Is there a geometric interpretation of the product integral?

Question

Riemann's "way to the Integral" is loosely speaking the limit of sums of this kind \begin{equation} \sum_if(x_i)\Delta x_i \end{equation} Now, if we replace the sum with a product and the multiplication by $\Delta x_i$ with exponentiation, we are led to the idea of a "product" integral: \begin{equation} \prod_if(x_i)^{\Delta x_i} \end{equation} The relation with the usual integral and the "product" integral is the following: \begin{equation} \prod f(x)^{dx}=\ e^{\int \ln f(x) dx} \end{equation} My question is: is there a geometrical (or measure-theoretical) interpretation of the product integral? As we all know the usual integral (in one variable) is the signed area under the graph of $f$.

Answer 1

I'm admittedly not familiar with product integrals, but I think the geometric interpretation route is in vain.

I'm trying to couch this in terms of units: if $f$ and $x$ are both measured in feet, the Riemann sum gives us what we'd hope for our interpretation--ft×(ft-ft)->ft^2

Moving to the product integral, I can't think of anything geometric where the exponent has units (especially when the end behavior of the exponents is to be infinitesimally small--taking NNNth roots). To me, exponents in geometry are related to areas/higher dimensional volumes--but in such formulas, the exponents are unit-free.

Answer 2

Actually the area under the graph is defined by the integral not the other way around.

By swapping the operations arbitrarily one can not expect the old concepts to map to something that is still meaningful. This is one specific generalisation that could also be applied with differentiation, how ever the new constructs do not map to a meaningful familiar concept.

Consider that the more intuitive generalisation of integration and differentiation under fractional calculus do not have a geometric interpretation for non integer values (the geometric interpretation is an open question).

Try changing the metric of space and see how that modifies definitions of integration or differentiation.

Answer 3

I have been thinking about this lately and for the user who commented on units, you can think in unit terms. Consider the limiting finite case; you want to compute the product of all the terms of the function $f(x)=x$ from $0$ to $1$.

You can start by choosing one point, the mid-point and you take the first root (root of unity) of this sum and you arrive at $1/2$. Now use two points; the one-third and two-thirds points. Multiply together and take the square root. You get $2^{0.5}/3$ or about $0.47$. Continue on, using finer and finer subdivisions, and always taking the $n$-th root. In the limit, this equals the result of the integral identity provided above, namely, $e^{\int(ln(f(x))dx,0,1)}$ which is $1/e$. And sure enough, the sequence of values you arrive at by considering $n$-points approaches $1/e$ as $n$ tends towards infinity.

With this in mind, you can see that the units will be the same as the units of $f(x)$. It’s interesting and I have not thought on it more at this time as to what, if any significance, the units would hold.

At first glance, it would seem to be the unit dimension of some infinite dimensional cube of a certain volume, but that doesn’t seem correct in that exponentiating the result by infinity would tend towards zero rather than a finite value (at least for the function $f(x)=x$ since $1/e$ is less than one).