What is the antiderivative of $e^{-x^2}$ I was wondering what the antiderivative of $e^{-x^2}$ was, and when I wolfram alpha'd it I got $$\displaystyle \int e^{-x^2} \textrm{d}x = \dfrac{1}{2} \sqrt{\pi} \space  \text{erf} (x) + C$$

So, I of course didn't know what this $\text{erf}$ was and I looked it up on wikipedia, where it was defined as: 

$$ \text{erf}(x) = \dfrac{2}{\sqrt{\pi}} \displaystyle \int_0^x e^{-t^2} \textrm{d}t $$ 

To my mathematically illiterate mind, this is a bit too circular to understand. Why can't we express $\int e^{-x^2} \textrm{d}x$ as a 'normal function'? Also, what is the use of the error function?
When $f$ is a continuous function on the interval $[a,b]$, we can find a function $F$ defined on $[a,b]$ such that $F'(x)=f(x)$ for all $x\in[a,b]$. This is the “fundamental theorem of calculus”; just consider
$$
F(x)=\int_{a}^{x} f(t)\,dt
$$
There are other functions with the same property, precisely those of the form $F(x)+c$ where $c$ is a completely arbitrary constant.

Sometimes this function can be expressed with the so-called “elementary functions”, that is, polynomials, rational functions, exponential, logarithm, trigonometric functions and any algebraic combination thereof. Some (actually many) functions do not admit an antiderivative expressible in this form; it's the case of $e^{-x^2}$ and it can be proved, although not easily.

Think of a simpler example: if all we have available as “elementary functions” are polynomials or, more generally, rational functions, the function $1/x$ wouldn't admit an “elementary antiderivative”, but it still would have one:
$$
\int_{1}^{x}\frac{1}{t}\,dt
$$
Since this is a “new” function, we give it a name, precisely “$\log$” and we have extended the tool set. The same happens with “$\operatorname{erf}$”, which has many uses in probability theory and statistics, being related to normal distributions.
Generally speaking , no function of the form $f_n(x) = e^{\pm\ x^n}$ possesses an anti-derivative whose expression can be parsed in terms of combinations of ‘normal’ functions, unless the value of $n$ is either $0$ or $1$. However, that does not stop us from deriving the following beautiful identity: $$n! = \mathcal{G}\left(\frac1n\right) \qquad,\qquad \mathcal{G}(n) = \int_0^\infty e^{-x^n} dx$$ where $n! = 1\cdot2\cdot3\cdot\ldots\cdot n$ , and the above formula can be used to extend the definition domain of the factorial function to other arguments than the natural numbers $\mathbb{N}$ , which yields the famous result $$\int_{-\infty}^\infty e^{-x^2} = \sqrt\pi$$