivalence rules based on linearity:
$int_a^c f(x) dx=int_a^b f(x)dx+int_b^c f(x)dx$
$int_a^b (f(x)+g(x)) dx=int_a^b f(x)dx+int_a^b g(x)dx$
$int_a^b c f(x) dx =c int_a^b f(x) dx$
$int_{infty}^{a} f(x) dx=int_a^infty f(x) dx$
 There is one complicated Laplacetransform based rule:
$int_0^infty f(x)dx=int_0^inftymathcal{L}_t(t f(t))left(xright)dx=int_0^inftyfrac1xmathcal{L}^{1}_t( f(t))left(xright)dx$
 There is a rule that allows to represent divergent integrals of polynomials via the most basic divergent integral $tau=int_0^infty dx$:
$int_0^infty x^n dx=frac{left(tau +frac{1}{2}right)^{n+2}left(tau frac{1}{2}right)^{n+2}}{(n+1)(n+2)}$
 Following Laplace transform, there is a similar rule (for $n>1$):
$int_0^infty frac1{x^n} dx=frac1{(n1)!}int_0^infty x^{n2} dx=frac{left(tau +frac{1}{2}right)^{n}left(tau frac{1}{2}right)^{n}}{(n1)n!}$
 There is the opposite rule, converting in the opposite direction:
$tau^n=B_n(1/2)+nint_0^infty B_{n1}(x+1/2)dx$
That said, one can use these rules to multiply divergent integrals of polynomials.
Example.
$int_0^infty left(2x^33x^2+x4right) dx cdot int_0^infty left(2x^23x+1right) dx=left(frac{2 tau ^3}{3}frac{3 tau ^2}{2}+frac{7 tau }{6}frac{1}{8}right)
left(frac{tau ^4}{2}tau ^3+frac{3 tau ^2}{4}frac{17 tau
}{4}+frac{23}{480}right)=frac{tau ^7}{3}frac{17 tau ^6}{12}+frac{31 tau ^5}{12}frac{83 tau ^4}{16}+frac{5333
tau ^3}{720}frac{4919 tau ^2}{960}+frac{1691 tau }{2880}frac{23}{3840}=int_0^{infty } left(frac{7 x^6}{3}frac{17 x^5}{2}+10 x^4frac{41 x^3}{3}+frac{1007
x^2}{60}frac{63 x}{10}frac{113}{120}right) dx+frac{127}{420}$
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.