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I am very uncomfortable with indefinite integrals, as I have difficulty giving them a precise sense that matches how they are written and the usual meaning of other symbols.

For example, when one writes $$ \int \sin(x) \,\mathrm{d}x = -\cos(x) + k$$ then the status of both $x$ and $k$ is pretty unclear (which quantifier is in front of each of these variables?)

Of course, I know how to translate this sequence of symbols into a proper mathematical sentence, but for students, it seems utterly challenging to give a precise meaning to this, in particular at the stage when we try to explain the distinction between a function and its value at a point, or when we consider functions of several variables.

In my experience, this kind of notation tends to reinforce the student's habit of seeing mathematical notation as a kind of voodoo formula that can be manipulated using certain incantations: no one probably knows what the incantation means, but using the wrong magic is forbidden for some reason (maybe it will summon an efreet?). On the contrary, I would like to show them the meaning behind everything we teach them.

For this reason, I try never to use indefinite integrals, relying instead on moving bounds, e.g.: $$ \forall a,x \quad \int_a^x \sin(t) \,\mathrm{d}t = -\cos(x)+\cos(a).$$

Questions: what possible issues are there in avoiding completely indefinite integrals? Is there any pedagogical advantage to using them? Is there a third way to go?

Edit: Let me add another issue with the notation $$ \int \sin(x) \,\mathrm{d}x = -\cos(x) + k$$ On the right-hand side, $x$ is implicitly a variable (as opposed to the parameter $k$). Still, on the left-hand side, it is both a global variable and a local (mute) variable of integration. Given the (already somewhat weird) role we give to the integration variable in definite integrals, this confusion bothers me a lot. Does anyone even imagine writing something like $$ \sum_n n^3= \frac{n^2(n+1)^2}4+k?$$

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    $\begingroup$ Just explain that it is a weird notation for the antiderivative, that the class might want to change, but you won't be able to go against the mathematical establishment $\endgroup$
    – vonbrand
    Commented May 18, 2014 at 13:28
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    $\begingroup$ Additionally, $+ C$ is not always sufficient. For example, the general antiderivative of $1/x$ is $$\begin{cases} \ln(x) + C_1 & \text{when $x > 0$,} \\ \ln(-x) + C_2 & \text{when $x < 0$.} \end{cases}$$ $\endgroup$ Commented May 18, 2014 at 14:36
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    $\begingroup$ Even if one takes the trouble to develop a self-consistent, coherent, optimized notational and conceptual system, there is no enforcement mechanism (well, ...) to make people behave sensibly in this or any other way. In particular, I guess we find ourselves needing to teach people how to cope with ambiguity or amorphousness, rather than telling them that there are absolutely-reliable universal conventions that will never change, etc. We even find ourselves forced (!) to deal with self-inconsistent, misleading, annoying conventions and "definitions". Dang. Hm. $\endgroup$ Commented Apr 14, 2016 at 23:23
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    $\begingroup$ Long ago, as a teaching assistant, I taught with a professor (a specialist in PDEs and harmonic analysis) who insisted that all integrals be written with limits. We were teaching a calculus course directed at students in biological science ("pre-meds"). He found that this approach, carefully enunciated, helped these students understand the ambiguities (notice the ambiguities don't vanish - rather they are absorbed into the sometimes indeterminate choice of limits). My sense was that the approach was quite successful. Other approaches can be successful too. $\endgroup$
    – Dan Fox
    Commented Oct 28, 2018 at 17:50
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    $\begingroup$ @FrancoisG.Dorais Technically you can still write $+C$; is just that in general $C$ is only guaranteed to be locally constant, not globally constant. $\endgroup$
    – J.G.
    Commented Nov 9, 2018 at 17:21

14 Answers 14

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On quizzes, homeworks, and tests, I repeatedly ask questions like this:

Find three different functions that have derivative equal to $x^2 + x$.

Forcing them to do antiderivatives and deal with the quantifier on the +C without staring at the notation helps some of them separate the +C from the voodoo magic.

I do a similar thing in college algebra classes to deal with unpleasant quantifiers:

Find three different polynomials with variable x that have roots at $x=2$ and $x=3$.

