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In this post, Qiaochu Yuan remarks that 'it is convenient but misleading to write $$ \int f(x) \, dx=g(x) $$ [where the derivative of $g$ is $f$]'. This sentiment seems to be shared by many contributors here, and I don't understand why. To me, both definite and indefinite integration are both valid operations you can perform on a function, and there is nothing suspect about indefinite integration.

I know about the fundamental theorem of calculus, which (as far as I understand) explains the link between indefinite and definite integration. If by integration we mean computing the area under the graph, the fundamental theorem of calculus shows us that integration is the opposite of differentiation, since $$ \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) $$ This shows that every continuous function has an antiderivative. Since a clear link between integration and antidifferentiation has been established, we give the antiderivative the convenient label 'indefinite integral'. (This also explains why the definite and indefinite integration notations are so similar.) This label is fine, so long as we remember that integration is defined as finding the area under the graph, while antidifferentiation is defined as finding the inverse of the derivative.

Another result of the fundamental theorem of calculus is that $$ \int_{a}^{x}f(t) \, dt=\int f(x) \, dx $$ So obviously every indefinite integral can be rewritten in terms of definite integrals, but I don't understand the motivation behind this. If $F$ is an antiderivative of $f$, then why is it more correct to write $$ \int_{a}^{x} f(t) \, dt = F(x) \, , $$ compared to $$ \int f(x) \, dx = F(x) \, ? $$

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    $\begingroup$ I don't imagine that anyone is questioning the validity of integration. But writing the indefinite integral that way $A$. obscures the nature of $x$ as a dummy variable for integration and $B.$ incorrectly suggests that there is only one function $f$ that might be used there. Would you, say, write both $\int x\,dx = \frac {x^2}2$ and $\int x\,dx = \frac {x^2}2+1$ ? I assume you see the problem with that. Much better to write, e.g., $\int_a^x g(x)\,dx = f(x)$ or, better, $\int_a^x g(t)\,dt = f(x)$ $\endgroup$ – lulu Oct 23 '20 at 18:06
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    $\begingroup$ Saying $\int f(x)dx = g(x)$ is equivalent to saying that $\int f(t)dt = g(x).$ Then, you are left wondering: how can this be, since there is no relationship between $x$ and the dummy variable $t$. Then, you avoid the formality of $\int_a^x f(t)dt = g(x)$ by the very narrow interpretation of $\int f(t)dt = g(x)$ as only signifying that $g'(x) = f(x).$ $\endgroup$ – user2661923 Oct 23 '20 at 18:48
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    $\begingroup$ Suppose $f(x)=\sin(x) \cos(x)$. One person might find an anti-derivative of $\frac12 \sin^2(x)$. Another might find $-\frac12 \cos^2(x)$. A third might get $-\frac14 \cos(2x)$. These are obviously different (the first is non-negative and the second non-positive) but they are equally correct (or incorrect) and they differ by constants $\endgroup$ – Henry Oct 26 '20 at 1:44
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    $\begingroup$ I like to think that you can use that notation, but then you have to be careful to realize that "$=$" means something different. You might find this answer to a previous similar question useful. $\endgroup$ – JonathanZ supports MonicaC Oct 26 '20 at 13:09
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Basically, there's a type error: "$\int f(x)\,dx$" is a perfectly meaningful thing, but that thing is not a single function - rather, it's a set of functions.

The point is that a function doesn't have a unique antiderivative. For example, ${x^2\over 2}$ is an antiderivative of $x$ (with respect to $x$ of course), but so is ${x^2\over 2}-4217$. It's not the indefinite integral which is suspect, but rather the notation we use around it - specifically, the way we use "$=$." Properly speaking, $\int f(x)dx$ refers to a set of functions.

This is generally addressed by including a constant of integration, so that we write $$\int x\,dx={x^2\over 2}+C$$ to mean "The set of antiderivatives of $x$ is the set of functions of the form ${x^2\over 2} + C$ for $C\in\mathbb{R}$."

