Mathematicians have come up with many ways of classifying how "exotic" some numbers are. For example, the most ordinary numbers are the natural "counting" number, and the next most exotic numbers are zero and the negative integers. Next are the rational non-integers, which are ratios of integers. Then come the algebraic numbers, which can be expressed as roots of polynomials with rational coefficients. But there are still transcendental numbers, like $\pi$ and $e$, which cannot be expressed in any of these ways. However, $\pi$ and $e$ can both be expressed as integrals of elementary functions.

My question is this: Do there exist any numbers (transcendental, of course) that cannot be expressed as the definite integral of an elementary function? Of course, I don't expect anyone to prove something like this (perhaps disprove it?), but I would like to know whether something like this is likely. Clearly, all real numbers can be expressed as an infinite series (just by using egyptian fractions and the greedy algorithm), but whether all real numbers can be expressed as an integral is less clear.

My thoughts: I really have no idea, but if such a number did exist, I would expect it to be something with infinite irrationality measure (like Liouville's constant, or any other Liouville number).

EDIT: To clarify, my question isn't really about "what other types of pathological numbers are there". I really want to know if anyone can provide any computable examples of numbers that aren't exponential periods.

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    $x=\int_{0}^{x}1\,dx$ makes the question a little non-sensical. – Jack D'Aurizio Dec 20 '17 at 15:39
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    The algebraic numbers are countable. The numbers which are definite integrals of algebraic functions with algebraic bounds are countable (look up ring of periods). The real numbers are uncountable, so most of them cannot be so written. – ziggurism Dec 20 '17 at 15:41
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    What about Chaitin's constant? It is a number that can't be calculated! – rafa11111 Dec 20 '17 at 15:43
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    I think the worst case are 'undefinable real numbers' en.wikipedia.org/wiki/Definable_real_number but exact definition may depend on context. – Nik Pronko Dec 20 '17 at 15:44

Short answer: yes, there are numbers worse than $e$ and $\pi$. Almost all real numbers are worse. There are uncountably many reals, whereas any class of "nice transcendental numbers" you can think of will be only countable (and I propose exponential periods as a good class of numbers of the kind you describe).

The class of numbers which are roots of polynomials with rational coefficients are the algebraic numbers. The numbers which are definite integrals of algebraic functions with algebraic bounds are called the ring of periods. It is a larger class of numbers, including many familiar (suspected) transcendental numbers such as $\pi$, $\log 2$, $\zeta(3)$, and $\Gamma(p/q)^q$. See this nice primer by Kontsevich and Zagier.

Actually some common numbers like Euler's number $e$, the Euler-Mascheroni constant $\gamma$, and $1/\pi$ are still (suspected to be) missing in the ring of periods. So Kontsevich and Zagier go further and extend the ring to exponential periods, defined as integrals of products of exponentials of algebraic functions with algebraic functions. This gives a class of numbers that include "all algebraic powers of $e$, values of $\Gamma$ at rational arguments, values of Bessel functions, etc."

I assume that the exponential periods still don't include $\gamma$, because finally they claim that if you extend the class further by adding it, then you have "all classical constants in an appropriate sense" (whatever that means).

In this primer on exponential motives by Fresán and Jossen, according to Belkane and Brosnan, $\gamma$ is an exponential period, as witnessed by the integrals $\gamma=-\int_0^\infty\int_0^1e^x\frac{x-1}{(x-1)y+1}\,dy\,dx$ or $\gamma=-\int_0^\infty\int_1^x\frac{1}{y}e^{-x}\,dy\,dx$.

Anyway, the rational numbers are countable. The algebraic numbers are countable. The ring of periods is countable. The exponential periods are countable. And including $\gamma$ certainly still leaves you with a countable class of periods.

But the real numbers are uncountable, so it remains the case that most real numbers are not periods, not exponential periods, and cannot be written in any of these forms.

So to answer the question, yes, real numbers do get more exotic than numbers in the ring of periods or exponential periods like $e$ or $\pi$.

My understanding of statements of the form "$\Gamma$ takes values in the exponential periods at rational values" is that we would expect $\Gamma$ to take values at irrational arguments that are not exponential periods (although in general I expect proofs of such claims to be hard to come by).

So I would expect numbers like $\Gamma(\sqrt{2})$, $e^\pi$, $\zeta(\log 2)$ to be examples of computable numbers which are not exponential periods. But these would presumably not be considered “classical constants”.

