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.

This concept isn't really unique to $\int e^{-x^2}dx$.

If your "domain of knowledge" is just the rational numbers, and you were asked "what number, when squared, gives you, 2?", and after not knowing the answer you were told it was $\sqrt{2}$, the number that when squared, yields $2$, that explanation would seem circular.

$\operatorname{erf}$ is just some function. The fact that it can't be expressed in terms of "simpler" operations isn't much stranger than the fact that the function $x \mapsto \sqrt{x}$ can't be expressed in terms of "simpler" operations.

• The difference is that I know for $\sqrt{2}$ that it is an irrational number, almost equal to $1.414$ – Phaptitude Oct 13 '13 at 17:15
• You also know that $\operatorname{erf}(2)$ is an irrational number, almost equal to $0.9953$. – Alec Oct 13 '13 at 17:50
• There's really nothing that makes $\log$ a "more correct" example of this than $x\mapsto\sqrt{x}$ as far as I know, other than the fact that logarithms are more "advanced" than squareroots, and that the $\log$ example also involves an integral. In terms of "why can't I write down an expression for this function?", $x\mapsto\sqrt{x}$, $\log$, and $\operatorname{erf}$ are really all fundamentally the same. – Alec Oct 13 '13 at 17:59

The integral you want doesn't have a nice antiderivative in terms of familiar functions. However, it's an important antiderivative, so mathematicians gave it a name.

As vadium123 said it is not possible to represent in familiar function. Here I try to answer your last question - What is $\text{erf}(x)$?
$\text{erf}(x)$ is the "error function" encountered in integrating the normal distribution (which is a normalized form of the Gaussian function).It can be represented in the form of infinite series as-
$$\text{erf}(x)=\frac{1}{\sqrt{\pi}}e^{-x^2}\frac{(2n)^{2n+1}}{(2n-1)!!}$$ It can also be defined in the term of Gamma function and incomplete gamma function as $$\text{erf}(x)=\frac{2(\Gamma(\frac{1}{2})-\Gamma(\frac{1}{2},x^2))}{\sqrt{\pi}}$$

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 $$\int_{-\infty}^\infty e^{-x^2} = \sqrt\pi$$

If you are familiar with double integration, another incredibly nice trick to derive the result Lucian mentions, i.e. that $\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}$, is the following:

Let $I = \int_{-\infty}^{\infty} e^{-x^2}dx$. Then one can show that $I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)}dx dy$. Now use polar coordinates $r^2 = x^2 + y^2$ and $dx dy = r dr d\theta$ to get

$I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} r e^{-r^2}dr d\theta = \pi$, from which the result $I = \sqrt{\pi}$ follows immediately.