The integration of e^(x^2)

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The integral of e^(x^2) does not have an elementary anti-derivative, a fact confirmed by multiple contributors referencing calculus literature. The discussion highlights that while e^(-x^2) is related to the error function (erf), no similar naming convention exists for e^(x^2). Attempts to find a solution through conventional methods or Laplace transforms are deemed ineffective, as they ultimately lead back to the non-elementary nature of the integral. Some participants suggest using series expansions or numerical methods, but these approaches do not yield a simple expression. Overall, the consensus is that the integral of e^(x^2) cannot be expressed in terms of elementary functions.
  • #31
HallsofIvy said:
I don't know how many times we have to say this- it's given in any good calculus book. The anti-derivative of e^{x^2} is not an "elementary" function.
My best guess, this number is aleph-naught: infinite, but countable.

I have assumed an infinite lifetime for the forum. :wink:
 
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  • #32
Redbelly98 said:
I have assumed an infinite lifetime for the forum. :wink:

lol :biggrin:

a nice point of view :wink:
 
  • #33
thepatient said:
No.


If u = x^2. du = 2xdx, not 2dx.

You're right. What a stupid error I made.

u = x^2, du = 2x
multiply du * e^u = 2x e^x^2.
 
  • #34
LTA85 said:
You're right. What a stupid error I made.

u = x^2, du = 2x
multiply du * e^u = 2x e^x^2.
\intex2dx If:

u = x2
du = 2xdx

You can't substitute dx by du, since dx = du/(2x), not just du. You would have dx in terms of two variables, you can't integrate x in terms of u. The anti-derivative of apples can't be oranges.In no nice way is this function integrable.
 

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