Going further, even the computable numbers are countable, so most real numbers are not even computable, let alone in the ring of periods.

  • Any chance that you know of any computable numbers satisfying this? – Frpzzd Dec 20 '17 at 18:36
  • @Nilknarf satisfying what? – ziggurism Dec 20 '17 at 18:36
  • A computable number that cannot be expressed as a definite integral of an elementary function. – Frpzzd Dec 20 '17 at 18:38
  • @Nilknarf As discussed in my post, $e$ is thought not to be in the ring of periods (the definite integral of an algebraic function over algebraic bounds). Another one is mentioned in the article, the Euler-Mascheroni constant. Both numbers are obviously computable. If you are looking for a computable number that is not an exponential period, well according to Kontsevich and Zagier, all classical constants are included. – ziggurism Dec 20 '17 at 18:40
  • @Nilknarf Kontsevich and Zagier's primer on periods says "all algebraic powers of $e$, rational values of the gamma function, etc" are exponential periods. I think we should infer that $e^\pi$ and $\Gamma(\pi)$ are not expected to be exponential periods for example (with the proviso that proofs of such claims are hard to come by). – ziggurism Dec 20 '17 at 18:52

Let us be more precise about definable numbers, to avoid common pitfalls.

Suppose we have chosen our favourite foundational system $S$, which is in modern mathematics ZFC. $S$ of course can be implemented by a computer program that will given any input theorem and purported proof will output "yes" if the purported proof is a valid one for the specified theorem, and will output "no" otherwise. Then we can say that every definable object (over $S$) is one for which there is some $1$-parameter sentence $P$ such that $S$ proves $\exists!x(P(x))$. In some sense $P$ is the definition of that object.

But to talk about definable reals, there is a problem. We cannot use the above definition, because we cannot ask whether there is a defining property for a particular real number, because there is no way to encode all real numbers into distinct finite strings, much less in a way that makes sense for us to talk about whether $S$ proves something about it. But here is one way to get around that issue. If $S$ (like ZFC) can talk about the collection of real numbers $R$, then we can define a definable real over $S$ to be a real $r$ such that there is some $1$-parameter sentence $P$ over $S$ such that every model $M$ of $S$ that has the same reals satisfies $P(r) \land \exists!x( x \in R \land P(x) )$, and we call this $P$ a defining property of $r$. Assuming that $S$ has a model, this definition implies that no two reals can satisfy the same defining property.

To emphasize again, this definition of "definable real" is not the same as "definable object that is a real" ('using' the earlier definition of "definable object"); the latter does not even make any syntactic sense, since a "definable object" is just a $1$-parameter sentence over $S$ and not an object. It is in fact impossible to talk (within $S$) about whether an object satisfies some $1$-parameter sentence over $S$ or not, for exactly the same reason as in Tarski's undefinability theorem.

Of course, for any reasonable $S$ (including ZFC), by Godel's incompleteness theorem we cannot prove that $S$ has a model (unless $S$ is inconsistent), but we can prove that if $S$ has a model with the same reals, then there are countably many definable reals. In such a case, under our definition of "definable real", uncountably many reals are undefinable. Even without restricting to integrals. Absolutely no way of defining reals will be able to capture all of them!

To preempt the objection that it is not really satisfying that we cannot say anything about definable reals without assuming a model of $S$ with the same reals, consider that we ought to accept $S$ as foundational only if we believe it is meaningful. And it would be meaningless if there is no model of $S$ with the same reals, because then by the semantic-completeness theorem $S$ would prove that fact, contrary to our belief that $S$ is meaningful! Hence we ought to believe (despite not being able to prove) that "definable real" is not a vacuous notion.


Furthermore, we can classify the definable numbers more finely. First we have computable reals, for which there is a program that on input $n$ will output a rational that is within $2^{-n}$ of the actual value. But the unsolvability of the halting problem can be used to show that there is no program that can given two computable reals (as program code) will output "yes" if they are equal and "no" if they are not.

So we can consider the next level in the hierarchy, which are programs that are allowed to get instant answers from the halting oracle (which is literally an oracle that will correctly answer whether any given program halts on the given input or not). Let us call these programs $1$-jump programs. Of course, there will be more reals that can be computed by $1$-jump programs. It turns out that the halting problem for $1$-jump programs cannot be solved by a $1$-jump program either. But again we can go to the next level, which is $2$-jump programs that are allowed to query an oracle for halting of $1$-jump programs. And so on.

