I still haven't come across any integrand such as $\int x/dx$ and now that I checked multiple calculators they revert the integral to $\int x dx$. So I think the closest I would have been to is the logarithm $ln(x)=\int_{0}^{x} dt/t$, isn't this supposed to be possible by substitution? What do I do if this infinitesimal appears in an integrand? Is it because of the infinitesimal rectangles in the Riemann sum defintion that $1/dx$ never occurs?
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3Does this answer your question? Usage of dx in Integrals – DatBoi Oct 01 '21 at 01:09
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2Strictly speaking $\int\frac{x}{dx}$ is not a well defined expression, and could be interpreted in a variety of ways. Classically, $dx$ is regarded as an infinitesimal, and thus you are essentially dividing by zero. – Graviton Oct 01 '21 at 04:23
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Satisfactory explanation of infinitesimals were given in non Standard analysis by Abraham Robinson. A new number system called hyperreal number system was introduced by him, which contains infinitesimals and infinities. According to that, An infinitesimal 'dx' is a number which is smaller than all real numbers but greater than 0 i.e dx is infinitesimally closer to 0. It's denoted as st(dx)=0(where st denotes the standard part function). Now if we consider the reciprocal of dx i.e 1/dx it's obviously something which is greater than all real numbers (As dx is smaller than all Real numbers). So it's something like infinity (An infinity in hyper real number system). Considering your function x/dx. So according to non standard Analysis st(x/dx) tends to infinity. So the required integral ofcourse diverges.
RAHUL
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Why downvoting this answer ? It is as absurd as upvoting this strange question by arclengths. The merit of the answer of Rahul is to give a context in which $x/dx$ could (?) make sense. – Jean Marie Oct 01 '21 at 08:52
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Provided that $f(x,0)=0$ and is continuous, one can always work backwards from the definition of the Riemann sum and - just like how certain definitions are assigned to delta functions - assign the following value to such an integral
$$\int_{x=a}^{x=b}f(x,dx) = \lim_{n\to\infty}\sum_{k=1}^n f\left(a+k\frac{b-a}{n},\frac{b-a}{n}\right)$$
This assignment is sensible in the sense that we get a computable limit that corresponds with the usual case of $f(x,y) = g(x)\cdot y$. But like any other notation, it has to have some sort of use to be practical or widespread. I would wager that most applications which could use this idea become overly complicated if you ditch their standard formulations.
However, the example you gave cannot be saved since $f(x,0)\neq 0$
Ninad Munshi
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