Does a function, $f(x)$, exist such that $\int f(x) dx $ can be found but $f' (x)$ cannot be found in terms of elementary functions.

For example, if $f(x)=e^{x^2}$, then the derivative is easily calculated by using the chain rule. However, there does not exist an anti-derivative in terms of elementary functions.

Does a function exist with the opposite property?

  • 5
    $\begingroup$ 18 people think this is well researched. Assuming by elementary you mean "nice", like polynomials and trig and such, note that $\frac{d}{dx}\left[\int^x_pf(t)dt\right]=f(x)$ (THE FUNDAMENTAL THEOREM OF CALCULUS!) if the integral is "elementary" then the derivative of that (f) must be, and the derivative of that must be too. $\endgroup$
    – Alec Teal
    Commented Dec 14, 2015 at 21:58
  • 9
    $\begingroup$ @AlecTeal Or they think it is useful and clear :P (Or they wanted a hat...) $\endgroup$ Commented Dec 14, 2015 at 23:00
  • $\begingroup$ @AlecTeal You don't need to assume what 'elementary' means; it is a long-ago established term with a well-known meaning (as opposite to 'nice'!). See Elementary function in Wikipedia. $\endgroup$
    – CiaPan
    Commented Dec 15, 2015 at 9:26
  • 1
    $\begingroup$ If you consider the Dirac Delta Function to be elementary, it meets the criteria; however this is unfair because it has an integral because it is defined to have an area under its curve despite having no width. $\endgroup$
    – Joshua
    Commented Dec 15, 2015 at 16:44
  • 2
    $\begingroup$ @Joshua: The reason why the Dirac delta "function" doesn't qualify is because, despite the widely used name, it's not really a function but a distribution. $\endgroup$ Commented Dec 16, 2015 at 2:59

6 Answers 6


If the antiderivative $F$ of $f$ is elementary, then so is $f' = F''$ (for any reasonable definition of "elementary function"). Thus, no such example can be found.


Here are some more details which were adressed in the comments and/or other answers:

  1. What I assume here is that for your favorite definition of "elementary function", the following is true: Every elementary function is differentiable and the derivative is again an elementary function.

    This is indeed fulfilled (on the respective domains) if you take as your elementary functions all functions which can be obtained from $\exp, \ln, \sin, \cos$ and polynomials by taking sums/quotients/products and compositions of these functions. This is a consequence of the chain rule.

    It is not fulfilled, however, if you also want to include roots, since e.g. $x \mapsto \sqrt{x}$ is not differentiable at the origin. But note that it is true if you only consider the roots as functions on $(0,\infty)$ instead of $[0,\infty)$.

  2. I assume that if your function $f$ has a continuous version (with respect to equality a.e.), you identify it with its continuous version.

    As noted in the answer of @RossMillikan, the Dirichlet function $f = 1_\Bbb{Q}$ is (Lebesgue)-integrable with "antiderivative" $x \mapsto 0$, but not differentiable. But note that $f = 0$ almost everywhere, which is elementary and has an elementary derivative.

    Finally, if $F(x) = \int_a^x f(t) \, dt$ is elementary, then (by Lebesgue's differentiation theorem) you have $f(x) = F'(x)$ almost everywhere. Hence, if $F$ is elementary (as outlined in point 1), then $F'$ is elementary and hence continuous, so that we get $f = F'$ almost everywhere. Since we agreed to identify $f$ with its continuous version, we get $f = F'$ everywhere, so that $f$ is differentiable with $f' = F''$ elementary, as claimed above.

  • 7
    $\begingroup$ Could you shed some light on how we know that the derivative of every elementary function is elementary? $\endgroup$
    – jpmc26
    Commented Dec 14, 2015 at 16:36
  • 1
    $\begingroup$ @jpmc26: Of course, this depends on what you mean by an elementary function. If, as in Ross Millikan's answer below, you consider $x \mapsto |x|$ to be elementary, this is false. But usually, one only considers functions like $\exp, \sin, \cos$, polynomials, quotients of polynomials, etc. These are always differentiable on the set where they are defined. $\endgroup$
    – PhoemueX
    Commented Dec 14, 2015 at 17:00
  • 2
    $\begingroup$ @jpmc26 if you are curious for an actual proof, I wrote a paper on the topic and give a link in my answer below $\endgroup$
    – ASKASK
    Commented Dec 15, 2015 at 2:10
  • 2
    $\begingroup$ "To rephrase, does a function, $f(x)$, exist such that $\int f(x)dx$ can be found but $f′(x)$ cannot be found in terms of elementary functions. $\endgroup$
    – ASKASK
    Commented Dec 15, 2015 at 2:17
  • 1
    $\begingroup$ @Anixx He asks for an integrable function whose antiderivative is an elementary function. The only scenario in which this can occur is if the function itself it elementary. $\endgroup$
    – ASKASK
    Commented Dec 15, 2015 at 2:20

Although the other answers say differently, I would put up the example of the Weierstrass function which is a pathological mathematical idea.

