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Gaussian integral using integration by parts

  1. Oct 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Show in detail that:
    [tex]
    \sigma_{x}^{2} = \int_{-\infty}^{\infty} (x -\bar{x})^{2} \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-X)^{2}}{2\sigma^{2}}} = \sigma^{2}
    [/tex]
    where,
    [tex]
    G_{X,\sigma}(x) = \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-X)^{2}}{2\sigma^{2}}}
    [/tex]


    2. Relevant equations
    [tex]
    \int u dv = uv -\int v du
    [/tex]



    3. The attempt at a solution
    There are some hints that are given. Replace [itex]\bar{x}[/itex] with [itex]X[/itex] (according to the text, this is because they are equal after many trials). make the substitutions [itex]x-X=y[/itex] and [itex] y/\sigma=z[/itex]. Integrate by parts to obtain the result.

    After following the hints I get the following integral:
    [tex]
    \frac{\sigma}{\sqrt{2\pi}} \int_{\infty}^{\infty} z^{2}e^{-\frac{z^{2}}{2}}dz
    [/tex]
    How would one go about using integration by parts for this integral?
     
    Last edited: Oct 24, 2013
  2. jcsd
  3. Oct 24, 2013 #2

    D H

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    What have you tried?

    BTW, your problem statement doesn't look right. That ##(x-\bar x)## should be ##(x-\bar x)^2##. You also have a problem in your final integral.
     
  4. Oct 24, 2013 #3
    If the problem is what you asked, you want to integrate by parts twice, taking down the ##x^2## one degree at each step.
     
  5. Oct 24, 2013 #4

    D H

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    You only need to integrate by parts once if you know the value of ##\int_{-\infty}^{\infty} \exp(x^2) \, dx## (or of ##\int_{-\infty}^{\infty} \exp(x^2/2) \, dx\,\,##).
     
  6. Oct 25, 2013 #5
    as it happens I don't know, altho I think I have done them over and over -- you'd think I would learn.
     
  7. Oct 26, 2013 #6
    Working with integration by parts I've tried some different combinations of [itex]u[/itex] and [itex]dv[/itex]:

    [tex]
    \frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z^{2} e^{-\frac{z^2}{2}} dz\\
    u=z^{2}\\
    du = 2 z dz\\
    dv = e^{-\frac{z^{2}}{2}}dz\\
    v = 2.5\\
    2.5 z^{2} |_{-\infty}^{\infty} - \int 2.5 *2z dz\\
    =0
    [/tex]

    Trying to recognize some symetry:
    [tex]
    \frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z^{2} e^{-\frac{z^2}{2}} dz\\
    \frac{2\sigma}{\sqrt{2\pi}} \int_{0}^{\infty} z^{2} e^{-\frac{z^2}{2}} dz\\
    u = z^{2}\\
    dv = e^{-\frac{z^{2}}{2}} dz
    [/tex]
    seems to blow up to infinity as does setting [itex]u=z[/itex] and [itex]dv = ze^{-\frac{z^{2}}{2}}dz[/itex]

    What I would like is a hint about choosing my [itex]u[/itex] and [itex]dv[/itex]. Or it seems like there's something I'm missing here.
     
  8. Oct 26, 2013 #7

    vela

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    Don't use limits when integrating dv. The latter choice of u=z and ##dv=ze^{-z^2/2}\,dz## is the way to go.
     
  9. Oct 26, 2013 #8
    [tex]
    \frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z^{2} e^{-\frac{z^{2}}{2}}dz\\
    u = z\\
    du = dz\\
    \\
    dv = z e^{-\frac{z^{2}}{2}} dz\\
    v = ?\\
    \\
    [/tex]
    working to find v and using a strange result that I found on the boards mentioned by nicksauce near the middle of the page link
    [tex]
    \int z e^{-\frac{z^{2}}{2}} dz\\
    u = z\\
    du = dz\\
    \\
    dv = e^{-\frac{z^{2}}{2}} dz \\
    v = \frac{\sqrt{\pi}}{\sqrt{2}}\\
    \\
    z\frac{\sqrt{\pi}}{\sqrt{2}} - \int \frac{\sqrt{\pi}}{\sqrt{2}} dz = 0\\
    [/tex]
    plugging back into above
    [tex]
    u = z\\
    du = dz\\
    v = 0\\
    dv = z e^{-\frac{z^{2}}{2}} dz\\
    \\
    [/tex]
    It seems like everything just keeps going to 0 or infinity.
     
  10. Oct 26, 2013 #9

    vela

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    Don't use limits when integrating dv. Calculate the indefinite integral.
     
  11. Oct 26, 2013 #10
    How would I handle [itex] \int e^{-\frac{z^{2}}{2}} dz[/itex] if its an indefinite integral?
     
  12. Oct 26, 2013 #11

    vela

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    You don't.
     
  13. Oct 27, 2013 #12
    [tex]
    \sigma_{x}^{2} = \int_{-\infty}^{\infty} (x-X)^2 \frac{1}{\sigma\sqrt{2\pi}} e^{-(x-X)^{2}/2\sigma^{2}} \\

    = \int_{-\infty}^{\infty} y^2 \frac{1}{\sigma\sqrt{2\pi}} e^{-y^{2}/2\sigma^{2}} \\

    = \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \frac{y^2}{\sigma^{2}}e^{-y^{2}/2\sigma^{2}} \\

    = \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty} z^{2}e^{-z^{2}/2} \\
    [/tex]

    then IBP and a u-sub to find [itex]v[/itex] by evaluating [itex]dv[/itex] as an indefinite integral.
    [tex]
    u = z\\
    du = dz\\
    \\
    v = -e^{-\frac{z^{2}}{2}}\\
    dv = z e^{-\frac{z^{2}}{2}} dz\\
    \\
    \\
    -ze^{-z^2/2} |_{-\infty}^{\infty} + \int_{-\infty}^{\infty} e^{-z^{2}/2}dz\\
    = \sqrt{2\pi}
    [/tex]
    going back up to the top
    [tex]
    \sigma_{x}^{2} = \sigma
    [/tex]

    Can you see where I'm missing the extra [itex]\sigma[/itex]? Also, what is the point of doing integration by parts if I just end up with the integral [itex]\int_{-\infty}^{\infty} e^{-z^{2}/2}dz=\sqrt{2\pi}[/itex], but I had to use wolfram to evaluate it, when at the very beginning I could have used wolfram to get [itex]\int_{-\infty}^{\infty} z^{2}e^{-z^{2}/2}=\sqrt{2\pi}[/itex].

    Is there a trick to evaluating [itex]\int_{-\infty}^{\infty} e^{-z^{2}/2}dz[/itex] by hand?
     
  14. Oct 27, 2013 #13

    D H

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    You left out the dx, dy, and dz in your integrals. Don't do that.
     
  15. Oct 27, 2013 #14

    Ray Vickson

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    Using ##u = z## and ##dv = z\: \exp(-z^2/2) \, dz## is exactly the way to go. I cannot get it to "blow up" as you claim.
     
  16. Oct 27, 2013 #15
    And there is my missing sigma! Thank you DH.
     
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