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    $\begingroup$ Also I find it helpful to do analogous problems completely graphically: sketch 2 different functions which have derivative $f$, where $f$ is graphed below. $\endgroup$ Commented May 18, 2014 at 21:22
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    $\begingroup$ A trick I've always liked: when teaching the antiderivative, tell student's it's "C, plus the other stuff." That is, make "C +" the first thing you do, rather than something you can forget to tack onto the end. (Obviously caveats for antiderivative of 1/x and such) $\endgroup$
    – bbayles
    Commented May 19, 2014 at 2:57
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    $\begingroup$ I like your idea "Find three different functions ...", which I don't believe I ever tried (or even had an awareness of trying). As for what I did do, see this 5 December 2008 AP-Calculus post at Math Forum. (If I get less busy later, I might LaTex the essence of it here unless someone else posts the same idea.) $\endgroup$ Commented May 19, 2014 at 15:51
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No, it is a bad idea to avoid indefinite integrals, the reason being simply that your students will encounter them elsewhere, and therefore need to be familiar with them. Calculus is a service course. The purpose of the course is to make science and engineering majors fluent in the language of calculus as used in their fields.

Rather than always using moving bounds, why not just tell your students that when we write $$ \int \sin(x) \,\mathrm{d}x = -\cos(x) + k$$ we're really describing a set of functions on each side of the equals sign, with an implied quantifier over $k$ on the right? In the US educational system, students are introduced to the notion of a "solution set" very early on, so this should be natural to them.

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    $\begingroup$ Your answer makes sense, but the concept of solution set is commonly a serious issue with first year students in France (at least in my experience). $\endgroup$ Commented May 18, 2014 at 19:06
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I go a step further than Thomas (see Henry Towsner's answer). In my view, $$ \int f(x) \ dx = \{ F(x) \ | \ F'(x)=f(x) \} $$ On a connected domain, it is true that $F'(x)=G'(x)$ implies $F(x)-G(x)=c$ hence, given an integrand which is continuous (or piecewise continuous, insert your favorite weakened set of functions here) we may write: $ \int f(x) \ dx = \{ F(x)+c \ | \ c \in \mathbb{R} \}. $ Then, I tell the students that nobody wants to write this all the time so we drop the $\{ \}$ and simply summarize it with a slogan: the indefinite integral is the most general antiderivative. This really means it is the set of all functions which form antiderivatives of the integrand. Moreover, I warn them, for this reason the usual rules of equality do not apply. In fact, $\int x \, dx = x^2/2+c$ and $\int x \, dx = x^2/2+42+c$ are the same answer.

Truth is, we are working on equivalence classes of functions as we study indefinite integration as the notion of equality has properly been replaced with congruence. Moreover, if we take function space and quotient by the subspace of constant functions then for some connected domain the indefinite integral and derivative operator are inverse operations. This I do not tell first semester calculus students, however, in a good semester of linear algebra I think it makes a nice quotient space discussion.

Obviously, the question remains, why on earth should we use the same symbol $\int$ for $\int f(x) \, dx$ and $\int_{a}^{b} f(x) \, dx$? These are radically different objects. The indefinite integral is a set of functions whereas the definite integral is a number. The answer is the FTC. That said, I think it important to make a point of emphasizing just how surprising it is that these two ideas have any connection at all.

Edit added 6/4/19 I see what Michael is saying in the comments about the error of me confusing a function with its value. The thing is, the notation $\int f(x) \, dx$ already indicates a variable for the functions in play. I'm not at peace with an answer which has $x$ on the LHS but not the RHS (say $\int f(x) \, dx = \{ F \ | \ F' = f \}$. If we are to go this route to talk about functions rather than their values then I'd probably adopt the notation $\int f$ for the indefinite integral of $f$ and keeping with the legalism of my current answer here I'd write: $$ \int f = \{ F \ | \ F' = f \} $$ I think I prefer my answer with its abuse, but I see why others would rather engage in the subtlety which Michael points towards.

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    $\begingroup$ Since in first years I try to enforce the rule of distinguishing the function $f$ from its value $f(x)$ at $x$, even considering $x^2/2+c$ as a set of function bothers me. There is almost no issue at all when one masters these notions, but I do not see how to avoid deep confusions for already struggling students. $\endgroup$ Commented May 18, 2014 at 20:47
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    $\begingroup$ @BenoîtKloeckner I share your frustration with getting students to distinguish $f$ and $f(x)$. But, I suspect their inability to get the concepts has much more to do with their reluctance to see math as more than problem solving. I can't fault them, they've been fed a steady diet of canned problems replete with their algorithmic solution mantras. It's a shock when they find that math is a story and you can choose your own adventure. $\endgroup$ Commented May 19, 2014 at 5:01
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    $\begingroup$ Many teachers believe it's the students fault if they can't distinguish $f$ from $f(x)$. That's a good point of view, which we should apply to every problem in teaching. But consider your suggestion $\int f(x)dx =\{F(x)\mid F'(x)=f(x)\}$: the RHS is probably $\mathbb{R}$, so maybe you meant $\int f(x)dx =\{F \mid F'=f \}$? Ok, now lets drop $\{\}$ and say $=$ denotes congruence. Then we should still correctly write $\int x dx = (x\mapsto \frac{x^2}{2})$. Now try doing any concrete computation, like integr. by subst. or parts with this notation, without confusing $f$ for $f(x)$. $\endgroup$ Commented May 28, 2018 at 8:35
  • $\begingroup$ Hey James, can we talk? $\endgroup$ Commented May 31, 2019 at 19:54
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Here's one problem with trying to replace indefinite integrals with definite ones.

