How to prove that for any real positive $a$ the integral $\int_0^a e^{-x^2}dx$ does not converge to a "familiar number"? I know that is possible to show that the integral
$$\int e^{-x^2} dx$$ 
has no closed formula, using Liouville theorem. But the function $f(a)$
$$ f(a) = \int_0^a e^{-x^2} dx $$
converges to $\sqrt{\pi}/2$ when $a$ tends towards infinity. But how does one show that for any real number the definite integral $f(a)$ doesn't converge to a "know" number, like $\pi, e, 2\sqrt2$ etc. $?$
Or does $f(a)$ converges to a familiar number when $a$ is real and positive?
Thanks.

This question arose when I was seeing some stuff about elliptic integrals and a I questioned if the arc length of $\sqrt{1-x^4}\:$ in the interval $(-1, t)$ given by
$$\int_{-1}^{t} \sqrt{ 1 + \frac{4x^6}{1-x^4}} \:\: dx $$
converged to some know number when $t \in (-1,1)$. But I don't think it does, but I don't know how to prove it.
 A: For $a\in\mathbb{R}$, define $f(a)=\int_0^a e^{-x^2}dx$. Then $f:\mathbb{R}\to\mathbb{R}$ is continuous. Also, since the integrand is an even function, $f$ will be an odd function. And since $f(x)\to\frac{\sqrt{\pi}}{2}$ as $x\to\infty$, we have than the range of $f$ is the interval $I=\left(-\frac{\sqrt{\pi}}{2},\frac{\sqrt{\pi}}{2}\right)$.
Hence for any familiar $b\in I$, there is an $a\in\mathbb{R}$ with $f(a)=b$.
To refine your question, it would help to be more explicit about what you mean by "familiar". Once you have a more explicit notion of what you mean by familiar, you can ask which elements of $\mathbb{R}$ will be sent to "familiar" numbers by $f$.
For example: we could ask, for which elements of $a\in\mathbb{R}$ is $f(a)$ rational? Or for which elements $a\in\mathbb{R}$ is $f(a)$ algebraic? Or for which elements $a\in\mathbb{R}$ is $f(a)=q\pi$ for some rational number $q$? 
Or you could ask, is there any rational $a\in\mathbb{R}$ with $f(a)$ rational? Or is there any algebraic $a\in\mathbb{R}$ with $f(a)$ algebraic? Etc.
I imagine that any of these questions would be both interesting and difficult.
