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^\infty\mathcal{L}_t[t f(t)]\left(x\right)dx=\int_0^\infty\frac1x\mathcal{L}^{-1}_t[ f(t)]\left(x\right)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)+n\int_0^\infty B_{n-1}(x+1/2)dx$

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


$ \int_0^\infty \left(2x^3-3x^2+x-4\right) dx \cdot \int_0^\infty \left(2x^2-3x+1\right) 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.