Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gaussian Integral Substitution

  1. Feb 17, 2013 #1
    If I had an integral

    [tex] \int_{-1}^{1}e^{x}dx [/tex]

    Then performing the substitution [itex] x=\frac{1}{t} [/itex] would give me

    [tex] \int_{-1}^{1}-e^\frac{1}{t}t^{-2}dt [/tex]

    Which can't be right because the number in the integral is always negative. Is this substitution not correct?

    Sorry if I am being very thick but I can't figure out why I can't evaluate this simple integral with this change of variables.
     
    Last edited: Feb 17, 2013
  2. jcsd
  3. Feb 17, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi pcvrx560! :smile:
    the substitution is fine, but the limits are wrong …

    as x goes up from -1 to 1,

    t (= 1/x) goes down from -1 to -∞, and then from +∞ down to 1 …

    you'ld need to write ##\int_{-1}^{-\infty} + \int_{\infty}^{1}## :wink:

    (or ##-\int^{-1}_{-\infty} - \int^{\infty}_{1}##)​
     
  4. Feb 17, 2013 #3

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Isn't the integral of e^x w.r.t. x simply e^x + C?
     
  5. Feb 17, 2013 #4
    Thanks, tiny-tim! That cleared it up for me.

    I didn't think about the range I was integrating over, I was just mindlessly plugging numbers into 1/t.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gaussian Integral Substitution
  1. Gaussian integral . (Replies: 8)

  2. Gaussian Integral (Replies: 1)

Loading...