  • That said, blindly adding a constant of integration still doesn't always fix the problem: let $f(x)=-{1\over x^2}+1$ if $x>0$ and $-{1\over x^2}-1$ if $x<0$; what's the derivative of $f$, and does $f$ have the form $-{1\over x^2}+C$ for some fixed real number $C$?
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  • $\begingroup$ Hi Noah. Thank you for posting this answer. Could you please clarify that I understand it correctly? I'll do my best to summarise your points. If we interpret the statement $$\int f(x) \, dx = F(x)$$ literally, then it is false: the LHS represents a set of functions, whereas the RHS refers to one particular function. Even if we write $$\int f(x) \, dx = F(x)+C$$ we still have to interpret the RHS as referring to all possible values of $C$, rather than just one value. $\endgroup$ – Joe Oct 23 '20 at 18:49
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    $\begingroup$ @Joe Yes. And even then a single constant of integration might not be enough; see the end of my answer, or see Randall's answer. $\endgroup$ – Noah Schweber Oct 23 '20 at 18:53
  • $\begingroup$ Oh yes. I forgot about that. Also, it occurred to me that I might be abusing notation in another way. Technically, $f$ is the function, whereas $f(x)$ refers to $f$ evaluated at a particular point $x$. So, if we were being really pedantic, $$\int x^2 dx = \frac{x^3}{3}+C$$ means that the set of functions, where each function is defined on connected interval(s) by $\frac{x^3}{3}$ plus some constant, have a slope of $x^2$ at the point $x$ for all $x$. Is that right? $\endgroup$ – Joe Oct 23 '20 at 19:30
  • $\begingroup$ @Joe No - in my experience when we write "$\int f$" with no further decoration, we're looking at antiderivatives which are defined on as large a domain as possible. The set you're describing is the union of the sets of the form $\int_If(x)dx$ for $I$ a connected interval. $\endgroup$ – Noah Schweber Oct 23 '20 at 20:10
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    $\begingroup$ I like this example calculation: $$ \int \frac1x \, dx = \int 1 \cdot \frac1x \, dx = x \cdot \frac1x - \int x \cdot \left( -\frac1{x^2}\right) \, dx = 1 + \int \frac1x \, dx $$ It looks like this gives $0=1$ after subtraction of $\int \frac1x \, dx,$ but this calculation rather exemplifies that the equal signs should rather be equivalence signs; $u \equiv v$ if $u'=v'$. $\endgroup$ – md2perpe Oct 25 '20 at 1:02
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The other answers have made good points about constants of integration but this is not actually what I meant, although it is related. What I meant is what lulu says in the comments: writing antiderivatives this way misleads you about the relationship between the $x$ on the LHS (which is a dummy variable) and the $x$ on the RHS (which is not). The "real" $x$ on the LHS is one of the bounds of integration, which is being suppressed in the notation.

The sense in which this is misleading becomes clearer once you start considering double integrals, which is the context of the question you link to. If it makes sense to write $\int f(x) \, dx = g(x)$, then surely it also makes sense to write $\int g(x) \, dx = h(x)$, right? Then does it make sense to write

$$\iint f(x) \, dx \, dx = h(x)$$

or not? What do you think?

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    $\begingroup$ Another way of looking at this might be: when we write $f(x)=g(x)$ there's an implicit quantifier that this is true for all $x$. So we have $f(1)=g(1)$, $f(7)=g(7)$, $f(\pi)=g(\pi)$, and so on. But when we write $\int x\,dx = x^2/2+C$ we wouldn't then conclude that $\int 3\,d3=9/2+C$... $\endgroup$ – Carmeister Oct 24 '20 at 14:59
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    $\begingroup$ @Joe: the $x$ in $\int f(x) \, dx$ is a dummy variable because it's been integrated over; the "real" $x$ is the upper bound of integration, which is why it's more accurate to write this integral as $\int_a^x f(t) \, dt$ (and $t$ can be replaced with any other dummy variable). This issue already arises when you consider definite and indefinite sums; the partial sum $\sum_{k=0}^n f(k)$ is a function of $n$ because $n$ is the upper bound of the sum, and $k$ is a dummy variable; after it's been summed over it no longer appears. $\endgroup$ – Qiaochu Yuan Oct 24 '20 at 19:46
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    $\begingroup$ @Joe: it's exactly the same as the summation example. Maybe it's clearer if we change the "real" variable: $\int_0^y f(x) \, dx$ is a function of $y$, not $x$, because $y$ is what appears in the bounds of the integral. The $x$ has completely disappeared. Yes? $\endgroup$ – Qiaochu Yuan Oct 24 '20 at 20:06
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    $\begingroup$ @Joe: please pay attention to the words I'm using; you've claimed I said indefinite integration is "suspect" or "inaccurate" and that's not what I said, I said "misleading" and I meant "misleading." It's misleading precisely because the LHS does not appear to contain a dummy variable, but in fact it does. The LHS is shorthand for an integral of the form $\int_a^x t \, dt$ where $a$ is a constant and $t$ can be replaced by any other dummy variable. People implicitly understand this and it's mostly fine for single integrals but gets confusing for double integrals. $\endgroup$ – Qiaochu Yuan Oct 24 '20 at 20:28
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    $\begingroup$ @Joe: A helpful way of seeing whether a variable is a “dummy” is asking whether it would make sense to plug in a value for it. Does it make any sense to say “take $\int x\,dx$, then plug in $x=3$, to get $\int 3\, d3$”? No! If you’re thinking of $\int x\, dx$ as a function (or, better, a set of functions), and you want to see what the function does at 3, you don’t plug 3 in for $x$, you put 3 in as the upper limit of integration. So $x$ is not the variable representing the input of the function. $\endgroup$ – Peter LeFanu Lumsdaine Oct 25 '20 at 9:43
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For one example, the familiar old "formula" $$ \int \frac{1}{x} \ dx = \ln|x| + C $$ is false (unless you define the indefinite integral VERY carefully). This purports to say that any antiderivative of $f(x) = \frac{1}{x}$ must take the form $F(x) = \ln|x|+C$ for some fixed constant $C$. But this is only true over a connected interval. For example, the function $$ G(x) = \begin{cases} \ln|x| +1, & x < 0\\ \ln|x|-1, & x > 0\end{cases} $$ satisfies $G'=f$, even though it is not expressible in the form $\ln|x|+C$. Done right, we should only define indefinite integrals over intervals (this is due to the Mean Value Theorem).

