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The motivation of this question is finding an axiomatization of Euclidean geometry. I consider Tarski's axioms a satisfactory axiomatization of those parts of Euclidean geometry that do not include arc length of curves, (surface) areas and volumes, so now I'm looking for axiomatizations that include curved lengths, areas and volumes. In the hopes of making this somewhat vague question more precise this is a question which I think should be true if this axiomatization is at all possible:

Take the real numbers $\mathbb{R}$. Take a pair of families of bounded sets $(S_a)_a$ and $(T_b)_b$, definable in the theory of real closed fields in $\mathbb{R}$ without parameters. Consider the relation $R_n (a,b)$ which holds if $S_a$ and $T_b$ have the same $n$-dimensional content. (I believe there are several different ways of defining $n$-dimensional content in geometric measure theory, which do not agree for wild sets. I'm assuming they all agree in this context.) Is the first order theory of $\mathbb{R}$ in the language of fields expanded with the relations $R_n$ decidable? (A no answer to this would be evidence against the existence such an axiomatization.) Is it o-minimal? (A yes answer would be evidence for the existence of what I am looking for, but it is not necessary.)

And less precisely, is there some axiomatization of this theory or some expansion, by axioms that are simple and intuitively true geometrically?

Edit: I see that there's a partial answer already, which is very nice. Unfortunately I did some abuse of notation common in model theory which might have caused confusion, I apologize. The relations $R_n$ are really of the form $R_{n,\phi,\psi}$ where $\phi(x,y)$ and $\psi(x',z)$ are formulas in the theory of real closed fields without parameters. The family $(S_a)_a$ is defined as $S_a$ being the set of $x$ such that $\phi(x,a)$ holds and $T_b$ is the set of $x'$ such that $\psi(x',b)$ holds. $S_a$ and $T_b$ are assumed bounded (with some bound that may depend on $a, b$, (this is just a guess on my part on what the right statement should be).) This does imply Tarski's question of the decidability of $\mathbb{R}_{exp}$ which I did not know was unconditionally open, although I did know it was o-minimal by Wilkie's theorem.

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    $\begingroup$ I believe that decidability of this theory would imply decidability of equality of algebraic periods, which as far as I know is an open problem. $\endgroup$ Commented Mar 15 at 5:10
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    $\begingroup$ @JamesEHanson it also solves Tarski's exponential problem. The natural log can be defined as the area under a hyperbola. $\endgroup$ Commented Mar 17 at 19:32
  • $\begingroup$ @MattF. Hmm the question mentioned "families" of sets. What are the families indexed by? $\endgroup$ Commented Mar 20 at 22:47
  • $\begingroup$ If we could quantify over algebraic functions on $[-1,1]$, we could use Legendre’s equation to define the integers: $$n\in\mathbb{N} \equiv \exists f\ (1-x^2)f’’-2xf’+n(n+1)f=0 \land f(1)=1$$ So an answer allowing quantification over bounded semialgebraic sets (which would otherwise be appealing) would not be decidable or o-minimal. $\endgroup$ Commented Mar 24 at 4:34
  • $\begingroup$ Worth pointing out that decidability of $\mathbb{R}_\mathrm{exp}$ is not just open but considered to be basically impossible, we don't know nearly enough about transcendental number theory. In fact, Lou conjectured that $\mathbb{R}_\mathrm{exp}$ was o-minimal as a sort of tractable replacement for Tarski's question, geometric tameness of definable sets as a replacement for decidability, see "Remarks on Tarski's problem concerning $(\mathbb{R},+,\cdot,\exp)$". $\endgroup$ Commented Aug 16 at 20:16

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${\bf Edit:}$ The answer below is unnecessary as the post only discusses definable families of bounded sets, in which case the parameterized integrals are in $\mathbb{R}_\mathrm{an}$, as @MattF points out. However, I think that it's actually more natural to allow unbounded sets as the motivation is to find a theory of volumes in Euclidean geometry, and Euclidean geometry certainly involves unbounded sets. My original answer is below.

If I understand correctly, o-minimality of your structure should follow from "Nature log-analytique du volume des sous-analytiques" by Comte, Lion, and Rolin. In fact the structure should be a reduct of $\mathbb{R}_\mathrm{an, exp}$.

