ONE WEIRD INTEGRAL! Introducing the PRODUCT INTEGRAL intuitively!
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange askers cite it as the source of the "$\int x^{dx}-1$" trick and then ask whether manipulating dx this way is rigorous or has any geometric meaning.
Recommended by 0 people across 0 different months, 2020 to 2022.
What it is
Papa Flammy introduces the product integral and uses it to evaluate the odd-looking expression "integral of x^(dx) - 1". He builds up from the limit definition of the natural logarithm and Riemann sums, ending with x ln x - x + C.
What people said
Cites Kyan Cheung · 2020-04 · Mathematics Stack Exchange · on “Evaluating integrals of the form $\int f(x,dx)$”
I watched this video [this video] which showed how one could evaluate $\int x^{dx}-1$ : $$\int x^{dx}-1$$ $$=\int \left(x^{dx}-1\right)\frac{dx}{dx}$$ $$=\int \frac{x^{dx}-1}{dx}dx$$ By Riemann Integration we have $$\int^b_a f(x)dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i)\Delta x$$ Where $x_i=a+i\Delta x$ and $\Delta x=\frac{b-a}{n}$ . Thus $\lim_{n\to\infty}\Delta x=dx$ . Therefore, we can rewrite the integral as $$\int \lim_{\Delta x\to0}\frac{x^{\Delta x}-1}{\Delta x}dx$$ (Note that $\lim_{\Delta x\to0}\frac{x^{\Delta x}-1}{\Delta x}=\ln x$ ) $$=\int \ln xdx$$ $$=x\ln x-x+C$$ I am wondering about what would happen if instead of integrating $x^{dx}-1$ we integrated $f(x,dx)$ for some continuous fu
Cites Eduardo Magalhães · 2020-05 · Mathematics Stack Exchange · on “$\int x^{dx}-1$”
If you go to Flammable Maths [LINK] '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 [this video] .
Cites zcxqwezxc · 2021-04 · Mathematics Stack Exchange · on “Integral of $\ln(dx+1)$”
I want to evaluate the following integral integral of ln(dx+1) [LINK] , yes the integral is inside the ln, NOT outside The inspiration came from this video integral of x^dx-1 [this video] , where the integral was divided and then multiplied by dx, so when evaluating my integral I wanted to use the same trick to get this integral [LINK] . Then I took the limit as dx approaches 0 to get infinity [LINK] . So finally I ended up with an integral of infinity * dx which is just equal to infinity [LINK] . I know the question doesn't make any mathematical sense, so I'm just asking if the question can be solved "symbolically"? Am I doing any mistakes in my calculation of the inegral? Is there another way t
Cites user905694 · 2022-03 · Mathematics Stack Exchange · on “Is $\int x^{dx}-1$ well defined?”
My question is precisely the question here [LINK] . A quick summary of the algebraic process to "solve" the integral follows (taken from miniprime1 [LINK] 's answer): $\int x^{dx}-1 = \int \frac{x^{dx}-1}{{dx}} {dx} = \int (\lim_{h \to 0}\frac{x^h-1}{h}) {dx} = \int \ln x {dx} = x\ln x-x+const.$ The reason I ask this question (again) is that to me, none of the answers to the linked question actually answer the question, particularly what the geometric meaning of these manipulations are. I can't wrap my head around thinking about how this integral relates to Riemann sums or Taylor Series. I have some further problems/questions as well: How well-defined is the operation of multiplying the inte