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    $\begingroup$ It can be repaired by allowing $C$ to be a locally constant function and not just a constant. This is a very simple instance of de Rham cohomology, in this case the zeroth de Rham cohomology $H^0_{dR}(\mathbb{R} \setminus \{ 0 \}, \mathbb{R}) \cong \mathbb{R}^2$. $\endgroup$ – Qiaochu Yuan Oct 24 '20 at 8:17
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    $\begingroup$ And, of course, things get even worse if you try to extend this antiderivative to the complex plane. $\endgroup$ – Michael Seifert Oct 24 '20 at 13:41
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    $\begingroup$ @QiaochuYuan Discussed in detail here. $\endgroup$ – J.G. Oct 24 '20 at 20:58
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I think one clear way to see this is that the most pedantic notation would be something like this:

$$ f(x) \in \int f'(x)\,dx $$

You just have to think of $f$ as some element in a vector space, e.g. $f \in C^1(a,b)$, the set of continuous functions with continuous first derivative. So the integral operator is a linear transformation, a map of the kind $\int_a^x: C(a, b) \rightarrow C^1(a, b)$. This makes evident that the writing $$F(x) = \int f(x)\,dx,$$ where $F'(x) = f(x)$, could easily be notation abuse, though most of the time it's not the case, as the others have pointed out. The writer knows what they're writing: they assumed $F$ as a stand-in for a whole set of functions, but this is not always clear to the reader. Another problem this brings to surface is that the integral operator $\int_a^x$ as it has been written above is ill-defined, as it should point one element $f \in C(a,b)$ to one element $g \in C^1(a,b)$, not to a whole set of them. How to define this integral operator soundly is a question above my paygrade.

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If we define a function $F(x)$ as the antiderivate of the function $f(x)$ then by FTC we see that: $$\int_a^xf(t)\,dt=F(x)-F(a)\neq F(x)\forall f(x)$$ The problem lies in that for a given known derivative, $F'(x)=f(x)$, there are many different equations which satisfy $F(x)$ which differ by a constant, often notated by${}+C$. This can be seen in many cases where people neglect this "constant of integration": $$\int\frac{1}{ax}\,dx=\frac{1}{a}\int\frac 1x\,dx=\frac1a\ln|x|+C_1$$ $$\int\frac1{ax}\,dx=\frac1a\ln|ax|+C_2$$ which at first seem completely different but when the correct values of $C_1,C_2$ are chosen these are the same, see?

I think the point that the post is trying to make is that while in many scenarios it is easy to define a new function as the antiderivative of another, some may find this as misleading by thinking there is only one function $g(x)$ which satisfies these conditions, when in reality it is what we call a "family" of similar functions

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Setting the math of it aside, there is a real pedagogical/linguistic problem.

New students (reasonably) think the "Definite" and "Indefinite" integral are two variations on the same phenemonenon, whereas in fact the former is the phenemonon and the latter is notation that makes the former easier to compute.

In particular, the Fundamental Theorem of Calculus looks tautological the first time students see it, since they already "know" that integrals are anti-derivatives.

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  • $\begingroup$ Do you mean that the only reason antiderivatives are useful in the first place is that they help us compute areas? If so, that would make sense: both derivatives and integrals tells us about slopes and areas, but the antiderivative, in of itself, doesn't seem to be very useful. $\endgroup$ – Joe Oct 25 '20 at 18:05

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