$\int x^{dx}-1$

Question

If you go to Flammable Maths's YouTube channel and scroll through some of his videos you see him solving the following integral:

$$\int x^{dx}-1$$

he explains that this is a Product integral. My questions are the following:

1 - What is the geometric meaning of a product integral?

2 - does it make sense to have:

$$\int f(x,dx)$$

and if $f(x,dx) = g(x)dx$ then it's just a regular integrals and if $f(x,dx) = g(x)^{dx}$ it's just a product integral?

I'll leave the link to the video here.

Answer 1

Comparing Taylor series with $df=f^\prime dx$ gives $dx^2=0$. Note that$$g(x)^{dx}-1=\exp(\ln g(x)\cdot dx)-1=\ln g(x)\cdot dx+O(dx^2)=\ln g(x)\cdot dx,$$so your first example is $\int\ln xdx=x\ln x-x+C$.

Answer 2

$\int x^{dx}-1$ = $\int \frac{x^{dx}-1}{{dx}} {dx}$ = $\int (\lim_{h \to 0}\frac{x^h-1}{h}) {dx}$ = $\int \ln x {dx}$ = $x\ln x-x+const.$

$ \therefore \int x^{dx}-1 = x \ln x - x + const.$