Can you integrate without a $dx$

Question

Long ago I realized that manipulation of derivatives was possible using algebraic quantities. One could take a differential instead of derivatives $$ d[\sin x]=\cos x\ dx $$ $$ \frac{d[\sin x]}{dx}=\frac{\cos x\ dx}{dx}=\cos x $$ My question is about the analog of this process with integrals without a $dx$ for example $$ \frac{\int\sin x}{\int x} $$ Here $\int$ is the inverse of the $d$ operator. There is deliberately no $dx$ so the integrals cannot be evaluated in the traditional sense. This is more the foundation for an idea than anything explicit.

Has this been looked at by anyone in the past? Does anyone know of anything similar to this?

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edited remarks

To clarify, consider $\int \sin x$. Because the input is not a differential but a finite quantity, it is expected that the integral would diverge to infinity. I desire to define how the integral diverges in a way like $$ \int \sin x=f(x)\int x $$

which is analagous to differentials (e.g. $d[\sin x]=\cos x\ dx$) and therefore the fraction $\frac{\int \sin x}{\int x}$ could be evaluated as a finite quantity $f(x)$.

I'm not looking for a simple "you can't do that" or the definition of an integral in the conventional sense. I want to know if anyone has explored this unconventional expression or has any insight into how it might be evaluated or even what it means.

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Marked as a duplicate, but probably because my new idea goes against what we were all taught in calculus class.

Answer 1

The general notation for integrals does not requires to write $\mathrm d x$. If $f$ is a function one usually defines $$ \int f = \int f(x)\mathrm d x $$ while for a more general measure $\mu$ one often finds $$ \int \mathrm d\mu = \int \mu(\mathrm d x) $$ So this is just a notation but it is as you say a useful notation since it emphasises the fact that the Lebesgue measure as nice properties under change of variable.

Then, I think your problem is not about the notation but more the problem of defining a continuous sum of non-infinitesimal quantities. Unfortunately, this only works if the quantity you want to sum is $0$ almost everywhere. So I would reserve a notation without $\mathrm d x$ to discrete measures. For example $$ \int_{x\in\mathbb{R}} f(x)\mathbf{1}_{x=c} = \int_{\mathbb{R}} f(x) \delta_{c}(\mathrm{d x}) = f(c) $$

Answer 2

The real definition of, for example, Riemann integration is based on a partition of $[a,b]$, $P = \{ x_0 = a, x_1 , ... , x_{n-1}, x_n = b \}$ with $x_{i+1}>x_i$, and the lower sums $$ L(f,P) = \sum_{i=0}^{n-1} \inf_{z \in [x_i,x_{i+1})}f(z)(x_{i+1}-x_i) $$ and upper sums $$ U(f,P) = \sum_{i=0}^{n-1} \sup_{z \in [x_i,x_{i+1})}f(z)(x_{i+1}-x_i). $$ If for all $\varepsilon >0$, there exists a $\delta >0$ such that for some $\mathcal{P}_\delta$ satisfying $ \min_{i} (x_{i+1}-x_i) < \delta$, $U(f,P_\delta) - L(f,P_\delta) < \varepsilon$, then the integral exists, as the common limit of the upper and lower sums as $\varepsilon \rightarrow 0$. There's lots of other versions of the integral you can motivate the same way, like Stieltjes or Lebesgue.

There's no $dx$ without $\int$, because the symbols are just ornaments. The integral is a limiting process as the norm of the partition goes to zero. The symbols have no magic to them, they're entirely ornaments that suggest something about how the limits work to the reader, with the understanding that the reader knows they are symbols gesturing towards the limiting process. Attaching meaning to the $dx$ in the integral or derivative is a mistake, because it leads you astray from understanding the limiting processes themselves, and why they work or don't.

Answer 3

The $ dx $ is necessary. It defines the integration variable. for one variable function, we can forget it, but for multivariate functions, we must specify the integration variable.

For example $$\int x^3=\frac{x^4}{4}$$ But $$\int(x^3+y^2)=???$$ $$\int (x^3+y^2)dy=x^3y+\frac{y^3}{3}$$