They show that if $\varphi(x,y)$ is a formula in $\mathbb{R}_\mathrm{an}$ (hence in particular in the real field) with $x,y$ tuples of variables then the volume of the set defined by $\varphi(a,y)$ is an $\mathbb{R}_\mathrm{an, exp}$-definable function of $b$. In fact, they show that it has a special "log-analytic" form that is much simpler then a general $\mathbb{R}_\mathrm{an, exp}$-definable function. Note that you need the logarithm to handle the formula $\varphi(t, x, y)$ given by $(1 < x < t) \land (0 < y < 1/t)$.

The CLR result has been generalized by Cluckers and Miller who show that the class of "log-analytic" functions is also closed under parametrized integrals in "Stability under integration of sums of products of real globally subanalytic functions and their logarithms". This latter paper is a better reference for those of us who do not read French.

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    $\begingroup$ How does this use the paper by Comte, Lion, and Rolin [2000], which does not seem to mention o-minimality? Could we say instead “All these integrals with parameters are restricted analytic functions, and therefore by Van den Dries, Macintyte and Marker [1994], their theory is o-minimal”? $\endgroup$ Commented Aug 17 at 4:49
  • $\begingroup$ Ah, I missed "boundedness". $\endgroup$ Commented Aug 17 at 16:50
  • $\begingroup$ The bounds depends on the parameter, so you do need exp. Unbounded volumes are legitimate geometric objects but (in my opinion) are better seen as ind-definable sets instead of definable sets. I believe this gives an answer to my question, I will look at these articles later. $\endgroup$ Commented Aug 17 at 21:16
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The question asks for a first-order theory to decide statements like $$\int_0^14x(x^2+1)^3 dx= \int_1^4 x\, dx.$$ The first-order theory $T$ described below will prove an equivalent of this claim.

$\DeclareMathOperator\deriv{deriv}$Update: For the parameterized version of the question, the answer below should be parameterized: instead of adding constants $c_{d,r}$ for all the periods, we should add functions $f_{d,r}(a)$; similarly $\deriv$ and $\phi$ and $\psi$ should all be allowed to contain parameters $a$; and the axioms should be universally quantified over $a$. I will leave the rest of this answer in an unparameterized form for simplicity, with new comments at the end in italics.

The theory $T$ can compute all algebraic integrals to any desired accuracy. Furthermore, if Kontsevich and Zagier's Conjecture 1 holds on decidability of equality for periods, then $T$ will decide all equalities among algebraic integrals.

So let $I_{d,r}$ be the $r^\text{th}$ bounded conjunction of polynomial inequalities with integer coefficients in $d$ variables. By results of Viu-Sos, all periods are the volumes of regions determined by such inequalities $I_{d,r}$.

Now let the language of $T$ be the language of real-closed fields augmented by infinitely many symbols $c_{d,r}$, intended to represent the volumes satisfying those inequalities.

Define $\deriv(\phi,\psi)$ to mean \begin{align} \forall x\,\forall\epsilon>0\,\exists \delta>0\, \forall h[&\phi(x,u) \land \phi(x+h,v) \land \psi(x,z) \\ &\implies h^2>\delta \vee (v - u - hz)^2 < \epsilon h^2] \\ \land\,\forall xyzw(\phi(x,y)\land\phi&(z,w)\land x<z\implies y<w). \end{align} When $\phi(x,y)$ defines an increasing algebraic function, $\psi(x,z)$ will define its non-negative derivative.

Now the axioms of $T$ will be the axioms of the theory of real-closed fields plus the following axioms for all positive integers $d$, $e$, $r$, $s$, $t$ and all bounded conjunctions $\phi(x,y)$ and $\psi(x,y,z)$ of polynomial inequalities. The added axioms mostly translate Kontsevich and Zagier's rules for manipulating integrals:

  • [addition of regions] \begin{align} \forall x[(I_{d,r}&(x) \vee I_{d,s}(x) \equiv I_{d,t}(x)) \\ \land \ (I_{d,r}&(x) \wedge I_{d,s}(x) \equiv \bot)] \\ \implies c_{d,r} &+ c_{d,s} = c_{d,t} \end{align}
  • [addition of functions] \begin{align} &\forall xyz[ \phi(x,y) \implies I_{d,r}(x,z) \equiv I_{d,s}(x,y+z))] \\ &\implies c_{d,r} = c_{d,s} \end{align}
  • [change of variables] \begin{align} &\deriv(\phi,\psi)\\ &\land \forall uvx[I_{d+1,s}(u,v,x)\\ &\phantom{\land \forall uvx[\ }\equiv\exists yz\,(\phi(x,y) \land \psi(x,z) \land I_{d,r}(u,y) \land 0 < v < z)] \\ &\implies c_{d,r} = c_{d+1,s} \end{align}
  • [Newton-Leibniz] \begin{align} &\deriv(\phi,\psi) \\ &\land \forall xyz[ \psi(x,z) \\ &\phantom{\land \forall vxy[}\implies I_{d,r}(x,y) \equiv (0<x<1 \land 0<y<z)] \\ &\implies \forall uv[\phi(0,u) \land \phi(1,v) \implies c_{d,r} = v-u] \end{align}
  • [multiplication of regions] \begin{align} & \forall xy[I_{d,r}(x) \land I_{e,s}(y) \equiv I_{d+e,t}(x,y)]\\ & \implies c_{d,r}\,c_{e,s}=c_{d+e,t} \end{align}
  • [positivity]$$c_{d,r} \ge 0.$$

Of these, Konstevich and Zagier did not mention and may not have needed the last two for their discussions of single-variable integral equalities.

As an example, the relevant change of variables for the left-hand side of the original equality is: \begin{align} \phi(x,y) &: y = x^2\\ \psi(x,z) &: z = 2x\\ I_2(u,y) &: 1 < u < 2(y+1)^3 < 16\\ I_3(u,v,x) &: 1 < u < 2(x^2+1)^3 < 16 \land 0 < v < 2x. \end{align} The conclusion of that axiom is $$\int_{x=0}^1\int_{v=0}^{2x}2(x^2+1)^3dx\, dv =\int_{x=0}^14x(x^2+1)^3dx =\int_{y=0}^12(y+1)^3dy.$$ This can further be simplified using Newton–Leibniz with \begin{align} \phi(x,y) &: 2y=(x+1)^4 \\ \psi(x,z) &: z=2(x+1)^3. \end{align} The conclusion of that axiom is $$\int_0^1 2(x+1)^3dx = \frac12(1+1)^4-\frac12(0+1)^4=\frac{15}2.$$ The right hand side of the original example is similar. So if $r$ and $s$ are the indices for $0<y<4x(x^2+1)^3<32$ and $1<y<x<4$, then $T$ gets the original equality in the form $c_{2,r}=c_{2,s}$.

Even with more complicated algebraic regions, for which we may not know algebraic integrals, we can still find piecewise-linear bounds. Then the positivity in $T$ allows it to establish the piecewise-linear volumes as bounds on the algebraic volumes, and in that way compute any volume to any desired accuracy.

On the questions at the end (for $T$ without parameters):

  • The Konstevich-Zagier conjecture implies that the quantifier-free theory of $\mathbb{R}$ with these constants is decidable, and if so the the rest of $T$ also seems likely to be decidable.
  • The theory $T$ is o-minimal, since any definable set of reals in its language is already definable with parameters in the language of real-closed fields, whose o-minimality establishes that the set is a finite union of points and intervals.
  • The theory $T$ has axioms that I find intuitive enough, but I wouldn't call them "simple and intuitively true geometrically". After all, the axioms include a version of the fundamental theorem of calculus, credited to Newton and Leibniz for their analytic version of that theorem. Barrow's earlier geometric version (from his Lectiones, Lecture X, Proposition 11, pp. 116-117) just makes me think the more intuitive approach is less geometric.

Update on the questions at the end, for the theory $T(a)$ with parameters:

  • $T(a)$ allows defining $\exp$ as the inverse of $\log$, so the decidability of $T(a)$ would settle Tarksi's exponential function problem, and I would not say that it seems likely.
  • $T(a)$ may well be o-minimal, but that's not a trivial claim. E.g. $T(a)$ defines not just $\log$ and its inverse $\exp$, but also $\arccos$ and its inverse $\cos\!|_{[0,\pi]}$, where the definition of $\cos\!|_\mathbb{R}$ would destroy o-minimality.
  • Even if its axioms are not geometrically intuitive, $T(a)$ has the advantage of making many geometric statements, like Euclid's claims that the areas of circles are as the squares on their diameters, or that two pyramids on the same base with the same height have equal volumes, with intuitive formalizations.
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    $\begingroup$ So we can think of the Kontsevich-Zagier conjecture as something like Hilbert's third problem, saying that if two elements of some natural geometric class of objects have the same volume, then there is some "geometrically reasonable" way to see this. $\endgroup$ Commented Aug 17 at 22:29

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