With the indefinite integral, we can say

$$ \int \frac{\mathrm{d}x}{1+x^2} = \arctan x + C $$

where $C$ can be any constant. However, the only antiderivatives we can express as

$$ \int_a^x \frac{\mathrm{d}x}{1+x^2} = \arctan x - \arctan a $$

are those with $C \in [-\pi/2, \pi/2]$. (or $C \in (-\pi/2, \pi/2)$ if you don't want to allow $a = \pm \infty$)

Does anyone even imagine writing something like $$\sum_n n^3= \frac{n^2(n+1)^2}4+k?$$

Yes, actually; it is a very natural thing to do when what you're looking for is an anti-difference. Even if your problem actually called for a definite summation, keeping track of the boundary can be more complicated than just leaving things up to a constant which you solve for later by plugging in values.

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You can emphasize definite integrals, by treating them first -- before the indefinite integrals and even before the derivative. Apostol's calculus text was famous for this.

After enough definite integrals, students may both understand and appreciate the indefinite integrals better. This order corresponds to the history also: the earliest texts we have which look like calculus are calculations of areas by Archimedes.

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  • $\begingroup$ Hughes-Hallett applied (which is probably the opposite side of Apostol in terms of difficulty/rigor!) also does definite integrals before antiderivatives, and probably some other "reform" texts do too. $\endgroup$
    – kcrisman
    Commented Oct 24, 2015 at 20:58
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The textbook we use (a fairly standard one in the US, Thomas) is actually pretty careful about this: it says that an indefinite integral is a collection of functions, namely, all the antiderivatives. The resulting possibilities ("an antiderivative" versus "the indefinite integral") are a bit confusing for students just learning the topic, especially since "$\cos x+k$" could mean either "the indefinite integral, i.e. all the possible functions at once" or "a particular antiderivative with $k$ some constant", but I've found that Chris Cunningham's suggestion---emphasizing that this represents multiple functions by having them explicitly instantiate multiple cases---helps.

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Have you considered the following notation: $$ \quad \int^x \sin(t) \,\mathrm{d}t = -\cos(x)+k$$ emphazing that the integral is a function of $x$ ?

Abbreviations for integration and derivation can then be written in shorthand with $\quad \int^x$ and $d/dx$

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    $\begingroup$ A problem with this approach is that it doesn't make $\int^x$ and $d/dx$ into inverses of each other: you don't get $\sin(t)$ out of $\frac{d}{dx}( \cos(x) + k )$. $\endgroup$ Commented May 19, 2014 at 18:32
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    $\begingroup$ Teaching students a nonstandard notation like this is a bad idea IMO. They need to be able to cope with how calculus is notated in the real world. $\endgroup$
    – user507
    Commented May 19, 2014 at 20:00
  • $\begingroup$ @IlmariKaronen, in some sense it does: $(d/dx)\int^x\sin(t)dt =\sin(x)$. In a more precise sense they are not inverse: $\int^x$ is applied not to the function $\sin$ but to the differential form $\sin(t)dt = d(-\cos(t))$ (see a relevant note by Terrence Tao). $\endgroup$
    – Alexey
    Commented Jul 20, 2019 at 17:41
  • $\begingroup$ @BenCrowell, the calculus is notated as desired by the author. If the "real world" notation is poorly designed and makes learning more difficult, a more appropriate notation should be used (with a remark that it differs from the one commonly used elsewhere). Notation is not the principal subject of calculus, but real numbers, functions, and operations on them are. $\endgroup$
    – Alexey
    Commented Jul 20, 2019 at 17:46
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    $\begingroup$ @tos, i approve this notation, and i think i've seen it before somewhere. Do you know by any chance of any well-written book that uses it? $\endgroup$
    – Alexey
    Commented Jul 20, 2019 at 17:47
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I think it's funny that you/your students have a problem with the indefinite integral but don't mind the mysterious placeholder $dx$ at the end of a definite integral.

