# Understanding the definition of "Indefinite integral".

Everywhere I have looked up, see here, the indefinite integral is defined as: $$F'(x)= f(x) \iff \int f(x) \, dx= F(x) + C$$

From what I understand if $$f$$ has an antiderivative $$F$$ then the set $$\{F(x) + C : C \in \mathbb{R}\}$$ is called "indefinite integral" of $$f$$.

But I don't understand the definition, $$F'(x) =f(x)$$... okay, where? What values of $$x$$? For every $$x$$ in the domain of $$f$$? Oh then I think we have a problem.

Let $$f(x)= 0$$ for $$x \in (0,1)\cup(1,2)= \text{Domain}(f)$$, we then go on to say $$\displaystyle\int f(x)\, dx =\int 0 \ dx = C = \{ F(x) = C, \forall x \in \text{Domain}(f): C \in \mathbb{R}\}$$, a set of constant functions... Sure but what if $$F:(0,1)\cup(1,2)\to \mathbb{R}$$ given by, $$F(x)= \begin{cases}1, \ x \in (0,1) \\ \\ 2, \ x \in (1,2) \end{cases}$$ then $$F'(x)= 0 = f(x)\, , \ \forall x \in \text{Domain}(f)$$ but $$F(x)$$ is not a constant function on $$\text{Domain}(f)$$ and so $$F(x)\notin \displaystyle\int f(x) \, dx$$ and yet $$F'(x)=f(x), \forall x$$. What gives? Surely, that definition is incomplete?

NOTE: I have not mentioned $$f$$ is Riemann integrable or not so writing $$F(x)=\int_a^x f(t) \, dt$$ is already a no-go.

• Commented Aug 28, 2022 at 8:24
• @MartinR Yes, but like you can't just use two different constant of integration for the same antiderivative? $F(x) +C_1$ and $F(x)+C_2$ are different antiderivatives. I didn't get that part. Commented Aug 28, 2022 at 8:31
• Tangential comment, but I avoid using the term “indefinite integral” entirely. I’ll say things like “let $F$ be an antiderivative of $f$”. Commented Aug 28, 2022 at 9:51
• I don’t think this is a clear question. Commented Aug 29, 2022 at 0:05
• You can just switch every $C$ to a locally constant function $C(x)$. Commented Aug 30, 2022 at 9:49

You make a good point. Indefinite integral notation and manipulation tends to be a little loose, in my opinion. And you’re also right that, on a disconnected domain, it’s perfectly allowable to have multiple different constants. Famously $$\int1/x=\ln|x|+C$$ can have a different $$C$$ on $$(-\infty,0)$$ to $$(0,\infty)$$. I wouldn’t worry too much: $$\int f$$ just stands for “some antiderivative of $$f$$, on some subdomain of the domain of $$f$$”.

For instance it’s common to perform u-substitutions and the like with indefinite integrals - but, without any bounds of integration, those substitutions might not be valid! You then end up with an antiderivative that works in some places, but not in others. And as far as indefinite integration cares, that’s ok. $$\int f$$ just notationally streamlines the integral calculus.