In this manner we get a whole lot of new real numbers, and still the halting problem for all finite-jump programs cannot be solved using any finite-jump program, so there is a next level beyond that, which we can call $ω$-jump programs. And so on and on and on.


[Update]

The question was subsequently edited to ask about computable numbers that cannot be expressed as definite integrals of elementary functions, where presumably the limits are also expressed using elementary functions. I am not sure what is the current state of the art, but given any algorithm to compute definite elementary integrals to arbitrary precision such that the time complexity is itself computable, we can easily diagonalize to produce a computable real that is not equal to any of the definite elementary integrals, because we can computably enumerate all elementary expressions. This paper seems to suggest that such an algorithm exists, though I have not looked at the details.

  • @user2357112: I have edited my answer, both to add in some additional information addressing the edited question, and to clarify based on your comment. Thanks! – user21820 Dec 22 '17 at 4:03
  • Comments removed. I submitted an edit to try to make the distinction between the two definitions of "definable" a bit more explicit; hopefully I didn't accidentally screw anything up. – user2357112 Dec 25 '17 at 17:53
  • @user2357112: Thank you for trying to make the distinction better. I've reworded it rigorously and added a new paragraph after that, to discuss the problem and why we should consider this definition of "definable real" meaningful. =) – user21820 Dec 26 '17 at 4:53

Expanding upon Nik Pronko's comment about definable reals, consider the set of finite sequences of characters over the Latin alphabet. We can count this set fairly easily: $a$, $b$, $c$, ..., $z$, $aa$, $ab$, $ac$...

The set of readable english documents is a subset of this set, so it must also be countable. The set of english documents that describe numbers is also a subset of that set, so the set of numbers that can be described in english is countable. The set of reals is uncountable, so there must be an uncountable infinity of numbers that cannot even be described in english.

For hopefully obvious reasons, I can't give you an example of any such number. The usefulness of this result is kind of debatable, but it does answer your question. For any given way of describing real numbers, there must always be an uncountable infinity of real numbers that cannot be described by that method.

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    "readable english documents" are not usually how numbers are described. Firstly they usually tend to use a lot of characters in other parts of Unicode, secondly "readable" as applied to maths papers is a very subjective criterion. (But of course, none of this invalidates your argument). – Martin Bonner Dec 21 '17 at 12:52
  • "The set of english documents that describe number is also [countable]" does not imply "the set of numbers that can be described in english is countable". – Zaaier Dec 21 '17 at 18:28
  • @Zaaier: Assuming axiom of choice, it does. – tomasz Dec 21 '17 at 22:26
  • What about an alphabet of uncountable size? Like, some ticks $\land$ with arbitrary angle. We could write down any real number, just couldn't read it back precisely enough... Think I'm gonna include that in a scify book about aliens when I write one. – Serge Seredenko Dec 22 '17 at 4:15
  • @SergeSeredenko: You first need to prove that there is an alphabet with uncountable size. Using physical notions does not work, because we do not even know that spacetime is continuous all the way down. – user21820 Dec 22 '17 at 5:34

Do there exist any numbers (transcendental, of course) that cannot be expressed as the definite integral of an elementary function?

The answer is simply no because, as Jack D'Aurizio pointed out in the comments, any real number $x$ can be written as $$ x = \int_0^x 1 \, dt. $$

Alternatively, we could write $$ x = \int_0^1 x \, dt. $$

  • Bah, this is a loophole. I meant a combination of elementary functions and rational numbers. – Frpzzd Dec 23 '17 at 18:17
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    @Nilknarf If this is a loophole, then the question needs to be stated more precisely. What do you mean by "a combination of elementary functions and rational numbers"? What is a "combination"? This is not a standard mathematical term, so it must be given a precise definition. Why would $\int_0^1 x \, dt$ not count as a "combination of elementary functions and rational numbers"? The function we are integrating is elementary (because it is a constant function), and the limits of integration are rational. – littleO Dec 23 '17 at 18:24
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    This is tautological. By the same logic 0*x + 0 = 0 implies that all numbers are algebraic. Or am I missing something here? – TLW Sep 9 at 3:03
  • only Jack D'Azurio is allowed to make this point – clathratus Oct 19 at 22:24

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