Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The elementary real-valued functions are not closed under integration. (Elementary function has a precise definition -- see Risch algorithm in Wikipedia). This means there are elementary functions whose integrals are not elementary. So we can construct a larger class of functions by adjoining all the integrals of elementary functions. You can repeat this process indefinitely. If I understand things correctly, the set of functions that is the countable closure of this process is closed under integration. Does any finite iteration of the process achieve closure under integration?

My guess is no. Has anyone thought about this?

share|cite|improve this question
Probably people who work in differential Galois theory have thought about this; someone familiar with the field should be able to give a straight answer. –  Qiaochu Yuan Aug 22 '10 at 3:16
Probably somebody who can explain this paper: to lesser beings like myself have a chance at resolving this question to everybody's satisfaction. –  Guess who it is. Aug 23 '10 at 7:29
Does the elementary integral of an elementary function always have an elementary integral? –  superM Apr 9 '13 at 19:12
I don't know what you mean by "elementary integral". If you mean "if the integral f of a given elementary function g is also elementary, is f elementary? The answer is no: e^(x^2) is the integral of the function 2x e^(x^2), but e^(x^2) is not elementary. –  SixWingedSeraph Apr 10 '13 at 22:29
See also…. –  lhf Jun 16 '14 at 21:12

3 Answers 3

I'm not entirely sure it can be closed; apparently more and more complicated integrals of compositions of elementary functions require further generalizations of the "hypergeometric function" to several variables; for instance, the integral of $\sin(\sin(x))$ requires a Kampé de Fériet hypergeometric function for its closed form, and the elliptic integrals (integrals containing square roots of cubic or quartic polynomials) require the Lauricella or Appell hypergeometrics. As to whether there is "the one hypergeometric to rule them all", I think that's an open question.


To clarify the little discussion Qiaochu and I had in the comments for other readers: the reason you should be interested in the hypergeometrics is that virtually all the elementary functions, and combinations thereof (sum, products, compositions, etc.) are expressible as hypergeometrics. As I have shown in the example I gave in the comments, integrating a hypergeometric function with p+q+1 arguments can require a hypergeometric function with p+q+3 arguments. If you imagine this procedure being applied by multiplying a hypergeometric function with each term of a power series, you get even more complicated hypergeometric functions (in that link, you have the multivariate generalization of the hypergeometric function known as Meijer's G function. Integrating a Meijer function can give you a Meijer function with even more arguments).

So, the question one should be asking, I think, is that if there exists "the one hypergeometric to rule them all"; or to (mis)use another colloquialism: "the mother of all hypergeometrics".


I guess all I have been trying to say, after all this elaboration: unless one would be satisfied with an "it's turtles all the way down" sort of answer, then I think there's no way to satisfactorily resolve this question.

share|cite|improve this answer
Yes, that's clear. What the OP is asking is, if you take all of those functions and allow sums and products and compositions, is the resulting set of functions closed under indefinite integral? If not, what if you took all of the indefinite integrals of those... –  Qiaochu Yuan Aug 22 '10 at 3:49
Qiaochu: I did essentially say that you keep getting hypergeometrics (with more variables!) when you integrate hypergeometrics. As to whether there's a ceiling... that's what I'm sure is open. –  Guess who it is. Aug 22 '10 at 3:57
To use the simplest example, consider… ; the integral of the product of a ${}_p F_q$ hypergeometric function with a power function requires a ${}_{p+1} F_{q+1}$ function; now imagine this being done termwise to a hypergeometric function multiplied by a power series. You keep getting hypergeometric functions with more and more parameters/arguments. Now, where does it end? –  Guess who it is. Aug 22 '10 at 4:10

In a differential Galois field, integration is the antiderivative operation and not the infinite summation, but I will still call it integration.

If we start out with simple elementary functions, say rational functions with integer coefficients, then there will be many functions that do not have integrals. If the original set of functions is finite, then only a finite number of functions need to be added. The process can then be repeated at the next level. The only catch is that at every stage one has to deal with constants (think of the roots of the denominators). Most functions force the addition of several integrals. The number of functions added grows geometrically with the levels, so I would expect there is no easy way to define a countable closure.

Are there initial sets for which the process stops? Yes, polynomials for example. Are there initial sets for which it goes on forever? Yes.

share|cite|improve this answer
More or less what I was trying to say. The integral of a hypergeometric is still a hypergeometric, albeit it can be more complex. Does it count as closure if the number of parameters can grow with each application of integration, or is there a "mother of hypergeometrics" that will cover all those integrals? –  Guess who it is. Aug 22 '10 at 22:27
I find this suggestive, but vague. Can you give more details and/or references to the literature? –  Pete L. Clark Aug 23 '10 at 1:48
Why don't we be extreme for kicks and giggles: are all solutions of Fuchsian DEs closed under integration? –  Guess who it is. Aug 23 '10 at 7:43
Formal power series are closed under antiderivatives and so is its subset, the poly-exp set of elementary functions. Can one find many other subsets? Yes. Can one characterize all the subsets? This may be impossible to do explicitly because it is very common to run into undecidability issues. If one takes a more analytic approach and worries about convergence of all these compositions, the common domain of all the functions will likely shrink with composition and end up with no opens. An interesting set of functions closed under antiderivatives seems like a cool research problem. –  papin Aug 23 '10 at 10:17

Alternate between applying the integral to each element and closing the set with respect to the elementary operations. Let's call the smallest number of such compound iterations necessary to construct a function its "rank". So exp has rank 0, but erf, erf(exp), and erf + exp all have rank 1.

The integral of a function with rank R may have rank R or R+1. The derivative of a function with rank R may have rank R or R-1, and the rank of the derivative of a function with rank 0 is always 0. If the derivative of a function with rank R has rank R-1, let's call that function "primitive" to distinguish between the two phases of the iteration. A non-primitive function can by definition be written in terms of primitive functions of the same rank (and possibly other functions of smaller rank). For example, erf is primitive, but erf(erf) is not, although they are of the same rank 1.

Suppose R is the largest rank. Then the integral f of any rank-R function f' is a non-primitive rank-R function. It is possible to express f in terms of functions that have a rank smaller than R or are primitive and have rank equal to R, and at least one function of the latter type is required. The sum of any two primitive functions remains primitive, so in the general case we have such a function appearing in a non-additive expression. For the same reason we can find a case where it is not the root of the expression tree. Now we can write f as a composition of a product:

f = g(h * j)

where j is a primitive function of rank R and g and h are some other functions. The derivatives of f are:

f' = (h * j)' * g'(h * j)

f'' = (h * j)'' * g'(h * j) + (h * j)' * g''(h * j)

f'' contains the same terms that define f' and is therefore also of rank R, so f' is not primitive. But there must be at least one such primitive function f', so the hypothesis of a largest rank is contradicted.

share|cite|improve this answer
I think my answer is still flawed because the argument doesn't actually require that R is largest. I'm going to leave it posted because I tried really hard to develop what I feel are the right concepts, maybe there will be a way to fix the problems. –  Dan Brumleve Aug 22 '10 at 9:56
Specifically the same argument applies to f(x) = erf(x) * x. f and f' have rank 1 but f'' has rank 0 and my argument says it can't, even without assuming 1 is the largest rank. –  Dan Brumleve Aug 22 '10 at 10:17
On a nicer note, Dan, your approach seems simpler. :) Maybe somebody who's actually seen the works of Appell and Kampé de Fériet and Meijer (and I should really stop name-dropping dead mathematicians at this point) might want to step in here and say something. All I know is that these guys have thought of these a long time ago. –  Guess who it is. Aug 22 '10 at 10:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.