• I think I understand now. However could you explain to me why Wikipedia says, $f(x)=F'(x)$ should be on open intervals or disjoint union of open intervals. Why not just any interval works? What happens if $F'(x) = f(x)$ on say $[1,2)\cup[3,4]$, can I not use different constants on each intervals $[1,2)$ and $[3,4]$? So why open intervals? Commented Aug 29, 2022 at 5:47
• @William That concern crops up in different places in analysis. There are some different conventions. The trouble is, to say $f(x)=F’(x)$ is to also say that $F$ is differentiable i.e. $\lim_{h\to0}\Delta F/\Delta h$ exists. But for that limit to make sense, we need $F$ to be defined at $x$ and at all $x+h$ for small $h$. So, that means the domain of $F$ should be open at $x$. However, a different convention would be to say this: $F$ is differentiable on $[a,b]$ means it is differentiable, in the normal sense, on $(a,b)$, and the one-sided derivative limits exist at $a$ and $b$. Commented Aug 29, 2022 at 5:56
• Thank you. So if I'm taking differentiability on endpoints to mean one sided derivatives and differentiability on interior points are the usual two sided derivatives, then I can summarise the whole drama like this: if $F'(x)=f(x), \ \forall x \in D$ then if $D$ is an interval (of finite or infinite length, open, closed, half open, doesn't matter) then the set of all antiderivatives of $f$ on $D$ is simply, $$\{F(x)+C, \forall x \in D : C \in \mathbb{R}\}$$ and (...continued) Commented Aug 29, 2022 at 6:49
• And if $D$ is a union of pairwaise disjoint intervals, like say $D=(a,b) \cup [c,d) \cup [e,f]$ where $(a,b)\cap[c,d)= (a,b)\cap[e,f]=[c,d)\cap[e,f]= \varnothing$ then the set of all antiderivatives of $f$ is $$\{g(x) =\begin{cases} F(x) +C_1, x \in (a,b) \\ F(x)+C_2, x \in [c,d) \\ F(x)+C_3, x \in [e,f] \end{cases} : C_1, C_2, C_3 \in \mathbb{R} \}$$ but this is way too cumbersome, we might as well leave it at $F(x)+C$ with an understanding that if $D$ is not an interval but union of disjoint pairwise intervals then we may just use different arbitrary constants on each of intervals of $D$? Commented Aug 29, 2022 at 6:50
• @William That’s all correct. Commented Aug 29, 2022 at 7:17

The $$+C$$ isn't just a constant, per se. It's a locally constant function. This is true for any defined integral and any domain. Over connected domains, a function is constant if and only if it's locally constant. Otherwise, it would just have to be constant over every connected component. To be a little more rigorous, we can think of the indefinite integral as the antiderivatives on a connected component of the domain.

For example,

$$F(x) = \begin{cases} -\frac{1}{x} + C_1, & \text{if x < 0} \\ -\frac{1}{x} + C_2, & \text{if x > 0} \end{cases}$$ is the "correct" general antiderivative of $$f(x) = \frac{1}{x^2}$$ on its usual domain $$\mathbb{R}\backslash\left\{0\right\}$$.

I like to think of $$+C$$ as "$$+$$expression whose derivative equals $$0$$," or $$+C$$ to represent an entire family of functions with a derivative of $$0$$. Also, we know $$F(x) + C_1$$ and $$F(x) + C_2$$ are two different expressions, but they are still elements of the set $$\left\{F(x) + C: C \in \mathbb{R}\right\}$$.

Here's another way of explaining all this. An indefinite integral is the equivalence class of functions under the equivalence relation of having the same derivative. So an indefinite integral is not one function, per se, but a set of functions.

I'm going to sidetrack here but in abstract algebra and linear algebra, The derivative is a linear transformation (homomorphism) that sends nonzero vectors (functions) to $$0$$, so we can only define its inverse (the indefinite integral) only up to adding something whose derivative equals $$0$$.

(Opinion) I do agree that people loosely define indefinite integral notation $$\int$$, but I think it was done to help keep people sane. I doubt any professor grading a Calculus 1 class would want to grade homework having the solutions that are something like $$\ln{|x+r|}$$, and the students are forced to write a piecewise-defined function. It's inconvenient. It's kind of like how some people say $$\sqrt{-1} = i$$. It's impossible to take the square root of a negative number, yet people still abuse the square root notation, so some people accept that $$\sqrt{-1} = i$$ even though others would argue it's wrong (not to mention that $$\sqrt{-1}$$ could equal either $$i$$ or $$-i$$ depending on which branch you choose, but that's a different story).

