Hi every body
I want to have the solution for the integration of exp(x^2)
Thanks
$\displaystyle \int_0^{\infty} dx \, e^{x^2}$ has a known value, but the indefinite integral doesn't have a "closed form." That is to say, the indefinite integral can't be expressed as a finite sum, product, etc. of polynomials, trig functions, etc. I can't tell you how to prove that, but I know it has been proven.
-Dan
The nice thing about mathematics is that when you encounter something, preferably something useful, but it isn't "possible" yet, you just define it.
The integral of e(x²) (usually with a minus-sign and some constants) appears to be very useful in e.g. statistics (normal distribution), but it doesn't have a closed-form primitive function, i.e. its primitive function cannot be expressed in terms of a finite composition of what we call 'elementary functions' (polynomials, trig, log/exp, ...).
Here comes the mathematician: we just give it a new name, we define it as a new function (search for the error-function, Erf(x)).
The name of the math that deals with these problems is called "differencial algebra" (I know not a creative name).
There is a theorem by Louiville:
If $\displaystyle f,g$ are rational functions and $\displaystyle g$ is not a constant, then,
$\displaystyle \smallint f\exp (g) $ is elementary if and only if there exists a rational solution to the differencial equation,
$\displaystyle f=h'+rg$
On some open interval.
It reminds me of a joke. A mathemation was asked to capture sheep for a farmer in smallest amount of fencing possible. What does he do? Him builds a fence around himself and says "I declare this the outside" *)
*)Hmmm... the Jordan Closed Curve theorem does not work here, interesting.