"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.