Quoting from Wikipedia:

In mathematics, the Weierstrass function is an example of a pathological real-valued function on the real line. The function has the property of being continuous everywhere but differentiable nowhere. It is named after its discoverer Karl Weierstrass.

Historically, the Weierstrass function is important because it was the first published example (1872) to challenge the notion that every continuous function was differentiable except on a set of isolated points.

In Weierstrass' original paper, the function was defined as the sum of a Fourier series:

$$f(x)=\sum_{n=0} ^\infty a^n \cos(b^n \pi x)$$ where $0<a<1$, $b$ is a positive odd integer, and

$$ab > 1+\frac{3}{2} \pi$$

The minimum value of $b$ which satisfies these constraints is $b=7$. This construction, along with the proof that the function is nowhere differentiable, was first given by Weierstrass in a paper presented to the Königliche Akademie der Wissenschaften on 18 July 1872.

The proof that this function is continuous everywhere is not difficult. Since the terms of the infinite series which defines it are bounded by $\pm a^n$ and this has finite sum for $0 < a < 1$, convergence of the sum of the terms is uniform by the Weierstrass M-test with $M_n = a^n$. Since each partial sum is continuous and the uniform limit of continuous functions is continuous, it follows $f$ is continuous.

As is evident from the functional form of this special function, it does have an antiderivative. So this can be considered a valid example.

  • 3
    $\begingroup$ For an example of the antiderivative of a Weierstrass function, see JimmyK4542's answer to What does the antiderivative of a continuous-but-nowhere-differentiable function “look like”? $\endgroup$ Commented Dec 14, 2015 at 19:17
  • 5
    $\begingroup$ +1 This is a great answer, but its non-acceptance is justified in that the question is looking for elementary functions. $\endgroup$ Commented Dec 15, 2015 at 19:22
  • 2
    $\begingroup$ @ToddWilcox: The only thing the question asks for is that the derivative should not be expressible in terms of elementary functions - so this answer should be OK, right? $\endgroup$
    – psmears
    Commented Dec 16, 2015 at 8:14
  • 5
    $\begingroup$ Great answer, +1, but I disagree with calling the Weierstrass function pathological because it is nowhere differentiable. To the contrary, this is the typical situation for a function just required to be continuous. Your link describes pathological as "an example specifically cooked up to violate certain almost universally valid properties" but there are no "almost universally valid properties" being violated here (and the link contradicts itself by citing this very example). Many real-world continuous processes behave like Brownian motion; the generic behaviour is highly non-smooth. $\endgroup$ Commented Dec 16, 2015 at 13:28
  • $\begingroup$ @MarcvanLeeuwen You are totally right. I myself had no idea why it should be called pathological but several of my friends and some of my mathematics professors prefer to call this pathological since they assume that the function was cooked up and constructed specifically to highlight this basic unusual property, same to the reason why Todd Wilcox above fails to call this an elementary function. And for the application of this function to describe Brownian motion, many consider it the Koch curve better to describe it. (contd...) $\endgroup$ Commented Dec 16, 2015 at 19:00

$f(x)=|x|$ is integrable, with integral $\frac 12x|x|$. You can take $f'(x)$ everywhere but zero.

The Dirichlet function, which is $1$ on the rationals and $0$ on the irrationals, is Lebesgue integrable (with value $0$) but has no derivative anywhere.

