Studying divergent integrals, I found a good formula for their multiplication:
$\int_0^\infty f(x)dx\cdot\int_0^\infty g(x)dx=\int_0^\infty D^2 \Delta^{-1} \left(\Delta D^{-2}f(x)\cdot\Delta D^{-2}g(x)\right) dx +D \Delta^{-1} \left(\Delta D^{-2}f(x)\cdot\Delta D^{-2}g(x)\right)|_{x=0},$
where $\Delta D^{-2}f(x)=\int_x^{x+1}\int_0^s f(t)\,dt\,ds$.
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)
Inactivate[
Integrate[f[x], {x, 0, Infinity}]\[CenterDot]Integrate[
g[x], {x, 0, Infinity}], Integrate] ==
Const + Integrate[Func, {x, 0, Infinity}] // Simplify // 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 fails only when Mathematica's discrete or conventional integrator fails to provide closed form answer. Thus, the formula even if fails to produce a closed form, still defines the product.
That said, I wonder, whether having the formula for multiplication we can derive a formula for division, particularly, for reciprocals of some divergent integrals, such as $1/\int_0^\infty dx$. Can we solve an equation
$W\int_0^\infty dx =1$
for $W$ based on this multiplication formula?
