Given a set $S$, which is a subset of the complex numbers, we can form the smallest field which contains $S$, which we will denote by $S_F$ by taking the intersection of all complex fields which contain $S$. We can, also given a field $F$ obtain its algebraic closure $\bar{F}$.  Given a set $S$ of complex numbers, we'll write $A(S)$ to be the algebraic closure of $S_F$.  Study of the algebraic numbers $\bar{Q}$ is very old. But many constants we care about are not in $\bar{Q}$.

This motivates the definition of periods which seeks to include a lot of constants which are transcendental but in some sense not too complicated compared to the algebraic numbers. A real number $p$ is a period if there is some $n$ such that there is a polynomial with rational coefficients $P(x_1, x_2 \cdots x_n)$, and a rational functional with rational coefficients $Q(x_1, x_2 \cdots x_n)$ such that $$p=\int_{P(x_1,x_2 \cdots x_n) \geq 0} Q(x_1,x_2 \cdots x_n) dx_1 dx_2\cdots dx_n.$$ We then say a complex number is a period if it has real and imaginary parts which are both periods.

All algebraic numbers are periods but so are some other things. For example, $\pi$ is transcendental but it is a period because $\pi = \int_0^1 \frac{4}{x^2+1} dx $. Here $n=1$, $x_1=x$, $P(x)=x(1-x)$, and $Q(x)=\frac{4}{x^2+1}$. 

One open conjecture is that Euler's constant $\gamma$ is NOT a period. (This is obviously a very difficult problem since we can't even right now show that $\gamma$ is irrational.) 

However, periods are in some sense a very limited set. For example, they conjecturally don't include $\frac{1}{\pi}$ or even $e$. (One suspects that resolving both of these will occur well before $\gamma$ is handled.) So if the periods are supposed to be a set of numbers which are in some sense handling most natural constants we care about, they aren't doing necessarily a great job.

One of the more obvious things to allow $P$ and $Q$ to be algebraic functions rather than just a polynomial and a rational function. However, this gives the same set of periods. 

Given $F$ a subfield of the complex numbers, we'll write $P_F$ as the same set of periods but where we allow coefficients instead of being in $\mathbb{Q}$ but in $F$. Here $P_{\mathbb{Q}}$ is just our usual set of periods. One obvious thing to do is to extend $P_{\mathbb{Q}}$ to a field and then take its algebraic closure. that is to look at $A(P_{\mathbb{Q}})$. But note that this is a field where we can do the same thing to again, use it to generate periods, and so on.  

Making this more precise, we'll set $T_0={\mathbb{Q}}$, and for $n>0$, set $T_n= A(P_{S_{n-1}})$. Now, we get a hierarchy $T_0, T_1, T_2 \cdots $.

Note that if $T_n =T_{n+1}$ for some $n$, then that will continue for all larger $n$. Note also that each $T_i$ is countable, and thus the union of the $T_i$ is countable, so most complex numbers will still not appear in this hierarchy.

**Question:** Is it true that for any $n$, $T_n$ is strictly contained in $T_{n+1}$? 

The obvious place to look for more on this is Kontsevich and Zagier's [survey on periods][1] but while they suggest an extension of periods to what they call ``exponential periods'' they do not have anything similar to the T hierarchy here.

My guess is that this question is much too difficult to be currently approached. But I'd be intrigued to see an example even if we can show that there is some $\alpha$ not in $T_1$ but contained in some $T_n$ for some $n>1$. [Yoshinaga constructed an explicit example of a computable number which is not a period][2] but that does not seem to be immediately helpful here.


  [1]: https://www.ihes.fr/~maxim/TEXTS/Periods.pdf
  [2]: https://arxiv.org/abs/0805.0349