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In discussing with my honors calculus class the fact that some continuous elementary functions do not have an elementary antiderivative, I realized I was unsure whether every discontinuous elementary function has an antiderivative at all.

The answer may depend on the precise definition of elementary function one uses. Note that the elementary function $\sqrt{x^2}/x$ equals +1 or -1 according to whether $x$ is positive or negative, and is undefined at $x=0$. Of course this particular discontinuous elementary function has an antiderivative on its domain. But once you have discontinuous functions like this, and you have trig functions, and you start forming compositions, you can get fairly nasty functions with infinitely many discontinuities, and it's not obvious to me that a function of this kind with domain $D$ necessarily has an antiderivative on $D$.

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  • $\begingroup$ Since I'm teaching the students from Stewart's book, let's use Stewart's definition as a starting point: elementary functions are "the polynomials, rational functions, power functions ($x^a$), exponential functions ($a^x$), logarithmic functions, trigonometric and inverse trigonometric functions, and all functions that can be obtained from these by the five operations of addition, subtraction, multiplication, division, and composition." $\endgroup$
    – James Propp
    Commented Mar 12, 2010 at 4:13
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    $\begingroup$ The exponential functions can be ornery beasts. (-1)^x is a terrible function on the negative real line, so you probably want to restrict yourself to a > 0. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 12, 2010 at 4:21
  • $\begingroup$ @Qiaochu: In fact Stewart's list did not include (-1)^x . @unknown: Unlike the "professional" definition, Stewart's definition did not include algebraic extensions. $\endgroup$
    – Gerald Edgar
    Commented Mar 12, 2010 at 13:07

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This shows that "elementary function" needs a good definition. We do NOT want to allow, for example $f(x) = 1$ when $x$ rational and $f(x) = -1$ when $x$ irrational. Even though $f^2 = 1$, this $f$ is not an algebraic function.

So, correctly defined, an elementary function is an analytic function on a domain in the complex plane, such that ...... [fill in the usual conditions]

Added later. My advice: For "elementary function" do not use the popularized form of the definition as in Wikipedia. Instead, use a definition from the actual mathematics papers. (Papers with proofs, not just quickie approximate definitions for the masses.)

For example

"Integration in Finite Terms", Maxwell Rosenlicht,
The American Mathematical Monthly 79 (1972), 963--972.
Stable URL: http://www.jstor.org/stable/2318066

Everything is carried out in differential fields ... In particular, every function involved is infinitely differentiable ... None of those "discontinuous elementary functions" mentioned in the question. Not even $|x| = \sqrt{x^2}$ is elementary.

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"Algebraic Properties of the Elementary Functions of Analysis", Robert H. Risch,
American Journal of Mathematics 101 (1979) 743--759.
Stable URL: http://www.jstor.org/stable/2373917

He also works in differential fields. Some quotes:

The elementary functions of a complex variable $z$ are those analytic functions that are built up from the rational functions of $z$ by successively applying algebraic operations, exponentiating, and taking logarithms. As is well known, this class includes the trigonometric and basic inverse trigonometric functions.

[Part II]
Suppose $\mathbb{C}(z, \theta_1, \dots, \theta_m) = \mathcal{D}_m$ is the abstract field, isomorphic to a field of meromorphic functions on some region $R$ of the complex plane, ...

==========

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    $\begingroup$ You give a semantic definition of 'elementary', whereas ever since Liouville, the usual definition is syntactic -- see Bruce Westbury's answer for the 'common' definition. $\endgroup$
    – Jacques Carette
    Commented Mar 11, 2010 at 23:21
  • $\begingroup$ When you change the meaning of "elementary function" you are misinterpreting the OP's question. I think it should be interpreted as: "does a function you could 'write down' in a typical calculus course always have an antiderivative?" $\endgroup$
    – Qiaochu Yuan
    Commented Mar 12, 2010 at 4:06
  • $\begingroup$ This is off topic (though interesting). The question is whether these functions have antiderivatives at all, not whether they have antiderivatives in closed form. $\endgroup$
    – David E Speyer
    Commented Mar 12, 2010 at 4:33
  • $\begingroup$ There is already a technical definition of "elementary function" that exists, and has existed since Liouville. The OP used it, rightly or wrongly. It's up to the OP to tell us if he erred in his usage or not. The answer to the question as you wrote it is simple: no, for trivial reasons -- it is trivial to write down everywhere undefined functions. If you restrict yourself to provably total functions, then the answer changes completely and becomes 'yes', but that is rather delicate to prove. $\endgroup$
    – Jacques Carette
    Commented Mar 12, 2010 at 4:36
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I assume you want $f$ to be defined everywhere on $D$. In that case, it's pretty clear for an "elementary" function like $\tan x$ that $D$ needs to have holes in it, and it's totally unclear what an integral "across" these holes should mean (from a real-variable perspective). You also probably don't want to consider functions whose integrals don't exist because they diverge, since you asked about discontinuities and unboundedness is a different reason for the integral not existing. That means you should really only consider the case where $D$ is a closed interval and $f$ is bounded.

In that case, there is a theorem due to Lebesgue which states that a function on a closed interval $[a, b]$ is Riemann integrable if and only if it is bounded and also continuous almost everywhere. This is true of essentially every reasonable "elementary" function I can think of; if you can write down an "elementary" function which is discontinuous on a set of positive measure then your definition of elementary is in trouble!

Edit: The functions you listed in your comment all have the property that they are continuous on intervals where they are defined, so they'll all have the above property and so will sums, products, and compositions thereof.

