I am currently researching divergent integrals.

**Definition.**An extended number is an expression of the form $\int_a^b f(x)\,dx$, where $a,b\in \overline{\mathbb{R}}$ and function $f(x)$ is defined almost everywhere at $(a,b)$. Generally (when Riemann or Lebesgue 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 Laplace-transform based rule:

$$\int_0^\infty f(x)\,dx=\int_0^\infty\mathcal{L}_t[t f(t)](x) \, dx=\int_0^\infty\frac1x\mathcal{L}^{-1}_t[ f(t)](x)\,dx$$

Divergent integrals, summable by Cesaro, Abel, Cauchy mean or other averaging technique are considered equal to their regularized values.

The changes-of-variables $\int_0^\infty f(a x)dx=\frac1a\int_0^\infty f(x)dx$ and $\int_a^\infty f(x)\,dx = \int_{a-b}^\infty f(x+b)\,dx $ are permitted only when $\lim_{x\to\infty}\overline{f(x)}=0$, where $\overline{f(x)}$ is the average value (by Cesaro, Abel, etc).

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 $n>1$):

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

- 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.

**Example.**

\begin{align} & \int_0^\infty (2x^3-3x^2+x-4) \, dx \cdot \int_0^\infty (2x^2-3x+1) \, dx \\[8pt] = {} & \left(\frac{\tau ^4}{2}-\tau^3+\frac{3 \tau^2}{4}-\frac{17 \tau}{4}+\frac{23}{480} \right)\left(\frac{2 \tau^3}{3}-\frac{3 \tau ^2}{2} + \frac{7 \tau }{6}-\frac{1}{8}\right) \\[8pt] = {} & \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} \\[8pt] = {} & \int_0^\infty \left(\frac{7 x^6}{3}-\frac{17 x^5}{2} + 10 x^4-\frac{41 x^3}{3}+\frac{1007x^2}{60}-\frac{63 x}{10}-\frac{113}{120}\right) \, dx+\frac{127}{420} \end{align}

Now, I have the following questions:

Can one obtain the formula for multiplication of integrals of polynomials in more simple or general form, for instance, involving matrices or binomial coefficients?

Can I in natural way generalize the formula to a larger class of the functions, particularly, rational functions?

11more comments