What is the indefinite integral of e^(-x^2) dx?

Question

No I am not asking for help with coursework, I just thought I would ask a nice question

Answer 1

It is a nice question, and the answer leads into what is called differential Galois theory (see my source). One says that an antiderivative F of your function is not "elementary". Elementary is a bit of a nebulous word, but one version is as follows.
F is "elementary" if it can be expressed in terms of trig functions, e^x, or polynomials by the use of addition, composition, subtraction, multiplication, division, exponentiation, or function inversion. [This may be somewhat redundant, or perhaps I could include more things. It's still true.] But you're not allowed to take an antiderivative.

Just as there is a sense in which most real numbers are irrational and transcendental, most antiderivatives are not elementary. And most power series. It's very easy to come up with elementary functions with nonelementary antiderivatives, as long as you avoid rational functions and trig polynomials.
For instance sin(x^2), 1/(log(x)), sin(x)/x, your example lack elementary antiderivatives. But it's incredibly hard to prove they are not elementary (see the reference).

It's not a big deal if a function isn't elementary. In your case you have what's called the "error function", which is a very well understood function. If you know a function's derivative and a particular value, you understand the function. See Mathworld, PlanetMath, wikipedia to see how people work with it.

I'd like to point out that before Euler, e^x would not have been counted as an elementary function. And 1/x did not have an elementary antiderivative. But then Euler *defined* log x to be the antiderivative of 1/x, inverted it, and that's where we get e^x (the "e" is for "Euler"). In some texts, sin(x) is defined as the inverse of the antiderivative of 1/sqrt(1-x^2), and one can develop trigonometry based on this.

I'd be impressed if this were coursework; most texts avoid this discussion of nonelementary functions, and only give you contrived things you can solve. This can be very frustrating if there's a typo.

Answer 2

This is the bell curve. It is not easy to integrate.

Suppose you want to integrate it from zero to infinity -- how do you do that?

It involves a trick. First, notice that the integral of e^(-x^2) dx is equal to the integral of e^(-y^2) dy -- so if you multiply them together, you get the square of the integral. Since x and y are independent variable, you can make it into a double integral

e^(-(x^2+y^2))dx dy

The next trick is to change to polar coordinates. This make the integral equal to:

e(-r^2) * r dr d(theta)

where the limits of integration are 0 to infinity for r and 0 to pi/4 for theta.

This integral is easy to solve. You let u = r^2 and do a u-substitution.

Answer 3

It cannot be solved by regular integration. Only by developing a Taylor series.

Answer 4

(1/2) SqureRoot (Pi) ErrorFunction(x)

Answer 5

Nice question and I got it.
It is lengthy that it takes time to write here.

Answer 6

Integration by parts gave me this mess:

Int(e^(-x^2))=x(e^(-x^2))-Int((-2x^2)(e^(-x^2)))

Don't blame me : I'm really only eleven...

Answer 7

-1/2x*e(-x^2) + c

Done by integration with substitution:

e(-u)dx
where u = x^2
therefore du/dx = 2x
so dx = du/2x
so:
e(-u)du/2x
then:
-e(x^2)/2x+c

Answer 8

Who are you?

Answer 9

how is THAT a nice question!? to make all us non-mathmatical wizards feel thick!!

Answer 10

are you kidding me?