Simplest thing to do, I think, is think of C as being locally constant, rather than globally constant. (After all, this is how one would need to generalise the fundamental theorem of calculus to disconnected domains anyway.)

Re: Reader Survey: log|x| + C

Right, that’s how Ithink of it. But are you saying that’s how you’d put it to beginning calculus students? Would you actually use the word “locally” and explain what it meant, or would you find some other way?

Of course, they don’t know what “disconnected” means. Connectedness is an easy enough thing to explain intuitively, but there’s a question of how much time can be spent on explaining just one integral.

My approach is basically what you called a cop out: to avoid integrating$1/x$ — or indeed any other function — over a disconnected domain. My justification for this is that the fundamental theorem of calculus, which most students at the level you’re talking about quickly start to believe is the definition of a definite integral, only holds over an interval.

Here’s a trickier cousin of your example to consider. Blindly attempting to apply the fundamental theorem of calculus, you might write$${\int }_{-1}^1\frac{1}{x^2}\mathrm{dx}=-\frac{1}{x}{]}_{-1}^1=-2.$$This is clearly wrong — if the integral exists at all its value must be positive. Searching for the mistake, if you’re not sophisticated enough to think you’d better check the hypotheses of the fundamental theorem, you might think you had the antiderivative wrong. And in fact you do have “the” antiderivative wrong — the real antiderivative of$1/x^2$ is$-1/x+C$, where$C$ is a locally constant function on$ℝ\setminus \{0\}$. By adjusting the values of$C$ on the two components of the domain, you can make the formal calculation above yield any finite value, but not the correct value of$+\mathrm{\infty }$.

As far as I can see the best solution at this level is to keep integration essentially restricted to intervals on which the integrand is everywhere defined.

Re: Reader Survey: log|x| + C

Nice! I’m sure I’ve seen that example before, but I’d forgotten it.

Re: Reader Survey: log|x| + C

I teach my students that$\log |x|$ isa anti-derivative of$\frac{1}{x}$. I rarely ask them to giveall anti-derivatives of a given function.

And I tell them that whenever the integrand becomes singular, this means that there are some hiddenlimits. Thus, I’d say that${\int }_{-1}^1\frac{1}{x^2}\mathrm{dx}$ isdefined as$${\int }_{-1}^1\frac{1}{x^2}\mathrm{dx}=\underset{ϵ\to 0+}{\lim }{\int }_{-1}^{-ϵ}\frac{1}{x^2}\mathrm{dx}+\underset{\delta \to 0+}{\lim }{\int }_{\delta }^1\frac{1}{x^2}\mathrm{dx}.$$ Now they can useany two anti-derivatives of$\frac{1}{x^2}$ in order to evaluate the definite integrals, and will always get the correct answer$\mathrm{\infty }$.

Interesting question! I don’t understand what the “cop out” option means, though. The question is about anindefinite integral, which doesn’t come with any domain at all. Right? If you’re talking aboutdefinite integrals, then I’ve never even tried (or seen any beginning calculus class that tried) to teach people to integrate over any domain other than a closed interval, and I think Mark is exactly right that then the integrand should be defined over the entire interval. But I don’t see how that solves the problem of what to tell them theindefinite integral of$1/x$ is.

Re: Reader Survey: log|x| + C

I suppose by “cop out” I meant sticking to definite integrals over subintervals of either$(0,\mathrm{\infty })$ or$(-\mathrm{\infty },0)$. It’s a cop-out because we’re deliberatelynot mentioning the indefinite integral of$1/x$. (Of course, cop-outs are sometimes pedagogically wise.)

As Mark says, students at this level tend to treat integration as the inverse of differentiation anyway. So to them, the difference between an indefinite integral${\displaystyle \int f(x)dx}$ and a definite integral${\displaystyle {\int }_a^{b}f(x)dx}$ isn’t so great.

To do the first, you (i.e. they) find some$F$ whose derivative is$f$, and then the “answer” is$F(x)+C$. (And only the most conscientious bother writing “where$C$ is the constant of integration”.) To do the second, you do the same thing but without mentioning a$C$, and then calculate$F(b)-F(a)$. I’d imagine that as far as they’re concerned, the main difference between indefinite and definite integrals is that in the second case, you have to do this extra little step of calculation.

Re: Reader Survey: log|x| + C

I don’t see how you avoid mentioning the indefinite integral of$1/x$. Are you saying that you just don’t talk about indefinite integrals at all, i.e. only write down the symbol$\int$ when it has an upper and a lower limit? I like that idea a lot, but it would be difficult to do with all the calculus textbooks that I’ve used.

