Integral of e^(x^2)*dx?

[Edge]

Gosh, it's been a long since I last visited.

Anyway, I was making a math study the other day on integrals and things like that. You know, you get to the point where you study complex ways of integrating functions...

However, I wonder... is there an integral for the function mentioned above...

Can we integrate every function no matter how weird is it?

I tried integrating that equation:

e^(x^2)dx ... and man, that was frustating... you only end up repeating the same process again and again...



so, I ask... is there a solution to this one? If not, why not?

[sanctus]

If you are looking for a Primitive to my knowledge there is none but the defined integral has values if you integrate over an ininite interval:
A way to see this is to start from the integral squared (change from x to -x, bounds do not change because you get one minus sign from the variable change and one for leaving the bounds as they are) and then to pass in polar coordinates:


hence the integral is

.
.
.
.
.
.
.
.
.
.

[Nootropic]

It is of an interesting note that in general the vast majority of functions cannot be easily integrated into a closed form, by this, I mean NOT a power series expansion, which is how we define the "imaginary error function" as noted by lambus. As noted by Sanctus, it is certainly possible to integrate this function over the entire real line, or any infinite interval. Interestingly enough, you can integrate a function of the form f(x) = x^5 * e^(x^2). Before ripping your hair out, make a rationalizing substiution and don't be afraid to do make another substiution and apply parts more than once. It's a fun integral, and interesting, I might say. I have actually meant to take time to look at functions of the form f(x) = x^m * e^(x^k), but I have yet to get around to that.

And the answer to your question is a resouding no, no we cannot integrate any function. The function e^(x^2) is of a class a functions who antiderivatives are defined as transcendental (neither elementary nor algebraic) and are not expressed in terms of a "normal" function. And the reason for there be no explicit formulae (as opposed to an infinite chain of polynomials) for certain integrals is rather complicated one, and I think I need some time to brood over that one (when it's not six in the morning). But interesting question, you'd never believe how many people get their doctorates in mathematics and never ask such simple questions.

And Sanctus, I've seen the derivation of this integral, but the one thing I never understood is why the integral squared is equal to the integral times the integral with respect to y. This is probably due to my unfamiliarity with multivariable calculus.

[sanctus]

The way I see it Nootropic is: the defined integral is just a number (including infinities, so maybe it gets abit more complicated) and a the root of a number squared is just a number. About y, again in the definite integral it is just a kind of summation variable (there was an expression for these variables in english, but I don't remember now, maybe "dummy variable"?) so you can call it whatever you want.
Also maybe the derivation of the integral the way I showed is complete only if you first show that the the definte integral not squared converges.
Hope you understand what I want to say.

[Nootropic]

Differential galois theory!

[sanctus]

???

.
.
.
.
.
.
.
.
.
.

[Nootropic]

Guess that works, Lambus. Differential galois theory allows one to look at a function and determine whether or not its antiderivative is elementary. Of course I could have explained this in my post, but who bothers to explain?

Check it out Differential Galois theory - Wikipedia, the free encyclopedia

.
.
.
.
.
.
.
.
.
.