TITLE: Is there a formula for "power integration," with integrand $f(x)^{dx}-1$, rather than $f(x)dx$?
QUESTION [8 upvotes]: I was joking around with integrals, thinking, "what else can I do to a function (such as $f(x)dx$) that results in something I can sum infinitely as in an integral?" The first thing I came up with was $f(x)^{dx}-1$. Loosely speaking, since $dx$ is infinitely small, this expression is infinitesimal, so it could be added as though it were a Riemann integral. This may be something new, but I have stumbled across something new, like integration, but defined differently. Similar to a Riemann integral adding up an infinite number of "slices" of a function, represented by $f(x)dx$, this construction adds up an infinite number of these: $f(x)^{dx}-1$. I'm not sure how to intuitively describe it (like the "slices" description on a Riemann integral). It is similar to a product integral in that $dx$ is in the exponent, but different in that the terms of the product integral are added rather than multiplied (subtracting 1 after each term so that the sum doesn't diverge). The notation would be something along the lines of this:$$\int_a^b\left(f(x)^{dx}-1\right)$$A more formal definition, similar to the definition of a Riemann integral, would be as follows:$$\int_a^b\left(f(x)^{dx}-1\right)=\lim_{n\to\infty}\sum_{k = 0}^{n-1} \left[-1+f\left(a+\frac{b-a}{n}k\right)^{\frac{b-a}{n}}\right]$$ Note: since raising a negative number to an irrational power is undefined on the real numbers, the argument $f(x)$ must be positive on the interval $(a,b)$. There may be a way around this, but I haven't found it because evaluating limits of sums is very difficult when the indices are in the exponents.
I have discovered several interesting results from this "power integral," by experimentation. None of these have I been able to prove, but by approximating these sums (up to 500 terms) on Desmos, I have been able to conclude at least some properties: $$\int_a^b\left(\left(f(x)^{-1}\right)^{dx}-1\right) = -\int_a^b\left(f(x)^{dx}-1\right)$$or, equivalently, $$\int_a^b\left(f(x)^{-dx}-1\right) = -\int_a^b\left(f(x)^{dx}-1\right)$$ "Power integrals" do not add by their intervals,$$\int_a^b\left(f(x)^{dx}-1\right)\neq\int_a^c\left(f(x)^{dx}-1\right)+\int_c^b\left(f(x)^{dx}-1\right)$$ nor do they preserve linearity, as a result of the differential being an exponent rather than a factor.$$\int_a^b\left(\left(f(x)+g(x)\right)^{dx}-1\right)\neq\int_a^b\left(f(x)^{dx}-1\right)+\int_a^b\left(g(x)^{dx}-1\right)$$Examples of some things I tried and their outputs:$$\int_0^x\left(C^{dt}-1\right)=0$$
$$\int_0^x\left(\left(Ct\right)^{dt}-1\right)=x\left(\ln(Cx)-1\right)$$
$$\int_0^x\left(\left(t^k\right)^{dt}-1\right)=kx\left(\ln(x)-1\right)$$
$$\int_0^x\left(\left(e^t\right)^{dt}-1\right)=\frac{1}{2}x^2$$

REPLY [2 votes]: I agree with  @pregunton
Learn about "product integration".  For example the book 
Dollard, John D.; Friedman, Charles N., Product integration with applications to differential equations. Foreword by Felix E. Browder. Appendix by P. R. Masani, Encyclopedia ot Mathematics and its Applications 10. Cambridge: Cambridge University Press (ISBN 978-0-521-17737-5/pbk). xxii, 253 p. (2011). ZBL1217.34018.