In my Calculus course, I take a differentials approach (see here), so in that sense $\int$ is the inverse operator of $d$. However, since $d$ is a many-to-one operation, the $+k$ is required in the output of $\int$ in the same way that $\pm$ is required when undoing a square. When I first introduce indefinite integrals (which is after definite integrals which motivate the notational symbols), I'll do an easy example like $$\int 3x^2\ dx = \int d(x^3) = x^3+k$$ (which most of my students get pretty quickly, especially when compared to $\sqrt{9}=\sqrt{(\pm 3)^2} = \pm 3$). I'll also do a more advanced example: $$\int \frac{x}{x^2+3}\ dx = \int \frac{\frac{1}{2}d(x^2)}{x^2+3} = \frac{1}{2}\int\frac{d(x^2+3)}{x^2+3} = \frac{1}{2}\int d(\ln(x^2+3)) = \frac{1}{2}\ln(x^2+3)+k$$ (which all of them need to think about before understanding).

Anyway, my point is treat $\int$ as the inverse operator of $d$ and draw upon the analogy of square roots to explain the $+k$.

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    $\begingroup$ But $\sqrt{\cdot}$ is usually defined as the operator giving the positive square root of its argument, so as to make it single-valued. This makes it quite a bad analogy doesn't it? As for $\mathrm{d}x$, I agree that it is somewhat mysterious, but it is simply a weird notation. It does not conflict too violently with other notation. $\endgroup$ Commented May 19, 2014 at 18:24
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    $\begingroup$ Do you try to formalize the notion of $\int$ and $d$ as operators, e.g., specifying their ranges and domains? This seems a little thorny to me, since the notation was designed to work on variables and expressions, not functions. Say $a$ is the area of a circle. Then $da$ historically meant just an infinitesimally small change in $a$. There's no implication that $a$ is a function of anything. It could be a function of the radius of the circle, or it could be a function of time if the circle is growing. Stuff like $dv=dx\:dy\:dz$ also doesn't seem to fit the operator paradigm well. $\endgroup$
    – user507
    Commented May 19, 2014 at 22:03
  • $\begingroup$ @BenCrowell: Definitely not for the students. I have a way of thinking about it for myself (d maps functions to a vector space over functions with basis elements differentials). This is good enough to keep me from saying outrageously wrong things, but it's probably not entirely mathematically sound. $\endgroup$
    – Aeryk
    Commented May 20, 2014 at 11:24
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    $\begingroup$ @Ben: One can model this as variables being scalar fields on some implicit domain and $d$ is the exterior derivative (defined only for sufficiently smooth scalars). If you work exclusively with scalars and differentials (and equations involving them), then the domain itself never actually needs to make an appearance in the formalism, although you can imagine it being behind the scenes as encoding the domain of variation and the prior relations between the variables. $\endgroup$
    – user797
    Commented Oct 22, 2015 at 22:20
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Teach your students that $$ \int \sin x \,\mathrm{d}x = -\cos x + k$$ is simply (very convenient) shorthand for this precise but long-winded statement:

Suppose the function $f$ is defined on an interval and has an antiderivative.

If $f$ is defined by $f(x)=\sin x$, then its antiderivatives are exactly those functions $g$ defined by $g(x)=\cos x$.


  • The above statement is not "utterly difficult", is fairly precise, and can be understood by even average high school/introductory calculus students.

  • Repeat the above statement (in speech, examples, exercises, test questions) as often as is judged necessary.