  • 12
    $\begingroup$ Does the Dirichlet function have an antiderivative? $\endgroup$ Commented Dec 14, 2015 at 16:59
  • 7
    $\begingroup$ @ToddWilcox: that depends on the sense in which you read the integral sign. If it is a Riemann integral, no. If it is a Lebesgue integral, yes, and it is constant $\endgroup$ Commented Dec 14, 2015 at 20:27
  • 1
    $\begingroup$ @ToddWilcox: Good point. I guess in a 'weak'/almost-everywhere sense it does, but only thanks to the non-one-to-one nature of passing from a function $f$ to its indefinite integral $x \mapsto \int_0^x f(t) dt$ (all constants differentiate back to the zero function even if they came from a discontinuous function). Perhaps this is one of those instances where something different is meant by antiderivative (like how conventions vary regarding zero as a natural number or rings without a unit element). $\endgroup$ Commented Dec 14, 2015 at 20:28
  • 2
    $\begingroup$ Oh, I think I see. So its "Lebesgue Antiderivative" (nomenclature abuse is one of my hobbies) might be said to be $F(x) = 0$? Except we couldn't then say that $F'(x) =$ (the Dirichlet function). $\endgroup$ Commented Dec 14, 2015 at 20:45
  • 1
    $\begingroup$ @ToddWilcox: exactly. Two integrable functions that are equal except on a set of measure 0 have the same Lebesgue integral, so we should state the fundamental theorem of calculus carefully (e.g. assert continuity, although that's overkill) before slinging around words like "antiderivative" :-) For that matter the same goes with Riemann integration, and two functions that are equal except at a single point. $\endgroup$ Commented Dec 15, 2015 at 14:31

PhoemueX's answer is correct, but if you want a little more detail on the subject, I actually wrote a paper on the topic that goes on to prove that all elementary functions have elementary derivatives (and quite a bit more). All it requires is an understanding of high school and some basic experience with proofs.

Here is a link to the paper

I explain what it means for a function to be an elementary function in section 2 (page 2), and then I begin section 3 (page 6) by proving that all elementary functions have elementary derivatives.


The question leaves two details up to interpretation, and another that is precise may well be so by accident. This is important because the answer will be yes or no depending on how the details are made more precise. However, for most ways of making the details precise, the answer is yes.

  • What do you mean by function? Apparently you are referring to functions from the reals (or possibly an interval - this doesn't really matter) to the reals. When mathematicians say "function" they usually mean functions (often implicitly restricted to functions from the reals to the reals) in the set-theoretical sense, which is the most general practical notion. But it seems you may have in mind a much more restricted definition, such as continuous functions, functions that can be described piecewise or even globally by power series, or even elementary functions.
  • What do you mean by "can be found"? Some possible ways to make this precise: It is consistent with set theory that the antiderivative exists. It follows from standard axioms that the antiderivative exists. There is a constructive proof that the antiderivative exists. We can write down a power series for the antiderivative. The antiderivative is an elementary function.
  • What do you mean by antiderivative? There is a precise, standard definition, but given your alternative explanation in terms of an integral it appears that that's not necessarily what you have in mind. There are many different definitions of integrals of varying generality (i.e. a non-continuous function may have an integral w.r.t. one notion but not another), and for all of them it makes some sense to refer to them as antiderivatives (in a wider sense). Fortunately, this detail turns out not to be relevant for the ultimate outcome, under reasonable assumptions on how the previous two points are made precise.

It is very well known that there exist functions from the reals to the reals which have a derivative $f$ that is not even continuous. (This is why many theorems use the words "continuously differentiable" in their assumptions, rather than the more general "differentiable".) Since a non-continuous function never has a derivative, any such function $f$ is an example of a function with an antiderivative but no derivative. But you may not consider this a real example because you may be interested only in continuous functions.

I believe some of these functions can be found in the strong sense that you can give precise definitions of them (and of their antiderivatives) - though not as power series, and certainly not as elementary functions. And the derivative cannot be found - in the strong sense that it definitely does not exist.

If you are only interested in continuous functions, then the answer is still yes. It is very well known, and easy to see, that there are continuous functions $f$ that are not differentiable. But every continuous function has an antiderivative. Again, this argument works with all definitions of "can be found".

If you are only interested in elementary functions, then the answer is no. It is well known that the derivative of every elementary function is an elementary function, but that there exist elementary functions whose derivatives are not elementary.


It depends on the presentation of the function.

If a function is written as a power series or Fourier series, integration of the terms of the series smooths it and improves convergence, while differentiation makes it rougher and less convergent (or divergent). For instance, the Weierstrass nowhere differentiable function is expressed as a Fourier series, and can be integrated, but the coefficients get much larger upon differentiation, which spoils convergence.

This is related to the fact that diffusion equations such as heat flow can be solved forward in time, because the data get smoother, but solving the differential equation backward in time is an "ill-posed" problem.

If the function is represented by a finite algebraic formula then differentiation is always possible and integration often not.

  • 2
    $\begingroup$ What does "finite algebraic formula" mean here? $\endgroup$ Commented Dec 15, 2015 at 2:24
  • 1
    $\begingroup$ For example, the function is piecewise an element of the closure of the field of rational functions with real coefficients under (finite) algebraic and differential extensions. $\endgroup$
    – zyx
    Commented Dec 15, 2015 at 2:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .