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More integration

  1. Oct 19, 2006 #1
    More integration :)

    [tex] \frac{1}{4 \pi \sigma^2} \int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{2\sigma^2}} [/tex]

    we know that
    [tex] \int_{-\infty}^{\infty} e^{-\frac{x^2}{2\sigma^2}} = \sqrt{2 \pi \sigma^2} [/tex]

    and then differentiate both sides wrt sigma
    [tex] \int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{2\sigma^2}} = \sigma^3 \sqrt{2 \pi} [/tex]

    sib the third into the first

    [tex] \frac{1}{4 \pi \sigma^2} \sigma^3 \sqrt{2 \pi} [/tex]

    [tex] \frac{\sigma \sqrt{2 \pi}}{4 \pi} [/tex]

    something is wrong .. where did i go wrong ... pelase help :(
     
    Last edited: Oct 19, 2006
  2. jcsd
  3. Oct 19, 2006 #2

    quasar987

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    You forgot a "-" sign...

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

    Specifically, would you mind writing what you get after differentiating the integral that equals [itex]\sqrt{2 \pi \sigma^2} [/itex] wrt sigma?
     
  4. Oct 19, 2006 #3
    i got
    sigma times sqrt(2 pi)
     
  5. Oct 19, 2006 #4

    quasar987

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    What is it supposed to give?
     
  6. Oct 19, 2006 #5
    it gives me sqrt (2 pi)
    after differentiating
     
  7. Oct 19, 2006 #6

    quasar987

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    One things's for sure;

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

    If I differentiate the exponential, I get

    [tex]\frac{-x^2}{2}\frac{-2}{\sigma ^3} = \frac{x^2}{\sigma^3}[/tex]

    So

    [tex]\int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{2\sigma^2}}dx = 2\pi \sigma^3[/tex]

    And

    [tex] \frac{1}{4 \pi \sigma^2} \int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{2\sigma^2}} = \frac{\sigma}{2}[/tex]
     
    Last edited: Oct 19, 2006
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