Addition is to Integration as Multiplication is to ______
Addition is to Integration as Multiplication is to ______ ? Everyone knows that definite integration is "a way to sum continuum-many terms" in a rough sense. Can we "multiply continuum-many factors" in a similar sense?
Solution 1:
If $f\colon [a,b] \to (0,\infty)$, then the closest analogue to a multiplicative integral of $f$ is $$ {\prod}_a^b \,f(x)^{dx} \;\underset{\mathrm{def}}{=}\; \exp\left(\int_a^b \ln f(x)\,dx\right). $$ This has the following properties:
If $a<r<b$, then $\displaystyle{\prod}_a^b\,f(x)^{dx} = \left({\prod}_a^r\,f(x)^{dx}\right)\left({\prod}_r^b\,f(x)^{dx}\right)$.
If $f(x)$ is a constant funtion $C$, then $\displaystyle{\prod}_a^b\,f(x)^{dx} = C^{b-a}$.
$\displaystyle{\prod}_a^a \,f(x)^{dx} = 1$ for any function $f$.
$\displaystyle {\prod}_a^b\,\bigl(f(x)g(x)\bigr)^{dx} \;=\; \left({\prod}_a^b\,f(x)^{dx}\right)\left({\prod}_a^b\,g(x)^{dx}\right)$
$\displaystyle\left({\prod}_a^b \,f(x)^{dx}\right)^k \;=\; {\prod}_a^b \,\bigl(f(x)^k\bigr)^{dx}$ for any constant $k$.
The (Monte Carlo) geometric mean of $f(x)$ on $[a,b]$ is $\displaystyle\left({\prod}_a^b \,f(x)^{dx}\right)^{1/(b-a)}$.
If $f$ is continuous, then $\displaystyle{\prod}_a^b \,f(x)^{dx}$ is the limit of a "Riemann product": $$ \displaystyle{\prod}_a^b \,f(x)^{dx} \;=\; \lim_{\max \Delta x_i\to 0}\, \prod_{i=1}^n f(x_i^*)^{\Delta x_i}. $$
If $\displaystyle F(t) = {\prod}_a^t f(x)^{dx}$, then $e^{F'(t)/F(t)} = f(t)$. Similarly, for any $C^1$ function $f\colon [a,b]\to\mathbb{R}$, we have $$ {\prod}_a^b \bigl(e^{f'(x)/f(x)}\bigr)^{dx} = \frac{f(b)}{f(a)} $$
Solution 2:
Convolution is analogous to multiplication.
Solution 3:
Product Integral. Math With Bad Drawings has an article that mentions it.
I think the one presented by Jim Belk is only Type I. Here is Type II.