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    $\begingroup$ Well, then your question is similar to that of how we might teach 8-year-olds to remember how to spell a long and difficult word. There are different views on this, but how I'd do this is by (spaced) repetition, examples, exercises, homework, and questions on quizzes/tests. (The practice of most mathematicians would be to simply state the above long-winded statement at most once -- probably because that's all they themselves need to remember it. But good teachers know when and how much repetition is necessary.) $\endgroup$
    – user9798
    Commented Oct 22, 2018 at 23:12
  • $\begingroup$ You stated that "for students it seems utterly difficult to give a precise meaning to this". I don't think the above long-winded statement is "utterly difficult". And if taught, I think most students at the high school/introductory university level will be able to understand it and remember it. $\endgroup$
    – user9798
    Commented Oct 22, 2018 at 23:18
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    $\begingroup$ @dtcm840 do you treat functions and mappings as different kinds of objects? If so, what do you tell your students is the difference between them? $\endgroup$ Commented Dec 7, 2018 at 8:51
  • $\begingroup$ @MichaelBächtold: Formally, a function $f$ from $A$ to $B$ is a subset of the cartesian product $A \times B$ (with the property that if $x\in A$, then $\exists !y\in B$ such that $(x,y) \in f$). But informally and at a high school level, I teach that a function consists of three "components", namely the domain $A$, the codomain $B$, and the mapping. And so informally, the mapping is the "rule" that tells us how to assign each element in the domain $A$ to an object in the codomain $B$. To specify a function, one must specify all three "components", of which the mapping (rule) is only one. $\endgroup$
    – user9798
    Commented Dec 8, 2018 at 2:39
  • $\begingroup$ @dtcm840 But even in your formal definition of map, the set $A\times B$ together with its decomposition as a product (and the information which is the domain and which is the codomain) is part of the data that make up a map. So there is basically no difference between what you call a map and a function. I wonder if your students get any advantage from making the distinction you make. $\endgroup$ Commented Dec 8, 2018 at 10:57
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After having looked into the history of calculus and into original works by Leibniz and others, I have come to the conclusion that the schizophrenic way the "indefinite integral" is sometimes taught is a byproduct of the common schizophrenic mixture of Leibniz-time notation with Bourbaki-time foundations of calculus. I've found a satisfactory way to deal with the former, but not yet with the latter (I am working on it).

When trying to make sense of the traditional Leibniz notation in calculus, it is important to understand that even the notion of function in the 18th century was not what it is now. Now a function is something along the lines of "a set of pairs," "a functional relation," "a triple of a functional relation, a domain, and a co-domain," "a morphism in the category of sets." In the 18th century, to be a function was a property/predicate on variables: one variable could be a function of another, and also two variables could be functions of each other.

According to what I have understood, the notation of Leibniz, as used in the 18th century, heavily relied on the context. Expressions, like integrals, were not self-contained. In fact, even integration bounds were not written below and above "$\int$." Apparently, the first well-known text where the bounds of integration appeared at the integration symbol was Théorie analytique de la chaleur (1822) by Joseph Fourier -- see The history of notations of the Calculus (1923) by Florian Cajori.

I long believed that in the expression "$\int x^2dx$," the variable "$x$" is bound (is not free) because of "$dx$," but this is not at all the idea. One can have "$\int ydx$," "$\int xdy$," "$\int(xdy + ydx)$," and how this is to be integrates should be specified separately. For instance: $$ \int_{x=0}^1 x^2dx =\frac{1}{3},\qquad \int_{y=0}^1 x^2dx = 0, $$ where in the second integral the variable "$x$" represents a fixed quantity (fixed in the context of this integral), while $y$ varies from $0$ to $1$. (This is not how this would have been written in the 18th century, because bounds were not written by the integration symbol, but this is the idea.)

Now, I return to the question of the meaning of the "indefinite integral."

If we simply want to write identities like $$ \int (f(x) + g(x))dx =\int f(x)dx +\int g(x)dx $$ without worrying about the bounds, like this would have been done in the 18th century, it suffices to interpret the "indefinite integrals" that appear here as indeterminate definite integrals -- definite integrals with indeterminate bounds. The meaning of this identity then is that for all $a$ et $b$ such that both sides are defined, we have: $$ \int_{x=a}^b (f(x) + g(x))dx =\int_{x=a}^b f(x)dx +\int_{x=a}^b g(x)dx. $$ In modern terms, this amounts to an identity of functions of two variables, which take $a$ et $b$ and return the value obtained by integrations from $a$ to $b$. I believed having already seen such a definition of indefinite integral given by Paul Halmos, but I have just tried to find it, and I have no idea where I could have seen it. Perhaps I deduced it myself from the definition of indefinite Lebesgue integral given by Halmos (see below).

Here is the definition of an indefinite integral that I plan to use in my teaching from now on:

$\int_x f(x)dx$, written also as "$\int f(x)dx$" when it is clear from the context that the integration is with respect to $x$, is the function that takes a pair of reals $(a, b)$ and returns the value $\int_{x=a}^b f(x)dx$. We adopt the convention to not write expressions like "$(\int_x f(x)dx)(0, 1)$," but to write instead "$\int_{x=0}^1 f(x)dx$."

If teaching time permits, I may mention that historically there was another notion of "indefinite integral," which I may call, for example, indefinite integral in the sense of Cauchy (see my another partial answer), but that this historical notion is virtually informalizable in the modern mathematics.

