Background
The classical Riemann integral of a function $f : [a,b] \to \mathbb{R}$ can be defined by setting $$\int_{a}^{b} f(x) \ dx := \lim_{\Delta x \to 0} \sum f(x_{i}) \ \Delta x. $$ Here, the limit is taken over all partitions of the interval $[a,b]$ whose norms approach zero.
We can do something roughly similar with product integrals. They take the limit over a product instead of a sum, and can be interpreted as continuous analogues of discrete products.
There are multiple types of product integrals. Type I is often refered to as Volterra's integral. It is defined as follows:
\begin{align*} \prod_{a}^{b} \left(1+f(x) \ dx \right) &:= \lim_{\Delta x \to 0} \left(1 + f(x_{i}) \ \Delta x \right) \newline &= \exp \left( \int_{a}^{b} f(x) \ dx \right). \tag{1} \label{1} \end{align*}
However, this is not a multiplicative operator. As an alternative, there is also Type II, the geometric integral. It is defined as
\begin{align*} \prod_{a}^{b} f(x)^{dx} &:= \lim_{\Delta x \to 0} \prod f(x_{i})^{\Delta x} \newline &= \exp \left( \int_{a}^{b} \ln f(x) \ dx \right). \tag{2} \label{2} \end{align*}
This does amount to a multiplicative operator. A third type, the bigeometric integral, is also an operator with this property.
Question
I wonder whether something similar can be done with other kinds of expressions (infinite or finite). In particular, I am curious whether we can obtain a continuous analogue of the discrete simple continued fraction. In the discrete case, it is defined as:
$$\underset{i=a}{\overset{b}{\large{\mathrm K}}} \ \frac{1}{f(i)} = \cfrac{1}{f(a) + \cfrac{1}{f(a+1)+\cfrac{1}{\ddots+\cfrac{1}{f(b-1) + \cfrac{1}{f(b)}}}}}$$
In other words, I am looking for a way to complete the following table, by finding a definition of what is described in the bottom right cell:
\begin{array}{|c|c|c|} \hline & \text{additive} & \text{multiplicative} & \text{simple continued fraction} \\ \hline \text{discrete} & \sum_{i=a}^{b} f(i) & \prod_{i=a}^{b} f(i) & \underset{i=a}{\overset{b}{\large{\mathrm K}}} \ \frac{1}{f(i)} \\ \hline \text{continuous} & \int_{a}^{b} f(x) \ dx & \prod_{a}^{b} f(x)^{dx} & \underset{a}{\overset{b}{\large{\mathrm K}}} \ \frac{1}{f(x)} \overline{dx} \\ \hline \end{array}
I think we could make a start by setting
$$ \underset{a}{\overset{b}{\large{\mathrm K}}} \ \frac{1}{f(x)} \overline{dx} := \lim_{\Delta x \to 0} \large{\mathrm K} \frac{1}{f(x_{i})} \Delta x .$$
However, I am not sure how this would translate into a formula that is similar to \eqref{1} or \eqref{2}. Is there a way to obtain such a formula (or perhaps multiple, as in the case of the product integral), and if so, what does it look like?
I don't think this is likely to work. An important feature of both sums and products is that they're commutative and associative, so that when considering adding/multiplying more and more nearby numbers, the order of addition/multiplication doesn't matter. On the other hand, continued fractions make use of division as well as addition, and division is neither commutative nor associative. This isn't just an abstract complaint: iterated division associates the even-numbered terms with the numerator more than with the denominator—the overall expression is increasing in those parameters while decreasing in the odd-numbered parameters. So I don't see any reasonable way of doing that "more and more" with nearby points without radically changing the dependence on individual values.