The n-Category Café

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.)

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 $$\int_{-1}^1 \frac{1}{x^2} dx = -\frac{1}{x}\bigr]_{-1}^1 = -2.$$ 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 $\mathbb{R}\setminus \{ 0 \}$. 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 $+\infty$.

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.

Interesting question! I don’t understand what the “cop out” option means, though. The question is about an indefinite integral, which doesn’t come with any domain at all. Right? If you’re talking about definite 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 the indefinite integral of $1/x$ is.

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”)

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

Definition  A group 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 $x' \in G$ such that $x \star x' = e = x' \star x$.

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 be many 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.

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

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

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

Perhaps the real 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. (-:

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.

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 that the 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!

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 a shorthand or as a mnemomic.

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.

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

$$\int \cos(x) dx = \sin(x)$$

(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

$$\int_a^x \cos(t) dt = \sin(x) - \sin(a)$$

or one could write:

$$\sin(x) = \sin(a) + \int_a^x \cos(t) dt$$

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)

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)

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.

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 $f(x) \coloneqq \frac{1}{\frac{1}{x}}$ equal to? The naive answer typically provided in such classes is that $f(x)$ is the identity function on the real numbers $\mathbb{R}$, 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 $cos(x)$.