A different symbol for the indefinite integral/antiderivative?

Examples. An indefinite integral (or antiderivative) of $$\cos$$ is $$\sin$$:

$$\int \cos = \sin.$$

Edit: There has been much unexpected confusion with the above statement. I define the above statement to mean precisely that an antiderivative of the cosine function (which has domain $$\mathbb R$$) is the sine function (which has domain $$\mathbb R$$). Or equivalently, the derivative of the sine function is the cosine function. This notation though is not what the question is about, so I hope you will not dwell on this. If you want to question the merits of this notation, that is perhaps for another question. If it really bothers you, you can just pretend the above is statement instead $$\int \cos x = \sin x$$ or $$\int \cos x \, \textrm dx = \sin x$$ or $$\int \cos x \, \textrm dx = \sin x + C$$. (I personally assign to these latter statements slightly different meanings. And yes, I do explain all of this very clearly and repeatedly to my students.)

The definite integral (or "area function") of $$\cos$$ from $$0$$ to $$\pi/2$$ is $$1$$:

$$\int^{\pi/2}_0 \cos = 1.$$

These are distinct concepts. But by custom, we use the same symbol for them, because the Fundamental Theorem of Calculus tells us that

$$\int^{\pi/2}_0 \cos = \left(\int \cos\right)\left(\frac{\pi}{2}\right)-\left(\int \cos\right)(0) = \sin \frac{\pi}{2}-\sin 0 = 1- 0=1.$$

This often causes students a great deal of confusion (see e.g. below "Related (but distinct) discussions"). In particular, students

• don't see any difference between the indefinite and definite integrals;
• believe that integration is, by definition, the inverse of differentiation; and
• thus fail to understand that the significance of the FTC or see that it is more than a mere definition.

To reduce the above confusion, I have already tried to follow @PeteL.Clark's recommendation (in this answer) of using the term antiderivative (rather than indefinite integral) as much as possible.

I am now thinking of also trying the following approach:

1. First introduce antidifferentiation. Never once mention the word integral. Don't use $$\int$$ as our symbol for the antidifferentiation operator---instead, use some other symbol like $$\square$$. In which case, we'd write things like $$\square \cos=\sin$$.
2. Introduce integration (i.e. the definite integral) as usual.
3. Go through the FTC, explaining that integration turns out to be, in a certain precise sense, the "same thing" as antidifferentiation.
4. Explain that because of the FTC (and custom), we'll now discard our $$\square$$ symbol for the antidifferentiation operator and replace it with $$\int$$.

I don't think I've ever seen any writer using this approach (but please let me know if you have). So, what disadvantages might the above approach have?

Pete L. Clark also writes

At some point I slip up because after all we are using the same integral sign for both: I almost wish we didn't. (Using almost identical notation for two a priori incredibly different things which in the setting of the FTC become almost the same is one of the brilliant notational innovations of calculus.

I don't quite understand the above quote. I understand that today, after centuries of use, it would be difficult and foolish to go against tradition and attempt to completely replace the $$\int$$ symbol for the indefinite integral/antiderivative with something else altogether.

But why wouldn't it have been better if mathematicians had just decided to use a different symbol from the start? What is so brilliant about using the same symbol $$\int$$ for "two a priori incredibly different things"?

It seems that this has simply resulted in a great deal of confusion (at least among average students of calculus), without any obvious gain in convenience or understanding.

Related (but distinct) discussions:

• I am not familiar with the notation $\int\cos=\sin$. Do you perhaps mean something like $\int\cos(x)\,\mathrm{d}x=\sin(x)+c$? – Joel Reyes Noche Sep 9 at 6:01
• @JoelReyesNoche: $\int\cos=\sin$ just means that an indefinite integral (or antiderivative) of $\cos$ is $\sin$, or equivalently, that the derivative of $\sin$ is $\cos$. – iqntt1s Sep 9 at 7:16
• Priestley (1997) doesn't use a separate symbol, but she "completely outlaws" the use of $\int$ as a symbol for the antiderivative: When students first meet integrals they are often taught to write, for example, '$\int\sin x = -\cos x$' to mean 'the antiderivative of $\sin$ is $\cos$, that is, $(d/dx) \cos x = -\sin x$'. We outlaw this usage completely. It is at variance with the notation adopted in 5.6, and is apt to cause confusion. – user20311 Sep 9 at 7:21
• I'm confused, because I think that $\int \cos(x)dx$ and $\int_0^{\pi/2} \cos(x) dx$ are entirely different in terms of notation. Do you also believe the evaluation of improper integrals needs yet a third symbol/notation, so as to to confuse it with definite integration? – Namaste Sep 9 at 13:15
• I'm not an educator but an engineer, and after finding this question on the HNQ I would like to understand: why is it a bad thing to consider the definite and indefinite integrals to be two sides of the same thing? Perhaps I'm one of the students who's confused by your standards, but I don't see it as confusion as much as just being a different way of understanding the same fundamental thing. Whether the definite and indefinite integrals are two sides of the same thing or two different closely related things sounds like a question of philosophy, not mathematics. – Hearth Sep 9 at 14:32

