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I am trying to make sense of 'linearity' of indefinite integrals.

Let us restrict to the 1-dimensional case. My point is that $\int 0\, dx = C \in \mathbb{R}$, so I cannot really say that $\int$ is a linear operator. Indeed, linearity of $f \colon V \rightarrow W$ ($V,W$ vector spaces) implies $f(0) = 0$.

In order to define $\int \colon V \rightarrow W$ in a good way, I should introduce an equivalence relation $\sim$ on $W$ saying that two elements are in the same equivalence class if they coincide up to a constant. So $\int$ becomes linear as operator $\int \colon V \rightarrow W_{/\sim}$. Assume a primitive of $f$ is $F$. Then $$\int f(x)\,dx = [F(x)] \in W_{/\sim}$$ or, as usual, $F(x)+C, C \in \mathbb{R}$. So I would say that the result of an indefinite integral is actually a coset in some quotient vector space. This even solves the problem that $\int \colon V \rightarrow W $ is not a well defined function.

My question is: is this a good way to think, or is there a better one? I have never seen such a thing, neither in a course of Analysis, nor in the books I have read. I am wondering why. It seems a very natural thing to do when introducing indefinite integrals.

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    $\begingroup$ I wouldn't say, strictly speaking, that $\int 0 dx = C\in\mathbb R$, Strictly speaking, $\int 0 dx = (x\mapsto C)$ for any $C\in \mathbb R$. $\endgroup$ – 5xum May 10 '18 at 13:48
  • $\begingroup$ Yes, I agree. The point is that there are infinitely many functions $x \mapsto C$. $\endgroup$ – Gibbs May 10 '18 at 13:52
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    $\begingroup$ Yes, this seems too me to be the right way to think of it. "The" indefinite integral is not even well-defined unless you say its value lies in a certain quotient space... $\endgroup$ – David C. Ullrich May 10 '18 at 13:55
  • $\begingroup$ IMHO, indefinite integrals are a tool from the past and should be abandoned. That's why they are not studied with deep attention nowadays. Here's a recent example of their dangerousness. $\endgroup$ – Giuseppe Negro May 10 '18 at 14:29
  • $\begingroup$ @GiuseppeNegro I do not agree. We should be more careful when studying them exactly for reasons like these. $\endgroup$ – Gibbs May 10 '18 at 14:37
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Of course this is a good way to think. But we need some extra notation. Given an interval $J\subset{\mathbb R}$ call two functions $F : J\to{\mathbb R}$, $\>G:J\to{\mathbb R}\>$ equivalent if $F-G$ is constant on $J$. It is then easy to see that the equivalence classes $\langle F\rangle$ form a real vector space in the obvious way. Let $V$ be the subspace generated by the $C^1$-functions on $J$. Then $$D:\quad V\to C^0(J), \qquad\langle F\rangle\mapsto F'$$ is a linear isomorphism with inverse the undetermined integral: $$\int:\quad f\mapsto \int f(t)\>dt\ .$$ Thereby each "differentiation rule" generates an "integration rule" as follows: $$F'=f\quad\Longrightarrow\quad \int f(t)\>dt=\langle F(t)\rangle\ .$$ And on and on.

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  • $\begingroup$ Yes, it goes without saying that the problem related to the definition of the operator of derivation should be fixed as well. Thanks for the answer. $\endgroup$ – Gibbs May 10 '18 at 14:47
  • $\begingroup$ Maybe a good tool to formalise this stuff is de Rham cohomology. $\endgroup$ – Gibbs May 10 '18 at 15:37
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See the section on operator notation here: https://en.wikipedia.org/wiki/Linearity_of_integration

It should answer your question. We can treat the integral as the inverse of the differential operator that takes its values in some vector space modulo the constant functions as the whole kernel of the differential operator.

Remark: There are possibly better, or perhaps more complicated ways to think of the integral. Depending on what you need it for I guess. It sounds like you're definitely making sense of it.

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  • $\begingroup$ Yes, it seems like they do exactly what I am doing. I really wonder why generally people do not talk about indefinite integrals in this way in first courses of Analysis. $\endgroup$ – Gibbs May 10 '18 at 13:56

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