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For indefinite integrals whose solutions cannot express with elementary functions, special functions are often defined, such as those shown below.

$$ \mathrm{Si}(x) = \int_0^x\!\frac{\sin t}{t}\,\mathrm{d}t \qquad \mathrm{Li}(x) = \int_0^x\!\frac{1}{\ln t}\,\mathrm{d}t \qquad \mathrm{S}(x) = \int_0^x\!\sin(t^2)\,\mathrm{d}t $$

Through definining special functions as non-elementary integrals there exist cases where other functions also become integrable in terms of these newly defined special functions. An example is shown for the hyperbolic sine integral, whose solution is often given as a special function; however, this special function can be given in terms of the sine integral.

$$ \mathrm{Shi}(x) = \int_0^x\!\frac{\sinh t}{t}\,\mathrm{d}t = \frac{\mathrm{Si}(ix)}{i} $$

Let $E$ be the set of all possible functions which can be created through combinations of the elementary functions. Examples are shown below for members of the set $E$.

$$ \frac{\sin x}{e^x} \qquad e^{x^2}\ln x \qquad \cosh\left(\frac{x^2 + 1}{\log_{10}(x)}\right)$$

Does there exist a finite set $S$ of defined solutions to non-elementary integrals, such that any function in the set $E$ can have its indefinite integral expressed as a combination of the functions in $S$ and the elementary functions?

finlay morrison
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    Hi. I’ve edited your tags to replace mentions of “set theory” with “(differential) Galois theory”. I know absolutely nothing about differential Galois theory, other than that it’s probably exactly the kind of thing which can answer your question – FShrike Aug 23 '22 at 10:51
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    I see no reason to think there is such a finite set. Even the "ellpitic integral of the first kind" $$F(z,k) = \int_0^z\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}$$ is an infinite family of functions indexed by $k \in \mathbb C$. – GEdgar Aug 23 '22 at 11:57
  • Please see Risch algorithm. Also you may write $S(x)=\int_0^x \sin\left(t^2\right)dt$ etc – Тyma Gaidash Aug 23 '22 at 12:08

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