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Integrating exp(-x^2) and some other stuff

  1. Jul 9, 2005 #1
    Hello everybody
    I'm very much interested in the thread about "Feynmans Calculus" (having read the books, too). The problem is I don't understand quite some of the stuff, because I don't have the necessary fundamental knowledge.
    So I thought to confront you with some lower level questions:

    How do you solve this: [tex]\int\ e^{(-x^2)} dx[/tex] ?
    It seems that [tex]\int_{-\infty} ^\infty e^{(-x^2)} dx=\sqrt{\Pi}[/tex]. Why is that?

    I honestly have no clou.

    I would be very thankful if somebody could explain this to me (then I could also continue to see wether the rest of my questions still makes sense...).
    Best regards and thank you in advance
    cliowa
     
  2. jcsd
  3. Jul 9, 2005 #2

    dextercioby

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    By defining the error function

    [tex] \mbox{erf} (x)=: \frac{2}{\sqrt{\pi}}\int_{0}^{x} \exp\left(-t^{2}\right) \ dt [/tex]

    Because a certain trick is performed using Fubini's theorem and a change of variables in the parametrization of an arbitrary point in [itex] \mathbb{R}^{2} [/itex].

    Daniel.
     
  4. Jul 9, 2005 #3
    [tex]
    \[
    \begin{array}{l}
    {\rm let }J = \int\limits_{{\rm - }\infty }^\infty {{\rm e}^{{\rm - x}^{\rm 2} } } dx \\
    {\rm then }J^2 = \int\limits_{{\rm - }\infty }^\infty {{\rm e}^{{\rm - x}^{\rm 2} } } dx\int\limits_{ - \infty }^\infty {e^{ - y^2 } } dy \\
    {\rm by \: fubini's \: theorem:} \\
    J^2 = \int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {e^{ - (x^2 + y^2 )} } dxdy} \\
    {\rm now \: employing \: a \: common \: change \: of \: variables:} \\
    J^2 = \int\limits_0^{2\pi } {\int\limits_0^\infty {e^{ - r^2 } rdrd\theta } } \\
    {\rm this \: is \: now \: easily \: evaluated} \\
    J^2 = 2\pi \mathop {\lim }\limits_{b \to \infty } \int\limits_0^b {e^{ - r^2 } } rdr \\
    J^2 = - \pi (\mathop {\lim }\limits_{b \to \infty } e^{ - b^2 } - e^0 ) = \pi \\
    {\rm so }J = \sqrt \pi \\
    \end{array}
    \][/tex]
     
  5. Jul 9, 2005 #4
    I'm sorry to say this, but I've never heard of that before. Exactly how does this work? The whole thing (error function) is new to me. Could you explain this to me? Thank you very much.

    @Cinncinnatus: How is that "common change of variables" performed?

    Thanks to everybody helping me out.
    Best regards
    Cliowa
     
  6. Jul 9, 2005 #5

    LeonhardEuler

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    e^(x^2) has no elementary antiderivitve. In probability int[e^(x^2)dx] is important because the normal probability density function is in the form e^(x^2) (with some constants in there). To get the distribution function you have to integrate it, so people make tables of numerical approximations for the integral over various intervals and call it the error function.
    To compute the integral exactly over the whole real line you can do that trick someone else showed. The change of variables is moving the double integral from rectangular to polar coordinates.
     
  7. Jul 9, 2005 #6

    dextercioby

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    In probability we use the minus in the exponential.

    Daniel.
     
  8. Jul 10, 2005 #7

    James R

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    Define [itex]r^2 = x^2 + y^2[/tex]

    This is a change to polar coordinates.

    The integration from negative infinity to positive infinity in the (x,y) plane is equivalent to an integration from 0 to infinity over r and 0 to 2 pi in [itex]\theta[/itex], in the [itex](r,\theta)[/itex] plane.

    The area element in cartesian coordinates is dx dy.
    The area element in plane polar coordinates is [itex]r dr d\theta[/itex].
     
  9. Jul 10, 2005 #8
    James R. beat me too it, thats what I did.

    I'm sure someone else could explain better than I how it actually works though...
     