Re: Reader Survey: log|x| + C

I just meant not mentioning the indefinite integral of$1/x$ (or similarly problematic functions). Mentioning definite integrals of$1/x$ over intervals not including$0$, and mentioning indefinite integrals of nicer functions, should be OK. I’m not sure that’s a good solution; I was just presenting it as an option.

In fact, my role this semester has just been to teach exercise classes, not lecture, so I don’t get to decide how things are presented anyway. (We don’t tend to use books here to nearly the same extent as I gather is normal in the US. While that certainly has some disadvantages, it does have the advantage that the lecturer isn’t constrained by what the book does.)

Thanks for this post, I had never even realized this issue.

One sneaky way out would be to speak of “a” general antiderivative, rather than “THE” general antiderivative. log|x|+C is a general antiderivative of 1/x, just not the most general possible one.

To support this, look at why exactly we teach freshmen constants of integration in the first place. It’s not for the sake of calculating definite integrals– the constants cancel there, provided the fundamental theorem is actually applicable (as opposed to Stefan Keppeler’s brilliant example). Nor (apparently) is it for the sake of the deeper theory, since as this blog post points out, if that’s the purpose, we’re doing it wrong! We teach constants of integration to make initial-value differential equations solvable. For that purpose, only “a” general antiderivative is needed, rather than “the most” general one.

(At least provided the question makes sense, not like “A car’s velocity function is 1/t, at time t=-1 it has position 0, find its position at time t=1”)

Re: Reader Survey: log|x| + C

Ah yes, interesting point.

While I’m at it, here’s another piece of widely taught wrongitude:

Definition  Agroup is a set$G$ together with a binary operation$\star$, such that:

1. for all$x,y,z\in G$,  $(x\star y)\star z=x\star (y\star z)$
2. there exists$e\in G$ such that for all$x\in G$,  $x\star e=x=e\star x$
3. for all$x\in G$ there exists$x\prime \in G$ such that$x\star x\prime =e=x\prime \star x$.

This seems to be quite common. It’s the “definition” contained in the notes for a course I took over this year. It’s also the “definition” I was given as an undergraduate myself. I remember thinking at the time that this was the height of logical rigour. I only realized years later that it was, in fact, nonsense.

It’s (3) that doesn’t make sense. It refers to an element called “$e$”, but no element called$e$ has been defined. Presumably it’s meant to refer to “the” element$e$ satisfying the condition in (2), but in principle, there could bemany such elements$e$ — that is, many identity elements. In principle, (3) might be true for some identity elements and not others. Is (3) required to hold for all identity elements$e$? Or just at least one of them? Can the identity element(s) for which it holds depend on$x$? And so on: it’s just not a logically well-formed statement.

Another way to describe the problem: (2) is logically equivalent to:

there exists$f\in G$ such that for all$x\in G$,  $x\star f=x=f\star x$.

If we replace (2) by this statement (which we can, because it’s logically equivalent) then (3) is clearly nonsensical — the letter$e$ hasn’t even been used before.

I know two and a half ways to say it properly. The first is to pause after (1) and (2), prove that identities are in fact unique, then proceed to (3). The second is to say “a group is a set$G$ together with a binary operation$\star$ and an element$e$ satisfying…”. The two-and-a-halfth is the same, but including inverses as part of the structure too (rather than asserting the existence of inverses as a property). Whichever way you do it, you’re going to have to prove statements on uniqueness of identities and inverses sooner or later, anyway.

I don’t find this as pedagogically challenging as$\log |x|+C$, because in this case I can see several ways of presenting the material that I’m comfortable with.

Re: Reader Survey: log|x| + C

I am aghast that I never noticed this before. I believe I’ve seen textbooks that give the definition correctly, but Iknow I’ve seen textbooks that do it incorrectly, and that I’ve failed to notice the difference and taught the wrong version to students.

Re: Reader Survey: log|x| + C

Wow. When I read what you just wrote, I immediately thought “would any experienced mathematician actually write that?” But then I went and picked up the first abstract algebra book I owned (Dummit & Foote) and was horrified to discover that that’s the definition they give! (My first abstract algebra class didn’t follow the book, though, and I don’t think I still have the notes, so I can’t say what definition I wastaught.)

How can anyone write that and fail to notice that it is nonsense? Maybe mathematicians are actually constructive type theorists and don’t know it.

