"Indefinite" is an understatement

I have a question for all the mathematically inclined readers out there. You guys have probably already noticed that I’m a bit picky when it comes to mathematical notation, whereby “picky” I mean “anal-retentive to an alarming degree.” I’ve gone on about renaming the imaginary number or being precise with trig, and I’m a believer that $pi$ was the wrong choice. In keeping with that theme, I have a notational question. It should be a basic one, but I found it was surprisingly more nebulous than I originally thought. Here it is:

What, precisely, is an “indefinite integral”?

That is, what does the symbol $displaystyle int f(x) , dx$ mean?

Lest you think I am (more of) a complete idiot, let me be more specific:

Does this symbol refer to a single function?
If not, does it refer to a family of functions?
If so, what family?

Once you’ve made up your mind, then ask yourself this second question: which of the following responses

(a) $ln(|x|)$
(b) $ln(|x|) + C$
(c) $displaystyle ln(|x)| + C + D frac{x}{|x|}$

is the correct answer to the question

$displaystyle int frac{1}{x} , dx = ?$

I’ll give my answer to the first question after the jump, and explain how I’ve always been wrong about it.

Okay, on to the big question:

What, precisely, is an “indefinite integral”?

That is, what does the symbol $displaystyle int f(x) , dx$ mean?

I’ve always taken $int f(x) , dx$ to refer to the family of all antiderivatives of $f(x)$ . In particular, an indefinite integral is really a set of functions. More specifically,

$displaystyle int f(x) , dx = left{ F(x) : F'(x) = f(x) mbox{ for all } x right}$ ,

which (assuming it’s nonempty!) is always an infinite family of functions, since whenever $F$ is an element of $latexint f(x) , dx$, then $G = F + mathrm{constant}$ is as well.

If $f(x)$ is defined on some interval $I$ , then the Mean Value Theorem implies that these are in fact the only possible antiderivatives; that is, if $F , G in int f(x) , dx$ , then $F - G equiv mathrm{constant}$ . As a result, if $F$ is any function defined on that same interval $I$ with $F' = f$ , then

$displaystyle int f(x) , dx = bigg{ F(x) + C : C mbox{ is a constant} bigg}$ .

Of course, in the calculus classroom, we usually abuse the notation and drop the curly braces and just write

$displaystyle int f(x) , dx = F(x) + C$ .

Hence, when I write something like

$displaystyle int x^2 , dx = frac{1}{3} x^3+ C$ .

I’ve always taken it to mean that every antiderivative of $f(x) = x^2$ is of the form $F(x) = x^3/3 + C$ for some appropriate value of $C$ .

(As an aside, I typically state computational bit of the Fundamental Theorem of Calculus as follows:

If $f$ is continuous on the interval $[a,b]$ , then $displaystyle int_a^b f(x) , dx = int f(x) , dx bigg|_a^b$

Of course, under my standard interpretation of the indefinite integral, this is another abuse of notation, for the definite integral on the left is a number whereas the expression on the right is, a priori, a set of infinitely many numbers. However, by virtue of the Mean Value Theorem, it is in fact a singleton set consisting of a unique number, which is, to me, part of the magic of the FTC.)

Okay, so far, so good.

Now, what about $int frac{1}{x} , dx$ ? Every calculus book on the planet writes

$displaystyle int frac{1}{x} , dx = ln (|x|) + C$ .

Unfortunately, under the interpretation of an indefinite integral above, this formula is wrong! More precisely,

$displaystyle left{ F(x) : F'(x) = frac{1}{x} mbox{ for all } x right} neq bigg{ ln (|x|) + C : C mbox{ is a constant}bigg}$ .

Certainly every function of the form $F(x) = ln(|x|) + C$ is satisfies $F'(x) = 1/x$ . However, there are other functions $G(x)$ with $G'(x) = 1/x$ that are not of this form. In particular, any function of the form

$displaystyle G(x) = ln(|x|) + C + D frac{x}{|x|}$

with $C, D$ constants also satisfies $G'(x) = 1/x$ on the same domain as $F(x) = ln(|x|) + C$ . However, while $G(x)$ looks like $F(x) + D$ for positive $x$ , it looks like $F(x) - D$ for negative $x$ , so it is not a vertical translate of $F$ .

With the Mean Value Theorem, it is easy to show that every function $G$ with $G'(x) = 1/x$ does take the form above, so according to my understanding of the indefinite integral, we should write

$displaystyle int frac{1}{x} , dx = ln(|x|) + C + D frac{x}{|x|}$

…but we don’t.

Clearly I’m wrong about something here… but where? What gives? What does an indefinite integral really mean?

I’ll share what I learned about this with you next week, but I’ll give you some time to resolve this quagmire for yourself first.

Updates!

(1) Apparently when I updated WordPress, it changed the comment settings, and was only allowing “registered WordPress” users leave comments. Thanks to the folks who sent me email about this problem… hopefully it’s fixed now!

(2) Part 2 is now available.

"Indefinite" is an understatement

Updates!

Leave a Reply Cancel reply

Recent Posts

Archives

Categories