  10. Jul 10, 2005 #9
    Thank you all very much. It seems a lot clearer now.

    Now here's how I got to the problems mentioned above:

    How to solve this (for x?):
    [tex]\int_{-\infty} ^\infty e^{(-ax^2)} dx=\sqrt{\frac{\Pi}{a}}[/tex]

    Having read the Feynman thread I would try the differentiation under the integral. So I take the derivative with respect to a]. I get
    [tex]-\int_{-\infty} ^\infty x^2 e^{(-ax^2)} dx=-\frac{1}{2}\sqrt{\frac{\Pi}{a^3}}[/tex].
    And now? how do I continue?
     
  11. Jul 10, 2005 #10
    What do you mean "solve this for x"?
    Isn't x a dummy variable in that equation?
     
  12. Jul 10, 2005 #11
    @Cinncinnatus: I don't know where I got the problem from, but when I saw it and had no clou how to do it, I wrote it down, and here it is. What would you solve for?
     
  13. Jul 10, 2005 #12

    Curious3141

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    There is nothing to solve for. It is an identity.

    Do you know how to make substitutions to evaluate integrals ? Try substituting [itex]u = x\sqrt{a}[/itex] to evaluate the integral on the left hand side, and see what you get.
     
  14. Jul 10, 2005 #13

    matt grime

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    change the variable in the integral, say ax^2=y^2, than apply what you already know.
     
  15. Jul 10, 2005 #14

    lurflurf

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    There are lots of ways to do this integral.
    The most common is (as has been mentioned) to consider the 2-D integral and change to polar form.
    Here is one using differentiation relative a parameter
    [tex]F(x)=\displaystyle{(}\int_0^xe^{-t^2}dt\displaystyle{)}^2+\int_0^1\frac{e^{-x^2(t^2+1)}}{t^2+1}dt[/tex]

    [tex]F'(x)=0[/tex]
    hint: do a change of variable u=xt after differentiating
    then use
    [tex]\lim_{x\rightarrow\infty}F(x)=F(0)[/tex]
    to solve for the integral
    Another that is generally useful when finding an integral that would be easy were the integrant multiplied by x is to multiply two integrals together, then change the variable in one so that it is multiplied by x then perform integrations. In this case
    [tex]I=\int_0^\infty e^{-x^2}dx[/tex]
    so
    [tex]I^2=\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dx dy[/tex]
    let y=zx so
    [tex]I^2=\int_0^\infty\int_0^\infty x e^{-x^2(1+z^2)}dx dz[/tex]
    do x integral
    [tex]I^2=\frac{1}{2}\int_0^\infty \frac{1}{1+z^2}dz[/tex]
    do z integral
    [tex]I^2=\frac{\pi}{4}[/tex]
    hence
    [tex]I=\frac{\sqrt{\pi}}{2}[/tex]
     
    Last edited: Jul 10, 2005
  16. Jul 10, 2005 #15

    mathwonk

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    this famous computation is apparently due to liouville.
     
  17. Jul 15, 2005 #16

    NSX

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    Since [itex]J = \int\limits_{{\rm - }\infty }^\infty {{\rm e}^{{\rm - x}^{\rm 2} } } dx[/itex], why is [itex]J^2 = \int\limits_{{\rm - }\infty }^\infty {{\rm e}^{{\rm - x}^{\rm 2} } } dx\int\limits_{ - \infty }^\infty {e^{ - y^2 } } dy[/itex] instead of [itex]J^2 = \int\limits_{{\rm - }\infty }^\infty {{\rm e}^{{\rm - x}^{\rm 2} } } dx\int\limits_{ - \infty }^\infty {e^{ - x^2 } } dx[/itex]?

    [hm .. infinity doesn't show on the upper bound, but pretend it's there]
     
  18. Jul 15, 2005 #17

    lurflurf

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    You can use any variable for the integration, it does not matter. Different variables are used because in subsequent steps using the same variable is bound to lead to confusion. In particular we like to write a product of integrals as an iterated integral. When this is done different variables must be used to make clear which varible corisponds to which integration.
     
  19. Jul 15, 2005 #18

    NSX

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    ah!
    interesting, I had no idea

    I will try to read up on some double integrals and cartesian/polar co-ordinate integrals this weekend.

    Thanks
     
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