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Homework Help: Improper integral with e^(-x/2)e^(-x^2/2). Realling annoying.

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int_{-\infty}^{\infty} e^{x\over 2}e^{-x^2\over 2} dx [/tex]

    2. Relevant equations

    [tex] \int_{-\infty}^{\infty}e^{-x^2\over a} dx = \sqrt{\pi\over a} [/tex] [tex] a>0 [/tex]


    3. The attempt at a solution

    Can't seem to penetrate it, I thought about trying to isolate the second term with integration by parts.

    [tex]\int_{-\infty}^{\infty} e^{x\over 2}e^{-x^2\over 2} dx = e^{x\over 2}\int e^{-x^2\over 2}dx - \int \frac{d}{dx}e^{x\over 2} \left[ \int e^{-x^2\over 2} dx \right] dx [/tex]

    But I don't think there's any sensible way to put limits in on the RHS to eliminate those factors.
     
  2. jcsd
  3. Jan 7, 2010 #2

    rock.freak667

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    Homework Helper

    [tex]e^{\frac{x^2}{2}}e^{\frac{-x^2}{2}}=e^{\frac{x^2-x^2}{2}}=e^0=1[/tex]

    Are you sure you typed the integrand properly?


    EDIT: sorry I am blind.
     
    Last edited: Jan 7, 2010
  4. Jan 7, 2010 #3
    Yeah re-read it! One is squared one isn't.
     
  5. Jan 7, 2010 #4
    My last comment could look like "yeah re-read it [past tense]..." I still can't do this.
     
  6. Jan 7, 2010 #5

    Pengwuino

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    Gold Member

    Try completing the square in the exponent.
     
  7. Jan 7, 2010 #6
    [tex] \int_{-\infty}^{\infty} e^{x\over 2}e^{{-x^2}\over 2} dx = \int_{-\infty}^{\infty} e^{-\frac{1}{2}(x-\frac{1}{2})^2+\frac{1}{8}} dx [/tex]

    [tex] = e^{\frac{1}{8}} \int_{-\infty}^{\infty} e^{\frac{u^2}{-2}} du = e^{\frac{1}{8}}\sqrt{\frac{\pi}{2}} [/tex]

    Thanks!
     
  8. Jan 8, 2010 #7

    t!m

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    Careful, your 'Relevant equation' is wrong. For gaussian integrals,
    [tex]\int_{-\infty}^\infty e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}[/tex]
     
  9. Jan 8, 2010 #8
    That was a typo by me!
     
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