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"Integration in finite terms" uses an exact definition of the class of elementary functions. According to J. F. Ritt, $\exp$, $\ln$ and the algebraic functions are analytic almost everywhere, and therefore the elementary functions.

"Integration in finite terms" treats only formal antiderivatives. Clearly, the concrete antiderivative depends on the concrete domain of the function in the integrand. If you integrate functions by applying other methods, you have also make decisions about the domain of the functions.

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What about something like $\sqrt{cos(x)-1}$? This function is just a bunch of discrete points sitting at $(2n\pi,0)$, but it is a composite of functions most people would consider elementary.

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  • $\begingroup$ The theory of anti-derivatives of elementary functions is usually stated algebraically. The most natural interpretation is then over the complex rather than the reals, where the above function becomes defined everywhere (but is still extremely nasty). $\endgroup$
    – Jacques Carette
    Commented Mar 11, 2010 at 23:19
  • $\begingroup$ I thought it was clear from the initial post that the OP in functions over the reals, but it may be true that the most natural place to consider the problem is over the complex numbers. $\endgroup$
    – Steven Gubkin
    Commented Mar 12, 2010 at 0:26
  • $\begingroup$ This answer shows that there is a second subtlety to the OP's question: it's not clear how to define integrability (from an honors calculus point of view) for a function defined on a discrete domain. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 12, 2010 at 4:07
  • $\begingroup$ Moreover, Steven Gubkin's nice example shows that Stewart's assertion that every elementary function has an elementary {\it derivative\/} (see Chapter 6 Review, True/False Question 9(a), and the answer given at the back of the book) is false unless one adds some caveats about the nature of the domain. $\endgroup$
    – James Propp
    Commented Mar 12, 2010 at 4:18
  • $\begingroup$ - Jim Propp (OP) $\endgroup$
    – James Propp
    Commented Mar 12, 2010 at 4:19
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Just to reiterate: we need to know what you have in mind by an elementary function.

The algebraic approach is to take say polynomials, $\exp(x)$ and perhaps $\ln(x)$ and then to say a function is elementary if it can be written in terms of these using operations of linear combinations, multiplication, composition. Then there are elementary functions with no antiderivative. So the obvious thing to do is to extend the definition of elementary function to include these antiderivatives. The theorem is that this doesn't work.

There are algorithms which are implemented in computer algebra systems which given an elementary function will either find an antiderivative or will prove that no elementary function is an antiderivative.

Response This answers the question the OP asked with the definition of elementary that OP intended. This appears to have caused offence so I wish to state my position.

It is a fact of life that in the classroom and in computer algebra systems that functions are described by formulae. This leads to many problems and the (now defunct) symbolic algebra newsgroup was inundated with complaints from people who did not appreciate the limitations of this approach.

There are several interesting discussions we could have on the issues that arise from this but I don't think this would be welcome on this site.

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    $\begingroup$ So there are elementary functions with no elementary antiderivatives. The question is whether there exists an elementary function that is not integrable at all. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 12, 2010 at 4:00
  • $\begingroup$ That's not what OP says. The responses to this question are producing more heat than light. I have answered the question OP asked with the definition of elementary function that he had in mind. $\endgroup$
    – Bruce Westbury
    Commented Mar 12, 2010 at 6:46
  • $\begingroup$ Qiaochu (whose name really ought to have an "e in it somewhere, for completeness! :-) ) is correct about my intention, though I hate to say Bruce is "wrong"; most likely there was some ambiguity in my original posting. $\endgroup$
    – James Propp
    Commented May 7, 2010 at 0:34
  • $\begingroup$ @James The responses have been edited so we now have (more or less) a consensus on what is meant by an elementary function. I agree with Qiachou that you can define the Riemann integral of an elementary function (ignoring any caveats). My point is that there are elementary functions whose anti-derivative is not an elementary function. I took this to be a question about teaching rather than research. At a basic level "functions are defined by formulae". At this level a function is defined by a formula and you require the anti-derivative to be defined by a formula. It turns out that this fails. $\endgroup$
    – Bruce Westbury
    Commented May 7, 2010 at 10:18
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I haven't a concrete example but there is a theorem that says that the derivative of a function has the intermediate value property. This fact makes me think that could be a elementary function without antiderivative.

EDIT:

I am going to be more explicit:

If f is a elementary function, it is defined in the interval (a,b), and it is the derivative of another function (not necessary an elementary function) then f satisfies the intermediate value property inside (a,b).

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  • $\begingroup$ Darboux's theorem requires conditions on the function. So what you say is incomplete (sorry to nitpick, but the difference between "a reasonable-sounding function" and "any function" is vast). $\endgroup$
    – Yemon Choi
    Commented Mar 11, 2010 at 21:21
  • $\begingroup$ If a function is differentiable everywhere in an interval, then the derivative has the intermediate value property. Defined "everywhere except one point" is not good enough. This example is defined everywhere except one point. $\endgroup$
    – Gerald Edgar
    Commented Mar 11, 2010 at 21:22
  • $\begingroup$ The derivative of the integral of an integrable function need not exist everywhere. The obvious example is, for example, a step function. (The fundamental theorem of calculus as it is usually stated only applies to continuous integrands!) $\endgroup$
    – Qiaochu Yuan
    Commented Mar 12, 2010 at 4:02
  • $\begingroup$ I want to say that there is a necessary condition in order to be the derivative of a derivable function defined in the interval (a,b). This condition is the intermediate value condition. I never use what you say, maybe I just misstated because I am not a native English speaker. I don't understand what is wrong in my post. Could you be more explicit please? $\endgroup$
    – Quimey
    Commented Mar 12, 2010 at 12:33

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