1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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

    User Avatar
    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

    User Avatar

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Improper integral with e^(-x/2)e^(-x^2/2). Realling annoying.
  1. Integrating e^-ax^2 (Replies: 7)

  2. Integrate e^(8x^2) (Replies: 3)

Loading...