Definition
I want to define something called continued product, which is the analog of continued sum $\int$ but for product.
We will "upgrade" all the elementary operations by $1$, which means sums will become products, and products will become powers.
Let's denote the continued product of a function $f$ like this:
$${}^\Pi\hspace{-3mm} \int f(x)^{\mathrm dx}.$$
We will define it analogically of Riemann sums.
Indeed, we already have
$$\int_0^1 f(x)\mathrm dx =\lim_{n\to\infty} \frac 1n \sum_{k=0}^n f\left(\frac kn\right),$$
so we will say that
$${}^\Pi\hspace{-3mm} \int_0^1 f(x)^{\mathrm dx} :=\lim_{n\to\infty}\left( \prod_{k=1}^n f\left(\frac kn\right)\right)^{1/n}.$$
Examples
- For instance, let's compute
$${}^\Pi\hspace{-3mm} \int_0^1 x^{\mathrm dx}.$$
We have
$$\left(\prod_{k=1}^n \frac kn\right)^{1/n}=\left(\frac {n!}{n^n}\right)^{1/n}=\frac 1n (n!)^{1/n}.$$
Using Stirling formula, we get:
$$\left(\prod_{k=1}^n \frac kn\right)^{1/n}\sim\frac 1n\left(\sqrt{2\pi n}\left(\frac ne\right)^n\right)^{1/n}=\pi^{1/n}(2n)^{1/(2n)}\frac 1e\sim \frac 1e.$$
So
$${}^\Pi\hspace{-3mm} \int_0^1 x^{\mathrm dx}=\frac 1e.$$
- We could also show that
$${}^\Pi\hspace{-3mm} \int_0^1 \exp(x)^{\mathrm dx}=\sqrt e.$$
Questions
Is there an analog to the fundamental theorem of calculus to compute those kind of integrals quicker ?
Does this integral have a geometric meaning ?
What would be some possible applications of this kind of integral ?
A comment says that is it only $\exp\int \log f$. I don't understand why, regarding to the first example.
