The integration of e^(x^2)

  • #26
HallsofIvy
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Do a little research on the Chain Rule. This is how you solve these kinds of problems.
No, that is how you differentiate functions. This question was about integrating.
 
  • #27
arildno
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To Thepatient:

You ought to see that the above power series is only useful for tiny x's, i.e, when just the few first terms are strictly dominant in magnitude.

For large x's (say A and B), the following scheme might be used:

[tex]I=\int_{A}^{B}{e}^{x^{2}}dx=\int_{A}^{B}\frac{2x}{2x}e^{x^{2}}dx=\frac{1}{2x}e^{x^{2}}\mid_{A}^{B}+\int_{A}^{B}\frac{1}{2x^{2}}e^{x^{2}}dx=\frac{1}{2x}e^{x^{2}}\mid_{A}^{B}+\frac{1}{2C^{2}}I[/tex]
where C is some number between A and B.
(the existence of such a C is guaranteed by the intermediate value theorem)

Thus, we get:
[tex]I=\frac{\frac{1}{2x}e^{x^{2}}\mid_{A}^{B}}{1-\frac{1}{2C^{2}}}[/tex]

when C can be regarded as a big number, the numerator by its own is a fairly accurate estimate.
 
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  • #28
[tex] erf(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt [/tex]

when

[tex] \int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi} [/tex]
 
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  • #29
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You can use integration by substitution.

Problem: e^x^2

Solution: Set u = x^2, then find the derivative of u, du = 2 dx
solve for dx. dx = du/2
and multiply it by the problem. dx*e^x^2 = 1/2*e^x^2.

Answer: 1/2 e^x^2 + C (c is a constant)
 
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  • #30
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You can use integration by substitution.

Problem: e^x^2

Solution: Set u = x^2, then find the derivative of u, du = 2 dx
solve for dx. dx = du/2
and multiply it by the problem. dx*e^x^2 = 1/2*e^x^2.

Answer: 1/2 e^x^2 + C (c is a constant)
No.


If u = x^2. du = 2xdx, not 2dx.
 
  • #31
Redbelly98
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I don't know how many times we have to say this- it's given in any good calculus book. The anti-derivative of [itex]e^{x^2}[/itex] 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:
 
  • #33
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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
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You're right. What a stupid error I made.

u = x^2, du = 2x
multiply du * e^u = 2x e^x^2.


[tex]\int [/tex]ex2dx 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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