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I've been trying to condense the sequence of steps for arguing that there is no elementary form for the integral as follows:

Consider DE's over $\mathbb{C}(x)$ with the usual derivation.

Let the solutions to $L=y''+x^ny'=0 $ be $1$ and $z=\int{e^{-\int{x^n}}}$ (i.e. $z'=e^{-\int{x^n}}$ and $z''=-x^n\,e^{-\int{x^n}}$, satisfying the DE).

We aim to show this only has an elementary solution when $n=-1$ (i.e. that $G(\mathbb{C}(x)\langle1,z\rangle)/\mathbb{C}(x))$ is only soluble in this case).

A basis for the permutations of solutions could consist of mapping $1$ to $1$ along with $z$ to $c+dz$.

If $\sigma$ is a differential automorphism, $\sigma(z')=\sigma(z)'=(c+dz)'=dz'$.

The automorphisms can be represented by the set of matrices $\begin{pmatrix}1 & c \\ 0 & d\end{pmatrix}$ in $GL(2,C)$.

Two such matrices only commute in general if $d=1$ (that they should commute follows from the connected component of the identity being abelian if the Galois group is soluble).

So we have that $\sigma(z')=z'$ with $z'=e^{-\int{x^n}}$ as at the start.

This means that $z'=e^{-\int{x^n}}$ must be in $\mathbb{C}(x)$ as it is mapped to itself. [My main question - is this a valid inference?].

Hence $e^{-\int{x^n}} = c$ for some polynomial in $\mathbb{C}(x)$, i.e. $-\int{x^n}=ln(c)$ which only happens when $n=-1$ and $c$ only has an $x$ term.

If this is all valid then there is no elementary solution to $y''+xy'=0$ as $n=1$ in this equation, and hence there is no elementary form for $\int e^{\frac{x^2}{2}}\,dx$, i.e. none for $\int e^{x^2}\,dx$.

John1970
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    See How to determine with certainty that a function has no elementary antiderivative – Dave L. Renfro Aug 24 '19 at 12:30
  • Thank you - that is a very interesting article. It takes the approach of Liouville's Theorem rather than differential Galois theory however - while related I don't think they're quite the same (e.g. differential Galois theory seems to look at permutations of the solutions). I'd be very interested if anyone can confirm whether I'm on roughly the right lines with my summary of what I think the argument is using differential GT. – John1970 Aug 24 '19 at 16:05
  • You should make everything clear : what differential field you are looking at, how you define and find its differential galois group, if it is solvable, what does it imply about the solution from elementary operations.. – reuns Aug 24 '19 at 16:14
  • Thank you for the reply - it is really appreciated. I've edited the question to try and make it clearer, namely that the differential field is $\mathbb{C}(x)$ with the usual derivative and the argument looks to show that $G(\mathbb{C}(x)\langle1,z\rangle)/\mathbb{C}(x))$ is soluble only when $n=-1$. Thanks again - any further suggestions you have will be very welcome. – John1970 Aug 24 '19 at 16:57
  • Just to confirm that I'd still appreciate anyone's comments on whether this is a good summary - in particular whether the third to last line is valid (i.e. that starts 'This means that...' – John1970 Sep 06 '19 at 14:35

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It turns out that the above is only a good summary of why the anti-derivative of $e^{x^2}$ is not elementary if we are using A. Magid's definition of elementary in his 'Lectures on Differential Galois Theory'. As mentioned here: Elementary function defined by A Magid Magid does not regard this definition as capturing the usual, intuitive definition of elementary. So, while perhaps of interest, the derivation I originally posted here is not really the best explanation of why the anti-derivative of $e^{x^2}$ is not elementary, the better explanation being Liouville's original work.

John1970
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