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Many elementary integrals are well known. For example $\int \sin(x) = \cos(x) + C$.

Then there is $\int e^{-x^2}$ that cannot be expressed in terms of elementary functions. Therefore the set of elementary functions is not closed for the operator $\int$.

Now, let's say I add a new function $\operatorname{integ}(x) = \int_{0}^{x} e^{-t^2} dt$ and define that to be a new elementary function like $\sin(x)$ and I keep doing this. Will I ever get to a complete closed set of elementary functions such that their integral is another elementary function in the new set I defined?

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  • $\begingroup$ In a finite number of steps ? $\endgroup$ Dec 24, 2017 at 19:50
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    $\begingroup$ To within a constant multiple, your integ function already has a name: integ($x$) = $(\surd\pi/2)\,\mathrm{erf}\, x$. $\endgroup$ Dec 24, 2017 at 20:09
  • $\begingroup$ "And I keep doing this" implies either a finite number of steps or a countable number of steps. Is that what you meant? $\endgroup$
    – Jbag1212
    Dec 25, 2017 at 7:54
  • $\begingroup$ That is part of the question. Is it possible? And if so, can it be done in a finite number of steps? Or at least countable? $\endgroup$
    – user
    Oct 24, 2020 at 17:56

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Sure, but it's not going to be a very simple class of functions. For the purposes of this answer, let $E$ be the set of elementary functions according to the usual definition. You could define $E$ formally by starting with a class of functions, say $F_0 = \{\sin(x), \cos(x), \exp(x), \log(x), x, c\}$, where $c$ represents all constant functions, and then taking the closure of $F_0$ with respect to addition, multiplication, and composition. Rigorously, you would define $F_{n+1} = \{f + g, f \circ g, f\cdot g \text{ | } f,g \in F_{n}\}$, then $E = \bigcup_{n\ge0} F_n$ is the set of all functions that can be built in finitely many steps. If you want to have a set $E'$ which is also closed under integration, you would just define $F_{n+1}$ to also include integrals of functions in $F_n$, and with this modified definition, the union of the $F_n$'s would also be closed under integration.

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  • $\begingroup$ How do you prove that this process does not continue indefinitely? $\endgroup$
    – user
    May 20, 2018 at 20:10
  • $\begingroup$ Let $f \in E'$. By construction $f$ is in $F_n$ for some $n$, which means $f$ is constructed from at most $n$ steps of composition, multiplication, addition, or integration of the starting functions in $F_0$ $\endgroup$ May 20, 2018 at 20:39
  • $\begingroup$ If you include the integrals of those functions, they may not have their integral in the set, therefore it may be an infinite process because you may have to add them and then their integrals again and so on $\endgroup$
    – user
    Dec 1, 2020 at 12:21
  • $\begingroup$ Right, that's how infinite unions work. So if we define $F_{n+1} = \{f+g,f\circ g, f\cdot g, \int f dx \mid f,g\in F_n\}$, then $\bigcup_{n\ge 0} F_n$ is closed under integration. $\endgroup$ Dec 1, 2020 at 18:14
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    $\begingroup$ I guess I don't have any proof that it's not, but I really doubt that it is possible in a finite number of steps. $\endgroup$ Dec 10, 2020 at 14:35
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Take whatever function $f(x)$ you want (regular, good, et cetera...) and differentiate it.

Then your $f'(x)$ will be a certain weird strange function with an elementary integral.

Example:

$$f'(x) = x^{\sin \left(\log \left(x^2-x\right)\right)} \left(\frac{\sin \left(\log \left(x^2-x\right)\right)}{x}+\frac{(2 x-1) \log (x) \cos \left(\log \left(x^2-x\right)\right)}{x^2-x}\right)$$

Which gives you an elementary function, when integrated.

$$\int f(x)\ dx = x^{\sin(\ln(x^2-x))} + C$$

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