In addition to the proposed definition of an indefinite integral, I plan to use the following definition of an indefinite difference:

$[f(x)]_x$, written also as "$[f(x)]$" when there should be no ambiguity, is the function that takes a pair of reals $(a, b)$ and returns the value $[f(x)]_{x=a}^b = f(b) - f(a)$. We adopt the convention to not write expressions like "$([f(x)]_x)(0, 1)$," but to write instead "$[f(x)]_{x=0}^1$."

We have the following obvious property: $$ [f(x)] = [g(x)]\quad\Leftrightarrow\quad [f(x) - g(x)] = [0]\quad\Leftrightarrow\quad f - g\ \ \text{is constant}. $$

The second fundamental theorem of calculus can now be written as: $$ \int f'(x)dx = [f(x)]. $$

This interpretation is compatible with and can be applied to "non-oriented" integrals $\int f(x)|dx|$, Stieltjes integrals $\int f(x)dg(x)$, "non-oriented" Stieltjes integrals $\int f(x)|dg(x)|$, etc.

This interpretation also agrees with my recent idea to view measure-theoretic integration as a binary operation that takes a signed measure and a measurable function and returns a signed measure. Just now I've re-discovered that this is exactly the definition of indefinite Lebesgue integral given by Paul Halmos in his Measure theory. Thus, the indefinite Lebesgue integral $\int fd\mu$ is the signed measure that takes a measurable set $X$ and returns the number $\int_X fd\mu$.


I am adding some examples of calculations:

$$ \int_t te^t dt =\int_t tde^t = [te^t]_t -\int_t e^tdt = [te^t]_t -[e^t]_t = [te^t - e^t]_t. $$

\begin{multline*} \int_x\frac{dx}{1 - x^2} =\int_x\frac{1}{2}\left(\frac{dx}{1 - x} +\frac{dx}{1 + x}\right) =\frac{1}{2}\left(\int_x\frac{dx}{x + 1} -\int_x\frac{dx}{x - 1} \right)\\ =\frac{1}{2}([\ln |x + 1|]_x -[\ln |x - 1|]_x) =\frac{1}{2}[\ln |x + 1| -\ln |x - 1|]_x =\left[\frac{1}{2}\ln\left|\frac{x + 1}{x - 1}\right|\right]_x. \end{multline*}

In the second example, it is by default understood that the obtained identity is equivalent to $$ \int_{x=a}^b\frac{dx}{1 - x^2} =\left[\frac{1}{2}\ln\left|\frac{x + 1}{x - 1}\right|\right]_{x=a}^b $$ for all $a,b\in\mathbf{R}$ such that the integral exists, which excludes the possibility that $a$ and $b$ be separated by $1$ or $-1$.

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  • $\begingroup$ I like your definition of indefinite integral and had thought about it myself, but would you then still be able to justify equations like $\int dx =x +c$? It also seems that you could not use this indefinite integral to solve Ode's as is commonly done on paper $\endgroup$ Commented Feb 23, 2023 at 12:10
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    $\begingroup$ @MichaelBächtold, I find this equation absurd, there is nothing to justify IMO. It has little to do with my definition either. I do not use indefinite integrals to solve ODE. In any case, before using indefinite integrals, one needs to define them. So, to use them for ODE, you would need to define them first. However, it is possible to give solutions to ODE using definite integrals. $\endgroup$
    – Alexey
    Commented Feb 23, 2023 at 12:18
  • $\begingroup$ What would you use your indefinite integrals for then? $\endgroup$ Commented Feb 23, 2023 at 12:58
  • $\begingroup$ @MichaelBächtold, I suppose for those things they are customarily used for, where their use does not look absurd or schizophrenic. For example, where you are writing calculations or identities about integrals, but do not feel like writing some arbitrary integration bounds at every integral sign. Like Second Fundamental Theorem of Calculus, linearity of integration, integration by parts. In a more advanced course, indefinite integrals can be redefined or interpreted as signed measure and be used as such. Could you show how you mean to use indefinite integrals for ODE? $\endgroup$
    – Alexey
    Commented Feb 23, 2023 at 13:19
  • $\begingroup$ Indefinite integrals, as I see them, are mainly used for 'computing' antiderivatives, which are used for solving for instance initial value problems. Along the way you might do an integration by parts or a substitution. One could of course do that without having a notation for antiderivatives (as you show in your other answer). I haven't tried that and wonder how 'easy' it is in practice. $\endgroup$ Commented Feb 23, 2023 at 13:31
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Indefinite integrals are easy to avoid, it only takes a will to do so.