By the way, I think your first proper way to say it is inferior to the other one-and-a-half. As I’m sure you know, the identity, and perhaps the inverses too, should be considered structure rather than a property, even though they are unique as soon as they exist — because you want them to be preserved by homomorphisms. In the case of groups, by accident it happens that preserving the multiplication implies preserving the identity and inverses, so you can get away with considering the other two as properties — but it’s not a good habit to get into. (-:

Re: Reader Survey: log|x| + C

Re: Reader Survey: log|x| + C

In fact, alsoWikipedia gets it wrong. The unit axiom is awkwardly phrased in that it suddenly turns an indefinite article into a definite one (emphasis mine):

Identity element:There existsan element$e$ in$G$, such that for every element$a$ in$G$, the equation$e\cdot a=a\cdot e=a$ holds.The identity element […]

What do you think about inserting here, after the period, something along the lines of “Such an$e$ is called anidentity element and can be shown to be unique”?

Re: Reader Survey: log|x| + C

(a) the binary operation on G is associative(b) there is an element e in G such that  (b1) ea = a for all a in G  (b2) for each a in G there is some b in G with ba = e

Re: Reader Survey: log|x| + C

I know two and a half ways to say it properly. The first is to pause after (1) and (2), prove that identities are in fact unique, then proceed to (3). The second is to say “a group is a set$G$ together with a binary operation$\star$ and an element$e$ satisfying…”. The two-and-a-halfth is the same, but including inverses as part of the structure too (rather than asserting the existence of inverses as a property). Whichever way you do it, you’re going to have to prove statements on uniqueness of identities and inverses sooner or later, anyway.

There is another way, taken, I think, by van der Waerden; it is to replace (2) and (3) by:

($(2+3)\prime$) For all$x,y\in G$, there is a unique solution$z\in G$ to$x\star z=y$.

This is a very specific work-around thatbuilds in uniqueness, and thus doesn’t address the spirit of your question; but I rather like the way that it unifies the original (2) and (3).

Re: Reader Survey: log|x| + C

That definiton of group could easily be revised to the following:

A group is a set$G$ with an element$e:G$, a function$(-{)}^{\prime }:G\to G$, and a function$(-)\star (-):(G\times G)\to G$ such that the conditions (1), (2), and (3) are satisfied.

I’m still thinking about my answer to your original question, but I thought I would point out that the problem is not confined to$\int \frac{1}{x}$, but spreads out from there to infect other antiderivatives as well, sometimes even more virulently. For instance, we teach that

$$\int \frac{1}{\cos (x)}\mathrm{dx}=\log |\tan x|+C$$

but this is (if possible) even wronger than$\int \frac{1}{x}\mathrm{dx}=\log |x|+C$, since in this case the domain has infinitely many connected components and the constant could be differenton each one.

Perhaps thereal culprit of this problem is the reprehensible habit of implicitly assuming the domain of every function to be “the largest subset of the real line on which the given algebraic expression is defined”. But that’s probably not going to go away any time soon. (-:

Re: Reader Survey: log|x| + C

Yes, good point. We get an infinite-dimensional space of antiderivatives before we know it.

Seeing as I’m in a mood for going round obnoxiously pointing out other people’s mistakes (and thereby inevitably setting myself up for a fall), I can’t resist pointing out that you probably meant either

$$\int \frac{1}{\cos (x)}dx=\log |\sec (x)+\tan (x)|+C$$

or

$$\int \tan (x)dx=\log |\sec (x)|+C.$$

But the point you were making is unaffected.

Re: Reader Survey: log|x| + C

Blarg. One year goes by without teaching any calculus and I forget all my derivatives. That doesn’t bode well.

You could give the traditional definition given to students at this level, and then end up with saying, “What I just said is not technically correct. The real situation is actually more complicated than I described. If you are interested in learning more about it, you can see me after class”. This would be enough to pique the curiosity of the few students actually interested. If I was that age, I would be eager to ask the teacher why the supposed correct answer is actually wrong. That way, you would not be wasting the time, causing distraction or confusion, to all of the other students who could not care less.

Re: Reader Survey: log|x| + C

I suppose I have a resistance to saying things that are “not technically correct”, in that “not technically correct” means “wrong”. A large part of the battle at this level of teaching, I find, is to train students not to write down things that are wrong.

For example, I routinely apply red ink to expressions such as

$$\dots =\sqrt{2}=1.414,$$

telling them that$\sqrt{2}$ is no more equal to$1.414$ than$0$ is equal to$1$. I don’t want them to be able to come back to me and say “well, I know it’s not technically correct. Butyou sometimes write things that aren’t technically correct, so why can’t I?”