• As I see it, you seem to have went with $F'(x)=f(x), \forall x \in \text{Domain}(f)$ then how would you define the "indefinite integral"? Surely you can't define it, as the set $\{F(x)+C, \forall x \in \text{Domain}(f) : C \in \mathbb{R} \}$ because this set throws away your adding "locally constant functions" out the window. How would you define indefinite integral, then? Commented Aug 28, 2022 at 9:11
• @William I wasn't trying to give a rigorous definition of what an indefinite integral is. But to answer your question, I think I would define it how Wikipedia defines it (en.wikipedia.org/wiki/Antiderivative). I believe if $\text{Domain}(f)$ is non-disjoint, then $\left\{F(x)+C,\forall x \in \text{Domain}(f) : C \in \mathbb{R}\right\}$ should be the set of all antiderivatives of $f(x)$. I think for piecewise functions it's different, just like my example. Truth is, I don't think there is an agreed way to define the indefinite integral. Commented Aug 28, 2022 at 9:30
• @William I found this. Maybe this video could help you better than I can. youtube.com/watch?v=u4kex7hDC2o&t=451s Commented Aug 28, 2022 at 9:38
• Thanks, that was helpful. I'll check it out. However, could you explain to me why does Wikipedia say, "If the domain of F is a disjoint union of two or more (open) intervals, then a different constant of integration may be chosen for each of the intervals." Why open intervals in the bracket? Why can't this work for Domain being disjoint union of any interval (open, closed, half open)? Commented Aug 28, 2022 at 16:47

A definite integral is a number, an indefinite integral is a function.

In particular, a definite integral is the area enclosed between the function and one of the axes and the curve in the delimited interval.

The indefinite integral is the inverse functional of the derivative. Meaning that the function provided by the indefinite integral, when derived, should give as a result the original function that was integrated.

The interval where said function is defined is exactly that: the interval where the function is defined. In example $$F(x)=1/x$$ is defined in $$\mathbb{R} - \left\{0\right\}$$

I feel thate's confusion between the indefinite integral (functional returning a function) and the definite integral (functional returning an area). The simbology is similar, but they actually are 2 different functionals. That's what i was thaught.

• The indefinite integral maps a single function to a family of functions, not just "a function." That's central to the question, which is actually about what kind of family of functions you get for a disconnected domain. Commented Aug 28, 2022 at 22:20

The main problems here are the domain of definition and constant of integration. Both of these can be solved if you restrict to intervals (open/closed/clopen), which is how it is usually defined in texts.

Thus, for a Riemann integrable function $$f : (a,b)\to \mathbb{R}$$, the antiderivatives of $$f$$ is the set $$\{F(x)+C|F : (a,b)\to \mathbb{R}, F'(x) =f(x) , C\in \mathbb{R}\}$$

The indefinite integral is more of a constructive tool for students to learn and convenient notations for people to solve problems. Thus, although it is fun to wonder how one can make it rigorous, it may not be necessary.

• You say "Riemann integrable $f:(a,b) \to \mathbb{R}$" but there is no such thing as Riemann integral of a function on open interval? Commented Aug 28, 2022 at 9:02
• There is. Riemann integrability is defined for bounded functions on intervals of finite length. The nature of the interval does not matter, (open/closed/clopen). Commented Aug 28, 2022 at 9:41
• Well that's weird. I was always under the impression that Riemann integral makes use of the "endpoints" of the interval in the definition while partitioning... First we define $\displaystyle\int_{[a,b]} f$ and then extend to $\displaystyle\int_{(a,b)}f$. Not the other way around. Commented Aug 28, 2022 at 16:29
• @William You are right, the Riemann integral technically needs a compact interval to be properly defined. However, you can take an $f$ defined on an open (and finite!) $(a,b)$ and extend it to the closure $[a,b]$ by arbitrarily choosing values at the endpoints. Then the value of the Riemann integral as defined by Riemannian sums won't depend on the values chosen, so the whole construction of the Riemann integral can also be carried out on open intervals. One should however think of the Riemann integral to require compact domains! Commented Aug 28, 2022 at 18:29
• Nothing stops you from partitioning $(0, 1)$ into $(0,1/2)$ and $[1/2,1)$. It may be customary to define it on closed intervals then extend it; but does not have to be that way. Many texts define it on intervals without specifying whether end points are included or not from the get go. The end points do not matter in the definition of the integral. Besides, the point of the post is how to address the problems OP has brought up, not the nature of the intervals. Everything I said is still true if we replace the open with closed intervals, except you have to take one-sided derivatives. Commented Aug 29, 2022 at 0:45