What is special about $\int e^{-t^2}dt$?

Question

I know that there's a special Taylor series expansion for $$\int e^{-t^2}dt = \sum_{k=0}^\infty(-1)^k{x^{2k+1}\over k!(2k+1)}$$ and when I ask Mathematica to calculate this integral

Integrate[E^(-t^2), t]

the answer is

1/2 Sqrt[\[Pi]] Erf[t]

$\frac 12 \sqrt{\pi}\;Erf(t)$, where $Erf(t)$ is some kind of an error function. The question is what else special is there about that integral and how could one calculate it by hand?

What exactly do you mean with calculate it? Do you mean proving that an antiderivative for the integrand function is $t\mapsto \dfrac {\sqrt \pi \text{erf}(t)}2$? Do you know the definition of $\text{erf}$? — Git Gud, Jun 16 '14 at 10:51
Yes, that's exactly what I mean. No I don't know the definition — Dmitry Kazakov, Jun 16 '14 at 11:00
The error function is extremely important in probability and statistics, since it represents the normal distribution. See also Gaussian integral. — Lucian, Jun 16 '14 at 16:53

orion · Accepted Answer · 2014-06-16T12:16:20.517

This is precisely the definition of the error function. The integral is one of the countless integrals that simply isn't expressible as a finite expression involving radicals, exp and trig functions. Most of the integrals are that way (only a very small number of integrals can be written in terms of elementary functions). However, this particular integral (integral of the Gauss distribution) is so commonly used (statistics, physics, who knows what else), that someone named it the error function and put it in tables and computer software.

If you think about it, every function has to be defined and calculated in some way. Computing ${\rm erf}(x)$ numerically in the background when you push a button is not much different than what happens when you push a button for $\sin (x)$ or any other transcendental function. It's just our choice what functions are important (although one could argue that exp+trig are elementary because of their complex analytical properties).

Your question somehow mentions the Taylor expansion. There's nothing special about it and has no connection with the definition of ${\rm erf}(x)$.

is there a proof that this function does not have any antiderivative? — Bman72, Jun 16 '14 at 12:22
@Ale It does, just not in terms of elementary functions. What you want is the Risch algorithm: http://en.wikipedia.org/wiki/Risch_algorithm
It's by no means an easy question. You should also read this thread: http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral — orion, Jun 16 '14 at 12:35

What is special about $\int e^{-t^2}dt$?

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