Re: Reader Survey: log|x| + C

I suppose I have a resistance to saying things that are “not technically correct”, in that “not technically correct” means “wrong”.

Yeah, I’m with you. It’s okay to lie in a seminar talk, since you’re talking to mathematicians who know perfectly well what it means to be precise, but when we’re trying to teach people to think mathematically, it’s dangerous to allow ourselves to be sloppy.

I think my inclination is to go the “constant that varies” route. But I would not say “constant that varies” or “locally constant function”, or introduce the notion of connectedness — I would just say thatthe constant can be different on each interval in the domain, but that we still just write$C$ even when the domain consists of more than one interval. And emphasize, when we come to FTC, that it is only valid when the function is continuous over the whole interval (so that in particular the constant in any antiderivative must be the same at both ends).

I think this is consistent with the level at which we talk about most things in a calculus class; all subsets of the real line we think about are disjoint unions of intervals, and any theorems that we state explicitly are usually stated in terms of functions defined and continuous on intervals.

Now I want to hear what you (Tom) think!

Re: Reader Survey: log|x| + C

Well, I’m afraid I don’t think anything very interesting, and it’s been altered a bit by what people have said here. When I wrote the post, my opinion was that one should avoid as far as possible talking about the indefinite integral of$1/x$ (and$1/\cos (x)$ etc), but if necessary, giving the two-cases formula.

Now I’m not so keen on the two-cases formula. I’m thinking that if there’s time to get into the indefinite integral at all, the thing to do is to make something interesting out of it: to say “look, the hyperbola has two different branches, and we get a different constant on each”. (Maybe both called “$C$”.) Present it as something exotic and try to attract the interest of the more imaginative students, rather than making it just one more integral to learn or mumbling and sweeping it under the carpet. Even if the full story isn’t told (and it probably can’t be), my current inclination would be to try to turn this difficulty into a point of interest — maybe a teaser for future courses on differential equations.

Re: Reader Survey: log|x| + C

Although I had never appreciated this issue (thanks!), when I taught Calculus last year, my approach successfully sidestepped this issue, by restricting the domain of $\frac{1}{x}$ to $(0,\infty)$, then integrating it to log(x)+C. Later on, I also integrated $\frac{1}{x}$ to $(-\infty,0)$ to $\log(-x)+C$. Finally, in the lead-up to the test, I introduced $\int\frac{1}{x}=\log|x|+C$ as ashorthand or as amnemomic.

Accidentally, I think this turned out to be a really good idea- i.e. I was lucky. It avoids teaching something wrong, it’s clear and intuitive to students, and it avoids integrating over non-connected domains. I don’t think any students got annoyed or confused by it, either.

Re: Reader Survey: log|x| + C

I feel like something of a fuddy-duddy now, in that I really want to say that the function$x\mapsto \frac{1}{x}$ simply isn’t integrable on any interval that includes$0$. I think Daniel gets closest to this fact of non-integrability with his “sidestep”, and the emphasis (perhaps I’m reading it in?) that the integrals for$\frac{1}{x}$ are one-sided things. It’s all very well to say that all the functions$x\mapsto \log |x|+C+cH(x)$ have the same derivative function; but not a single one of them is of the form$x\mapsto (k+{\int }_a^x\frac{1}{t}\mathrm{dt})$ for any$k$ or$a$. And I’m more than happy to distinguish between a function having aprimitive orprederivative (“antiderivatve” is a term that will have to die!), as vs. a function being integrable on intervals. Certainly, one mustn’t suggest that all primitives areintegrals; the upshot of the Fundamental Theorems should be seen as

• assuming good circumstances, one has aconstruction of acontinuous primitive
• when applied to the derivative of a (necessarily, differentiable) function, this construction produces a vertical shift of that function.

It’s worth-while pointing out cases in which these theorems do not apply: as instigated this post, there are primitives that are more general than any integral; there are also integrable functions which aren’t the derivative of anything (toss in not-too-many-too-large removable discontinuities, e.g.)

Having read this and the comments (so far) I have to say that that’s what we get for trying to teach students about Fredholm operators before they are ready!

I think that my “solution” to this would be to try to avoid the problem altogether and never, if at all possible, to talk about “indefinite integrals”. I think that the possibilities for confusion would be much less if instead of

$$\int \cos (x)\mathrm{dx}=\sin (x)$$

(and it wasn’t until I’d written that that I realised that I’d also made the classic error of scoping of variables) one wrote

$${\int }_a^x\cos (t)\mathrm{dt}=\sin (x)-\sin (a)$$

or one could write:

$$\sin (x)=\sin (a)+{\int }_a^x\cos (t)\mathrm{dt}$$

which would be my favourite.

