What is the proof for $F(x)=\int_{a}^{x} e^{t^2}dt$ not being elementary? Simply as stated above: What is the proof, or how does one prove, $F(x)=\int_{a}^{b} e^{t^2}dt$ isn't elementary?
All I know is that it can be proven, but I couldn't find a proof for it.
 A: Let's use the result of Liouville https://math.stackexchange.com/a/163/442

Let $f, g$ be rational functions, $g$ not constant.  The indefinite integral
$$
\int f(x)e^{g(x)}\;dx
$$
is elementary if and only if there is a rational function $h$ so that
$f=h'+hg'\;$.

Take $f(x) = 1$, $g(x) = x^2$ there.  The theorem tells us that
the indefinite integral $\int e^{x^2}\,dx$ is elementary if and only
if there is a rational function $h(x)$ such that $f=h'+hg'\;$; that is:
$$
1 = h'(x)+2xh(x) .
\tag1$$
Now if ($1$) holds in any interval of the real line, then it
holds in all of $\mathbb C$.
In the linear differential equation with analytic coefficients ($1$), every point is a regular point.
So the solution has no poles.  The rational function $h$ is, in fact,
a polynomial.  Now let us study the behavior as $x \to \infty$.  I claim
$h(x)$ is bounded as $x \to \infty$.  Suppose not.
The Laurent series near $x=\infty$ is
$$
h(x) = c x^m+O(x^{m-1})\qquad\text{as }x\to\infty
$$
with $c\ne 0, m \ge 1$.  So
$$
h'(x) = cmx^{m-1}+O(x^{m-2})
$$
and
$$
h'(x)+2xh(x) = 2 c x^{m+1} +O(x^{m})
$$
But $m \ge 1$ and $2c \ne 0$, so this is not the constant $1$.
Summary: $h$ is a polynomial, $h(x)$ is bounded as $x\to\infty$.
Therefore $h$ is constant.  Which again contradicts ($1$).
A: More generally,

(A) Let $g$ be a polynomial. If $\int e^g$ is elementary, then $\deg{g}\le 1$.

This follows from a polynomial version of Liouville's theorem:

(B) Let $ f $ and $ g $ polynomials. If $ \int {f {e ^ g}} $ is elementary,
then $ \int {f {e ^ g}} = P {e ^ g} $, where $ P $ is a polynomial.

This in turn follows from a rational version of Liouville's theorem:

(C) Let $ f $ and $ g $ be rational functions, with $ g $ not constant. If $ \int f {e ^ g} $ is elementary, then $ \int {f {e ^ g}} = R {e ^ g} $, where $ R $ is a rational function.

Theorem B implies theorem A:
By theorem B, with $ f = 1 $, we have $ \int e ^ g = P {e ^ g} $, where $ P $ is a polynomial. Taking derivatives and canceling $e^g$ gives
$
1 = P '+ Pg'
$.
If $ \deg {g}> 1 $, then $ \deg {g'} \geq 1 $ and so $ 0 = \deg {1} = \deg (P' + Pg') = \deg {Pg'} $, since $ \deg {P}> \deg {P'} $.
Thus, $ Pg '$ is constant and so is $ g' $. But that implies $ \deg {g '} = 0 $ or $ g' = 0 $, a contradiction. So $ \deg {g} \le 1 $.
Theorem C implies theorem B:
By theorem C, we have $ \int {fe ^ g} = Re ^ g $, where $ R $ is a rational function.
Taking derivatives and canceling $e^g$ gives
$
f = R '+ Rg'
$.
Writing $ R = \frac {P} {Q} $ with $ P $ and $ Q $ relatively prime polynomials and $ Q $ monic, we have:
$$ f = \frac {P'Q-PQ '} {Q ^ 2} + \frac {P} {Q} g' $$
and therefore
$$ Q ^ 2f = P'Q-PQ '+ PQg' $$
and
$$ Q (Qf-P'-Pg ') = -PQ' $$
Since $ P $ and $ Q $ are relatively prime, $ Q $ divides $ Q '$ and therefore $ Q = 1 $ (since $ Q $ is monic).
So $ R = P $ and $ R $ is a polynomial.
