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Integrate e^(x^2)

  1. Apr 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Integrate e^(x^2)

    2. Relevant equations



    3. The attempt at a solution

    I have no idea at all..
     
  2. jcsd
  3. Apr 9, 2008 #2

    HallsofIvy

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    There's a reason for that. That function, like most integrable functions, does not have an anti-derivative that can be written in terms of elementary functions.
     
  4. Apr 9, 2008 #3
    But someone told me that it's important in Quantum Mechanics. (I'm still first year)
     
  5. Apr 9, 2008 #4
    You can solve it using a lambert function, but that is no elementary solution. I only know that as by a fluke someone asked me the same question yesterday. I already knew it was insoluble though with elementary functions.
     
  6. Apr 9, 2008 #5

    tiny-tim

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    Hi sai! :smile:

    e^(-x^2) is important in Quantum Mechanics.

    But it's usually integrated from 0 to ∞, for which there's a trick (from memory, the result is something like √π), and so an anti-derivative isn't needed. :smile:
     
  7. Apr 9, 2008 #6
    I see. one of my friends is very good at maths and he's trying to solve this one for days. I'll tell him.

    yeah i remember reading it somewhere. it's got something to do with the gaussian distribution if i'm not mistaken.
     
  8. Apr 9, 2008 #7

    tiny-tim

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    Yes, you're right … it is the gaussian distribution (aka the normal distribution)! :smile:
     
  9. Apr 9, 2008 #8
    Thanks :)
     
  10. Apr 9, 2008 #9
    [itex]\int_0^{\infty} e^{x^2}\;dx [/itex] where [itex]e^{x^2}={\frac{2}{x}}e^{x^2}[/itex]

    Is it something like this? I can't remember but I've seen that somewhere before? Reaching a bit now though, I wouldn't take my advice
     
    Last edited: Apr 9, 2008
  11. Apr 9, 2008 #10

    tiny-tim

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    No … it's more like ∫e^x^2dx = √[∫e^x^2dx∫e^y^2dy], and then changing to dr and dθ. :smile:
     
  12. Apr 9, 2008 #11

    HallsofIvy

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    And, once again, it must be
    [tex]e^{-x^2}[/tex]
    in order to converge.
     
  13. Apr 9, 2008 #12

    tiny-tim

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    :redface: oops! :redface:

    Thanks, HallsofIvy! :smile:
     
  14. Apr 9, 2008 #13
    Hehe in that case I don't know where that came from. :smile:

    Humour me could you show me how you would go about doing that in more detail? I'm not sure quite if I get that?

    Are you talking about spherical polars? If so how does that work? Or should I say why does that work?
     
    Last edited: Apr 9, 2008
  15. Apr 9, 2008 #14

    tiny-tim

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    Well, ∫e^-x^2dx and ∫e^-y^2dy are two different ways of writing the same thing, aren't they?

    So each is the √ of their product, ie:

    √∫∫e^-(x^2 + y^2)dxdy

    = √∫∫e^-r^2dxdy

    = √∫∫r.e^-r^2drdθ

    = … ? :smile:
     
  16. Apr 9, 2008 #15
    You are right! It is integrated from negative infinity to positive infinity. I get one method to do that, but have another problem..
    e^(x^2)=e^(-i)*e^(ix^2)=e^(-i)*(cos(x^2)+isin(x^2))
    I try to integrate sin(x^2) or cos(x^2) instead, however, i havent get it yet:confused:
     
  17. Apr 9, 2008 #16

    I don't know [itex]e^{i\pi}+1[/itex] you're the teacher. :smile:

    Please be gentle with me I haven't studied any of this in seven months, so I've forgotten most of it. :smile:
     
  18. Apr 9, 2008 #17

    D H

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    The goal is to evaluate

    [tex]I=\int_0^{\infty}e^{-x^2}dx[/tex]

    This is obviously a positive value (if the integral converges), so [itex]I=\sqrt{I^2}[/itex], or

    [tex]
    \begin{aligned}
    I^2 &= \left(\int_0^{\infty}e^{-x^2}dx\right)\left(\int_0^{\infty}e^{-x^2}dx\right) \\[6pt]
    &= \int_0^{\infty}e^{-(x+y)^2}\,dxdy &\text{physicists have no problem w/ this step}\\[6pt]
    &= \int_0^{\pi/2}\int_0^{\infty}e^{-r^2}r\,dr\,d\theta & \text{conversion to polar}\\[6pt]
    &= \frac{\pi}4 \int_0^{\infty} e^{-u} du & \text{with substitution $ u=r^2 $} \\[6pt]
    &= \frac{\pi} 4 & \text{and thus}\\[6pt]
    I &= \frac{\sqrt{\pi}}2
    \end{aligned}
    [/tex]
     
  19. Apr 9, 2008 #18
    Ah right I get it thanks DH. That's what I was looking for. :smile:

    [tex]\begin{aligned}\\[6pt]&= \int_0^{\infty}e^{-(x+y)^2}\,dxdy &\text{physicists have no problem w/ this step}\\[6pt]\end{aligned}[/tex]

    Are the Secret Algebra Squad (SAS) onto them or something? :smile:

    [tex]
    \int_0^{\infty}e^{-x^2}dx=0.88622692545275801364908374167057
    [/tex]

    Nice and irrational. :smile:
     
    Last edited: Apr 9, 2008
  20. Apr 9, 2008 #19

    tiny-tim

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    Welcome to PF!

    Hi Zachary! Welcome to PF! :smile:

    No … e^(x^2) doesn't equal e^(-i)*e^(ix^2).

    And ∫e^(ix^2)dx can be dealt with in the same way as ∫e^(-x^2)dx, except that you end up with √∫e^(iu)du, which doesn't converge. :smile:
     
  21. Apr 9, 2008 #20
    sorry.. I meant x^2=-(-1)*x^2=-(i^2)*(x^2)=-i*(i*x^2) so that i can use e^ix=cosx+isinx.
    But D H has provided the answer, that is quite nice.
     
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