TITLE: Explaining why we can't "find" an antiderivative of $f(t) = e^{t^2}$.
QUESTION [12 upvotes]: We can't find 
$$
\int e^{t^2} \; dt
$$
using basic tools from a calculus class. That is, we can't express an antiderivative of $f(t) = e^{t^2}$ using the basic operations. We can of course just define 
$$
F(t) = \int_{a}^t e^{s^2}\; ds.
$$
I am looking for a way to explain to a student why we can't express the antiderivative using basic operations (addition, subtraction, root, powers, etc.). In particular I am interested in getting at answering the "why we can't do that". I know that one could "just" prove it, but is there a simple argument or something that illustrates this?

REPLY [7 votes]: You can't prove it easily, but you can give some intuition :

derivative of $e^x$ is $e^x$
Trying to find $f$ such that $f'(x)=e^{x^2}$, we can guess that $f(x)=g(x)e^{h(x)}$
So $f'(x)=(g'(x)+g(x)h'(x))e^{h(x)}=e^{x^2}$
So we want to find a solution such that $h(x)=x^2$ and $g'(x)+2xg(x)=1$

Note the last point seems easier, but if they try, they will not find elementary function $g$ such that $g(x)=\frac{1-g'(x)}{2x}$. You can at least show this is not a polynomial function, nor a quotient of polynomials function, nor a usual trigonometric function.
But let $g(x)=\sum a_ix^i$, then $a_1=1$ and $(i+1).a_{i+1}+2a_{i-1}=0$, so you can at least give some expression to $g$ by solving $a_i$ (you can suppose that $a_0=0$).