Basic antiderivatives: $$ x^2 =\frac{d}{dx}\frac{x^3}{3},\quad \sin x =\frac{d}{dx}(-\cos x),\quad \frac{1}{x} =\frac{d}{dx}\ln|x|. $$

Antidifferentiation by change of variable: $$ e^x\sin(e^x) =\sin(e^x)\frac{d}{dx}e^x =\frac{d}{dx}\sin(e^x), $$

Antidifferentiation by parts: $$ x e^x = x \frac{d}{dx}e^x = x \frac{d}{dx}e^x + (\frac{d}{dx} x)e^x - (\frac{d}{dx} x)e^x\\ =\frac{d}{dx}(x e^x) - (\frac{d}{dx} x)e^x = \frac{d}{dx}(x e^x) - e^x\\ =\frac{d}{dx}(x e^x) -\frac{d}{dx}e^x = \frac{d}{dx}(x e^x - e^x). $$

Using differential forms, equivalent calculations can be carried out with simpler notation: $$ x^2\,dx = d\frac{x^3}{3},\quad \sin x\,dx = d(-\cos x),\quad \frac{dx}{x} =d(\ln|x|), $$ $$ e^x\sin(e^x)\,dx =\sin(e^x)d e^x =d\sin(e^x), $$ $$ x e^x\,dx = x d e^x = x d e^x + (dx)e^x - (dx)e^x\\ = d x e^x - (dx)e^x = d x e^x - e^x\,dx\\ = d x e^x - de^x = d(x e^x - e^x). $$

When I teach definite integrals (or try to teach indefinite integrals despite being fuzzy myself about their exact meaning), I prefer to carry out calculations with differential forms first and apply definite (or indefinite) integral at the end.

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  • $\begingroup$ I think preferable to avoid differential forms, because their meaning is often obscure to student. My worst deception was to have all students of class be able to compute dF (with $F(x,y,z) =x^2+y^2+z^2 -3xyz$ say) and utterly unable to compute correctly $dF_{(1,2,3)}(4,5,6)$. $\endgroup$ Commented Jul 21, 2019 at 13:37
  • $\begingroup$ @BenoîtKloeckner, at least to me differential forms are more clear than indefinite integrals and fit naturally. If the teacher understands himself what he is teaching, it is already a big advantage. When i use differential forms just to compute an integral, i do not spend much time on them and i do not ask students to compute $dF(1,2,3)[4,5,6]$. $\endgroup$
    – Alexey
    Commented Jul 21, 2019 at 14:47
  • $\begingroup$ Inspired by this note by Terrence Tao, I even started making a distinctions between $\int_a^bf(x)dx$ and $\int_{[a, b]}f(x)|dx|$. $\endgroup$
    – Alexey
    Commented Jul 21, 2019 at 14:49
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    $\begingroup$ Alexey: for you, I do not doubt it; and as said indefinite integral have an obscure meaning to students too. But the very point of my question was that I take issue with objects that do not make a precise sense in student's minds, so that using differential forms would only move the problem (this is an experimentally proven statement). $\endgroup$ Commented Jul 21, 2019 at 15:48
  • $\begingroup$ I agree that differential forms are rather obscure objects. I tried to use them with students because i found the notation for them simpler and less ambiguous than then one for indefinite integrals. I thought that since many students learn integrations rules as formal manipulations of symbols, using differential forms could not make it worse. I could not evaluate how useful this approach was. $\endgroup$
    – Alexey
    Commented Jul 21, 2019 at 20:13
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Historically, there was indeed a notion of "indefinite integral" used, for example, by Cauchy. At the time, the notions of "functions" and "variables" were unlike what they are now.

In Augustin Cauchy's Résumé des leçons données a l'École Royale Polytechnique sur le calcul infinitésimal -- Tome 1 (1823), the 26th lecture has the title "Indefinite integrals." I find the definition given there somewhat obscure, and to quote it here I would need to include half a page of the preceding context.

In Émile Picard's Traité d'analyse -- Tome I (1891), the second chapter has the title "Indefinite integrals." Here is the definition from page 38:

On donne souvent le nom d'intégrale indéfinie à une intégrale définie ayant pour limite supérieure la variable $x$ et une limite inférieure qui est une constante qu'on ne fixe pas. On représente simplement par $$\int\!f(x)\,dx$$ celte intégrale indéfinie, sans prendre la peine de marquer les limites. C'est simplement une manière de désigner une fonction ayant pour dérivée $f(x)$.