So one still gets the integral as a function (of its upper limit) but without the potential bizarreness of the indefinite integral.

(and it might mean that we stop saying things like$\int \cos (x)=\sin (x)$ which, although not technically wrong, falls into that trap of using the same label for two different variables)

Re: Reader Survey: log|x| + C

I think that my “solution” to this would be to try to avoid the problem altogether and never, if at all possible, to talk about “indefinite integrals”.

I would definitely agree wholeheartedly with this, except for the likelihood that in some future class (or maybe even in the real world?) someone will expect the students to know what the phrase “indefinite integral” means.

I don’t find these examples upsetting. I think that they reflect that an average class will be of mixed ability, some of whom will be happy when given a recipe, and (possibly) the basic idea; and some of them won’t be happy to find that they’ve been misled slightly (I’m in this camp). If we can live with a certain level of ambiguity in ordinary every-day language we use, surely its allowable to have that in textbooks?

Perhaps definitions should be marked with some sort of symbol to indicate what level of mathematical scrutiny they will survive:)

Anyway, isn’t this sort of ‘sloppy’ definitions fairly rampant in physics? ie The Path Integral.

Did we really make it to the end of this thread without anyone saying that, whatever you do for that integral, at least it is clear what

$\int 1/cabin d(cabin) $

is??

(A joke I read in a Pynchon novel, back when the world was young)

Re: Reader Survey: log|x| + C

Funny you should post this now - I just got back from a class where I covered this exactly!

My approach is to talk about antiderivatives graphically first, before ever writing down a single formula. I draw the graph of a function f on the board and ask the students to draw the graph of a function F whose derivative is f. I give the students a few minutes to do this.

Then I ask the students if there are any other functions besides the one they drew which would work. Some student will realize that shifting their function vertically will work.

Then I draw a function with an obviously disconnected domain, and repeat the process (NOTE: students think that (0,1)U(2,3) is a lot more disconnected than (0,1)U(1,2) - they will often have the insight that the two pieces can be vertically shifted independently with the first interval, but not with the second.)

I do one more example where the domain is an interval with a single point deleted.

At this point the students have already realized for themselves that antiderivates differ from each other by a locally constant function.

After we have been drawing graphs long enough, I start writing down formulas, but we always make sure the formulas make sense graphically (checking increasing/decrease behavior and concavity of the antiderivative). When we get to 1/x, the students provide the correct form without a second thought.

Re: Reader Survey: log|x| + C

When we get to$1/x$, the students provide the correct form without a second thought.

To coin a phrase, the proof of the teaching is in the learning. Forget the pedagogical theory, if it works, I’ll try it.

I’ve only read the question today, but I’d like to point at that, typically, in France, we apparently have a different approach to the whole integral thing. Teaching of mathematics in France has been very influenced by Bourbaki’s books, by the way.

I’ve never taught calculus myself, but if I recollect my school years correctly: from the start on we use the integral sign only for integration on an interval (is that what english speakers call definite integrals?) on which the function is defined. When we are asking about antiderivatives (which we call primitives) we spell it out (usually asking about an antiderivative, rather than the general form). We are also, usually, explicit about the domain of the input function.

That said, it might not prevent from learning the wrong theorem about the general form of antiderivatives. As students we are usually very inattentive, it might be valuable for the teacher to stress the conditions of the theorem, and even to give Tom’s example to illustrate why they matter.

Btw: on the definition of groups, it may be worth pointing out that the two-and-halfth version (with *, e and inverse being given as operations) is also the version advertised by universal algebra.

but is log|x| + C really wrong? you are suggesting to use$\int \frac{1}{x}dx=\mathrm{Log}\left[Cx\right]+D\mathrm{Sign}\left[x\right]$ instead, which means$\frac{1}{x}=\frac{1}{x}+\mathrm{D\; DiracDelta}\left[x\right]$ D=0?

Something similar happens at the level of secondary school algebra: what is$f(x)≔\frac{1}{\frac{1}{x}}$ equal to? The naive answer typically provided in such classes is that$f(x)$ is the identity function on the real numbers$ℝ$, but technically that is not true, because the identity function is defined at$0$, while the function$f(x)$ is not. Instead,$f(x)=x^2\frac{1}{x}$.

The identity function (as a partial function) technically does not have a multiplicative inverse, for the same reason that zero does not have a multiplicative inverse. The same goes for any function that intersects the x-axis, such as the trigonometric functions$\sin (x)$ and$\cos (x)$.