Integral with $x^{dx}$

Question

I was just watching this video, where he calculates $$I := \int x^{dx} -1.$$ The reasoning is that $$I =\int\frac{x^{dx} -1}{dx}dx,$$ and then he uses $$\lim_{h\rightarrow0} \frac{x^{h} -1}{h} = \ln x$$ to give that $$I = \int\ln x dx.$$

The argument makes sense to me, at least informally.

Is it standard to ascribe meaning to integrals of this form as the author has done in the video, or is it just a curio that he's come up with?

Is there some way to extend his method to a wider class of integrals of the form $\int f(x, dx)$?

Clearly the $f$ in his example has been chosen to give the required result. Are there applications where you would want to be able to calculate $\int f(x, dx)$ for a more generic $f$ than the one in the video? Is there some kind of geometric meaning to this expression?

Edit: I've been linked to a similar question in the comments. The answers to this question don't address any of the questions I've asked, and essentially only restate the working that I've written.

Answer 1

Riemann integrals $\int g(x)dx$ are limits of sums with ever narrower area strips approximated as with-$dx$ rectangles. This affords a generalization: $\int_a^bf(x,\,dx)$, if it exists, is $\lim_{n\to0}\sum_{i=0}^{n-1}f(a+i/n,\,(b-a)/n)$.

(We want the same limit to be obtained for arbitrary ways of splitting $[a,\,b]$ into $n$ strips, not just those where their widths are equal; the definition of Riemann integrals already encounters this subtlety, but we'll overlook it herein because it's more instructive for now to illustrate the limit in the special case than to verify the general case obtains it.)

In your example,$$\int_a^b(x^{dx}-1):=\lim_{n\to0}\sum_{i=0}^{n-1}((a+i/n)^{(b-a)/n}-1)=\lim_{n\to0}\frac{b-a}{n}\sum_{i=0}^{n-1}\ln(a+i/n)=:\int_a^b\ln xdx,$$where $u:=v,\,v=:u$ both mean "$u$ is defined as $v$". Or indefinitely, $\int x^{dx}-1=\int\ln xdx$ (which makes sense in light of $x^h-1=e^{h\ln x}-1\sim h\ln x$ for small $h$).

You can use this kind of reasoning to show that, if $f(x,\,h)\sim g(x)h$ for small $h$, $\int f(x,\,dx)=\int g(x)dx$. You may want to consider other examples, such as $\lim_{h\to0}f(x,\,h)=0$ or $f(x,\,h)\sim g(x)h^2$.