In English:

One often gives the name of indefinite integral to a definite integral that has for the upper limit the variable $x$ and a lower limit that is a constant that one does not fix. One represents simply by $$\int\!f(x)\,dx$$ this indefinite integral, without bothering to mark the limits. It is just a manner to designate a function that has $f(x)$ as the derivative.

I believe people are trying to include the notion of "indefinite integral" in the sense of Cauchy in modern textbooks. They invent interpretations like "an unknown primitive" or "the set of all primitives." But I believe to understand that for Cauchy the indefinite integral $\int f(x)\,dx$ was something like a variable that depends on the variable $x$, but the dependence is only determined up to an additive constant. So the identity of the form $$ \int f(x)\,dx = F(x) + C $$ was making sense.

Of course, if $\int f(x)\,dx$ is an unknown primitive of $f$ ($F$?) or the set of all primitives, this identity is not even false, it is absurd (there is a "type error" in it).

Good luck formalizing this in the modern mathematics!

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Question: What is the function, having $\sin(x)$ as its derivate?
Answer: There are many of them. Here they are (some of them):

enter image description here

Question: So many? But there are resemblances.
Answer: Indeed: they must all be (possible) vertical shifts of $-\cos(x)$, because taking the derivative will remove that vertical shift. Obviously, this vertical shift gives you the opportunity to choose a particular one, e.g. if you want to choose one for which $D^{-1}(\sin)(x)$ is $0$ for $x$ being $0$ (I have used $D^{-1}$ as the anti-derivative operator). As you can see on the drawing, this happens when $k=1$ (the blue one).

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    $\begingroup$ You seem to have missed the point of my question, which is not about understanding or explaining antiderivatives and their non-uniqueness, but about the indefinite integral notation. $\endgroup$ Commented Nov 3, 2023 at 20:30
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when one writes $$ \int \sin(x) \,\mathrm{d}x = -\cos(x) + k$$ then the status of both $x$ and $k$ is pretty unclear (which quantifier in front of each of these variables?)

  1. Yes, when an indefinite integral is specified like in the above (that is, as a general representation or as a function of the form $-\cos(x)+k,$ where $k$ is arbitrary), it doesn't really make sense to quantify $k,$ unless initial/boundary conditions are given, in which case we can assert that $$\exists k{\in}\mathbb R\;\forall x{\in}\mathbb R \int \sin(x) \,\mathrm{d}x = -\cos(x) + k.$$

  2. On the other hand, when the above indefinite integral is undersood as a literal set of antiderivatives, then:

    for each real $x,$ the indefinite integral of $\sin x$ is the set of functions such that each one, for some real $k$, equals $-\cos(x)+k.$


Appendix (excerpt from How to interpret the indefinite integral)

An indefinite integral is more a useful notational shorthand than a mathematically important object. It is conventionally called a family of antiderivatives; as such, if its integrand is of the form $g'$ and has an interval domain, then, literally, $$\int g'\,\mathrm dx=\{g(x)+C\mid C\in\mathbb R\},$$ where $C$ represents uncountably many values. How to understand the "indefinite integral" notation in calculus? contains several justifications for manipulating and adding indefinite integrals as actual sets by reading $\text‘=\text’$ as an equivalence relation such that writing \begin{align}x^2+C=x^2+3+C\quad&\implies0=3\\x^2+C=x^2+3+D\quad&\implies E=C-D=3\end{align} make sense.

More simply, we can frame an indefinite integral as the general representation of its integrand's antiderivatives, its specification containing one independent parameter (arbitrary constant) $C_i$ per maximal interval of its integrand's domain. Here, each instantiation of $$\int g'\,\mathrm dx=g(x)+C$$ has an unimportant value of $C.$ Even though this object is not a particular antiderivative, we manipulate it as if $C$ is merely undetermined.

In the context of solving differential equations with given conditions, $C$ becomes an unknown whose value is to be determined.

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    $\begingroup$ It seems you misread the question, which was "what possible issues are there in avoiding completely indefinite integrals? Is there any pedagogical advantage to using them? Is there a third way to go?". I know what indefinite integrals mean, and I assumed people interested in the question do too. $\endgroup$ Commented Feb 21, 2023 at 10:52
  • $\begingroup$ @BenoîtKloeckner I cast 3 upvotes on this page years ago, and belatedly quoting and addressing that fragment of your question post means that my supplementary response is specific to that aside about quantification; the appendix adds context to my main answer. Is that okay? $\endgroup$
    – ryang
    Commented Feb 21, 2023 at 10:59

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