I am currently researching divergent integrals.

  1. 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.

  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:

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

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

  2. 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).

  3. 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.


\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:

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

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

  • $\begingroup$ Do I understand correctly? An extended number really means a collection of maps, each denoted $f\mapsto \int_a^b f(x)\,dx\in E$, defined on the functions $f(x)$ you described above, mapping onto some ring $E$, where $E$ contains $\mathbb{C}$ as a subring. So there might be many choices of such maps and in particular of $E$. Is $E$ required to be commutative? Is $E$ required to really be a ring, or some weaker structure (semiring, ternary ring)? $\endgroup$
    – Ben McKay
    Sep 11 at 18:48
  • $\begingroup$ @BenMcKay well, yes, I envisage $E$ to be commutative and the multiplication procedure described here for polynomials is commutative (so, its generalization should be as well). I envisage $E$ to be an integral domain (stronger than ring but possibly not a field). No zero divisors, I think. $\endgroup$
    – Anixx
    Sep 11 at 18:57
  • 1
    $\begingroup$ This post has another example of the phenomenon described at math.stackexchange.com/questions/1433899/…. $\endgroup$
    – KConrad
    Sep 11 at 22:41
  • 1
    $\begingroup$ @KConrad по-русски мы его знаем как Лебега, по-этому при переходе на латиницу приходится учить имена заново. $\endgroup$
    – Anixx
    Sep 11 at 22:55
  • 1
    $\begingroup$ So an extended number is not just an expression, but an equivalence class of expressions. Does each expression denoting a divergent integral belong to a singleton equivalence class, or can there be equivalences among such expressions as well? (I guess there are your (1)–(6), but they don't even obviously to me imply equality of an expression representing a convergent integral with its value, so I am not sure how to understand it.) $\endgroup$
    – LSpice
    Sep 18 at 14:37

Answering my own question.

The operator looks like this:

$\int_0^\infty f(x)dx\cdot\int_0^\infty g(x)dx=\int_0^\infty D^2 \Delta^{-1} \left(\Delta D^{-2}f(x)\cdot\Delta D^{-2}g(x)\right) dx +D \Delta^{-1} \left(\Delta D^{-2}f(x)\cdot\Delta D^{-2}g(x)\right)|_{x=0}$

Here $\Delta D^{-2}f(x)=\int_x^{x+1}\int_0^s f(t)\,dt\,ds$.

The formula is unambiguous. Here is the Mathematica code:

f[x_] := x
g[x_] := x

Func := D[
    D[Sum[(DifferenceDelta[Integrate[Integrate[f[t], {t, 0, x}], x], 
         x]*DifferenceDelta[Integrate[Integrate[g[t], {t, 0, x}], x], 
         x]), x], x], x]
Const := (D[
     Sum[(DifferenceDelta[Integrate[Integrate[f[t], {t, 0, x}], x], 
         x]*DifferenceDelta[Integrate[Integrate[g[t], {t, 0, x}], x], 
         x]), x], x] /. x -> 0) 
 Integrate[f[x], {x, 0, Infinity}]\[CenterDot]Integrate[
   g[x], {x, 0, Infinity}], Integrate] == FullSimplify[
Const + Integrate[Func, {x, 0, Infinity}]] // Expand // Quiet // TraditionalForm

Out:=$\int _0^{\infty }xdx\cdot \int _0^{\infty }xdx=\int_0^{\infty } \left(x^3-\frac{x}{6}\right) \, dx+\frac{1}{180}$

The code works only with polynomials.

With other types of functions, it seems, it does not ($f(x)=\exp(-x), g(x)=\exp(x)$ gives a wrong constant term).

Thus, the question on how to generalize this still remains.


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