# Linearity of indefinite integrals

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.

• I wouldn't say, strictly speaking, that $\int 0 dx = C\in\mathbb R$, Strictly speaking, $\int 0 dx = (x\mapsto C)$ for any $C\in \mathbb R$. – 5xum May 10 '18 at 13:48
• Yes, I agree. The point is that there are infinitely many functions $x \mapsto C$. – Gibbs May 10 '18 at 13:52
• 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... – David C. Ullrich May 10 '18 at 13:55
• 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. – Giuseppe Negro May 10 '18 at 14:29
• @GiuseppeNegro I do not agree. We should be more careful when studying them exactly for reasons like these. – Gibbs May 10 '18 at 14:37

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.