First, yes, many teachers use that term, "anti-derivative" and "integral" only when a definite integral is in use.

For your examples of each, it seems to me that you offered a clear distinction between the two expressions, i.e. that the bounds is what what makes the integral definite. For me, that's where the explanation to students tends to reach a satisfying conclusion.

For your desire to introduce a new symbol, please, don't. No matter how well intended, no good can come of it. If you use a textbook, it will confuse students who really need written material to follow the lectures. If you teach 100% from notes, they will still have an issue when

(a) moving on to the next course, and

(b) going on line for any further assistance.

There are already multiple symbols used for derivative, with centuries of history. Best to not add to the confusion.

• This reminds me of Peirce's new symbols for $e$ and $\pi$, see e.g. jstor.org/stable/2299486 if you have access to the Monthly. – kcrisman Sep 12 at 1:27
• No offense to Peirce, but his “the existing symbols are inconvenient”? Interesting article. – JTP - Apologise to Monica Sep 12 at 1:43

I'll echo other responses with the same: Do NOT introduce made up notation.

I've made the effort in my Calculus courses to follow your outline while avoiding any new symbols. Similar to user20311's comment, when I first cover antiderivatives, all questions are phrased as either

• Find an antiderivative of $$\sin(x)$$,
• Find $$F(x)$$ such that $$\frac{d}{dx}F(x) = \sin(x)$$, or
• Find $$F(x)$$ such that $$F'(x) = \sin(x)$$.

In fact, using capitalization to indicate antiderivative was Newton's original notation. (In the examples I've seen, Newton usually was working with a polynomial $$a(x)$$ and its antiderivative $$A(x)$$.)

Then, as you indicate, after covering Riemann Sums, using the $$\int_a^b$$ sign for definite integration, and covering the FTC, then introduce the $$\int$$ for antiderivative (which was, of course, Leibniz's notation).

• I would recommend the question "Find three different antiderivatives of $\sin(x)$" to really go another step in this direction. See also the discussion at matheducators.stackexchange.com/a/2340/11 – Chris Cunningham Sep 10 at 18:27
• I agree. If I use the actual word "antiderivative", I frequently ask for the "general antiderivative", the "family of antiderivatives", "the antiderivative through a specific point," or, as you say, 3 examples of antiderivatives. – Aeryk Sep 10 at 18:49

Before inventing new notation, it is very important to learn the accepted notation and teach it to your students correctly. The question contains the following claim:

$$\int \cos = \sin$$

which is not a meaningful statement. It is either incorrect, or unclear.

This notational issue is not a nitpick: it is a critical issue that, if it goes uncorrected, will prevent your students from succeeding in future courses.

Following your notation, the student who attempts to evaluate this integral:

$$A = \int \cos(2x) \mathrm{d}x$$

is not only more likely to arrive at the incorrect answer -- they will also be unable to see how their actions are in conflict with the "fact" you have told them, namely that "$$\int \cos = \sin$$." And this is only the beginning.

I would urge you, therefore, to not only avoid inventing any new notation, but to think deeply about why you don't like the standard notation and ask questions to colleagues or this site about why the notation is what it is. Specifically in this case, you should write your fact as

$$\int \cos(x) \mathrm{d}x = \sin(x) + C$$

so that students who use the rule incorrectly can be pointed at the detail that the fact is very clear that the integral of cosine of $$x$$ $$\mathrm{d}x$$ is the sine of $$x$$ up to a constant, and that it makes no claims about the cosine of $$2x$$.

