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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.

Eduardo Magalhães
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    As you know, the integral is a limit of a sum, which is why we denote it with a long S (for sum). For a product integral this notation is confunsing, It would be better to write som product symbol. See https://en.wikipedia.org/wiki/Product_integral – Jens Renders May 21 '20 at 15:22
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    Nothing than a pocket player trick abuse of notation. – Michael Hoppe May 21 '20 at 15:38
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    Why should I look for over $12$ minutes at a video for explanations that could be given in print in a few lines? I'll would be ready then to discuss the content. – Christian Blatter May 21 '20 at 17:37

2 Answers2

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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 $\int\ln xdx=x\ln x-x+C$.

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

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

miniprime1
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    I do not understand why the other answer is accepted, but the OP does not seem to be asking for this computation. Also a remark, this is on the board in the first few seconds of the linked youtube video – Calvin Khor Jul 11 '21 at 04:42