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There are many elementary functions such as $e^{-x^2}$ which don't have an elementary antiderivative, but a definite integral of the same integrand has a closed-form value, e.g. $$\int_{-\infty}^\infty e^{-x^2} \mathrm{d}x = \sqrt{\pi}$$ I'm curious whether it is decidable whether or not such a definite integral has a closed-form value. In other words:

Does there exist an algorithm that will determine in finitely many steps whether or not the value of a given definite integral (with an elementary integrand) can be expressed in a closed-form expression?

a function of a single variable that is defined as taking sums, products, and compositions of finitely many polynomial, rational, trigonometric, hyperbolic, and exponential functions, [including their inverse functions.]

a mathematical expression expressed using a finite number of standard operations

where those "standard operations" are elementary functions, and the only allowed inputs are integers. (Note that rational numbers can be made using the function $f(x)=x^{-1}$ and certain well-known constants can be made such as $\pi = \arccos(-1)$ and $e = \exp(1)$)

I know that for indefinite integrals, the Risch algorithm does this, but I couldn't find anything about definite integrals.

Now for my follow-up question, define $S$ to be an arbitrary set of integrable functions and $R$ to be an arbitrary set of constants.

In general, is it decidable whether or not the value of a definite integral of a given function (not necessarily elementary) is expressible as a closed-form expression using the functions in $S$ and the constants in $R$ as inputs?

I would guess that the answer is no, but I couldn't find anything about it online.

Mathyland
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    $\gamma=-\int_0^{\infty}e^{-x}\log x,dx$ is the Euler gamma constant, and no one knows whether it is elementary or not. It hasn't even been proved that it's irrational. Similarly for $\zeta(3)={1\over2}\int_0^{\infty}{x^2\over e^x-1},dx$ (although that one is known to be irrational, it's not known whether there's a closed form). – Gerry Myerson Jun 07 '21 at 05:23
  • Any thoughts about my comment? – Gerry Myerson Jun 08 '21 at 12:24
  • @GerryMyerson $\zeta(3)$ is already a closed form, and any constant, incuding $\gamma$ is an elementary function. – Anixx Jun 23 '21 at 08:43
  • As to the question, no, it is not decidable. Particularly, it is not decidable whether a given integral is equal to zero. – Anixx Jun 23 '21 at 08:44

1 Answers1

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I wanted to correct your understanding of the Risch algorithm, but this is too long for a comment. We will say that a function is Richardson if it can be written by composing:

  • rational constants, $\pi$
  • $x$
  • $+, -, \cdot$
  • $\sin(x)$
  • $|x|$
  • Some function without a Richardson antiderivative, e.g. $e^{-x^2}$

Then it is undecidable if a Richardson function has a Richardson (or elementary) antiderivative. You can read more by looking up Richardson's Theorem: https://en.wikipedia.org/wiki/Richardson%27s_theorem

Where then, does the Risch algorithm go wrong? It does correctly and computably produce a list of anti-derivative candidates, such that the anti-derivative is on the list iff there is an elementary anti-derivative, but it turns out that checking equivalence of elementary formulas is undecidable (the main result of Richardson's Theorem), and thus checking if any of the candidates' derivatives are equivalent to the original function is non-algorithmic.

TomKern
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    Huh that's really interesting. I would not have guessed that checking equivalence of elementary formulas is undecidable. Though it looks like if we don't allow absolute values, it's unknown whether or not checking equivalence is undecidable (link). – Mathyland Jun 07 '21 at 06:25