• True, though for graduate students I stress the pointlessness of dummy variables... and the implied constant of integration... and many other best-treated-as-context-dependent things, rather than attempting a formalism that pretends to avoid awareness of context. – paul garrett Sep 10 at 21:33

An alternate notation that I made up for derivative and antiderivative of $$f$$ are $$f^{\downarrow}$$ and $$f^{\uparrow}$$. I experimented with them when trying to explain or write steps of tableau method or ladder method for multiple integration by parts. (This is hallowed ground so I will be offending many, but I go ahead and drop the variable and its differential, keeping in mind that this is for students of Calculus 1!) By mixing the notations one can write

\begin{align} \int (fg) &= f g^{\uparrow} -\int (f^{\downarrow} g^{\uparrow}) \\ \int (fg) &= f g^{\uparrow} - f^{\downarrow} g^{\uparrow \uparrow} + \int (f^{\downarrow \downarrow} g^{\uparrow \uparrow}) \\ \int (fg) &= f g^{\uparrow} - f^{\downarrow} g^{\uparrow \uparrow} + f^{\downarrow \downarrow} g^{\uparrow \uparrow \uparrow} - \int (f^{\downarrow \downarrow \downarrow} g^{\uparrow \uparrow \uparrow}) \end{align}

etc.

You can call them "down" and "up" operations due to what happens to the exponent of $$x^n$$ in the operation. So you have "double down" and "double up", and so on. Useful for students who might have a hard time with steps in $$\int x^n e^{ax} dx$$. You can push it and write $$f^{\uparrow n}$$.

• This is cute and I rather like it, but one would need to be very careful using it in an elementary classroom, as it is notation which is certain to not be understood in the wider world. – Xander Henderson Sep 16 at 23:29

It's the whole subject of the Bourbaki's book IV Integration.

First generalize your notion of derivative: a derivative is any function $$d:A \rightarrow A$$ of the space $$A=\mathcal{C} ^ \infty(\mathbb{R})$$ such that

• $$d(f+g) = d(f)+d(g)$$
• $$d(\lambda f) = \lambda f$$ ($$\lambda \in C^{te}) \cong R \subset A$$
• $$d(fg) = f d(g) + d(f) g$$

The first two mean the operator $$d$$ is linear, the third is called the Liebnitz rule. Please note that this definition is purely algebraic, and this was the goal of Weil: to be able to make differential geometry on $$Z/pZ$$ (or their topological completions).

The main result is to show that every derivative is of the form $$df = f'(x) d\mu(x)$$ for a suitable $$\mu$$ (note that $$\mu$$ is rarely the identity $$\mu(x)=x$$). More rigorously, the work aims to establish for which classes of algebras $$A$$ the above is true. But suffice to say here that it is at least true for $$A=\mathcal{C} ^ \infty(\mathbb{C})$$, thus for $$A=\mathcal{C} ^ \infty(\mathbb{R})$$.

This is true, because by a fundamental theorem any holomorphic function has an antiderivative, so it is possible (but not easy) to find $$\mu(x)$$ such that $$f = \int {d(f)(x) d\mu(x)}$$. Note that the measure $$d\mu$$ is found after the derivative $$d$$ is given, by globalizing (this is the hard part) along all $$f$$. Not moreover that $$\mu$$ is rarely a function $$\in A$$ (it is a distribution defined except on a countable subset), nor unique (the two distributions can differ on a countable subset).

However, it exists. So your approach is perfectly legal. You can use any distribution, in particular $$\mu(x)=x$$, so that $$d\mu(x)=dx$$, and you can make this assumption implicit: all your formulas will be correct, as long as they only involves only algebraic relationships.

In facts, you will see that calculus is often made easier and clearer by using only the above stated rules. This is especially true in $$\mathbb{R^n}$$.

When you'll have too look at the meaning of $$d/dx$$ or $$\int (\cdot) dx$$, you'll have many choices: Bourbaki's view of distributions generalizing functions, integral in the sense of Borel or Lebesgue, Liebnitz' infinitesimals, non-standard analysis (Robinson ultrapower or Nelson Internal Set Theory). There are all different semantics, which surface to the same syntax: any algebraic or calculus that can be made in one theory is feasible in the others and gives the same result(s).