1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Improper Integral with Infinity in Limits

  1. Oct 11, 2012 #1

    ∫xe^[-x^2] dx

    So basically I've solved for everything in this problem and it looks like it should be an indeterminate form and thus divergent. My book and Wolfram both say it's 0 and convergent though.

    I get it down into:
    lim [[e^(-t^2)] - e^0]/2 + lim [e^0 - [e^(-v^2)]]/2

    When I plug stuff in I get:

    [e^∞ - e^∞ - e^0 + e^0]/2

    I can see why it might be 0 from the stuff above, but e^∞ - e^∞ should be indeterminate rather than 0. Can someone please explain what I'm not getting?

  2. jcsd
  3. Oct 11, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The integrand is an odd function, so if you can show that [itex]\int_{0}^{\infty} x e^{-x^2} dx[/itex] is finite, then the integral from [itex]-\infty[/itex] to [itex]\infty[/itex] will be zero. Seems like a simple substitution will do the trick.
  4. Oct 11, 2012 #3
    I'm afraid I don't understand what you're saying.
    Last edited: Oct 11, 2012
  5. Oct 11, 2012 #4
    Well you know that sin(x) is an odd function right? An odd function means that f(-x)=-f(x) and the function just flips over when it crosses the origin like sin(x) does. So that if f(-x)=-f(x) then surely:

    [tex]\int_{-1}^{1} f(x)dx=0[/tex]

    if f(x) is odd. Same diff for any symmetric interval like -2 to 2, -100 to 100 even -infty to infty.

    So what about the function [itex]x e^{-x^2}[/itex]? Is that one f(-x)=-f(x)? Then it would be an odd function then so that:

    [tex]\int_{-a}^{a} f(x)dx=0[/tex]

    And would be likewise zero if a were infinity and the integral is finite in the interval

    [tex]\int_0^{\infty} f(x)dx[/tex]

    And I think you can compute:

    [tex]\int_0^{\infty}x e^{-x^2} dx[/tex]

  6. Oct 11, 2012 #5


    User Avatar
    Science Advisor

    You are trying to evaluate "[itex]e^{\infty}[/itex]" when you should be looking at "[itex]e^{-\infty}[/itex].
    Last edited by a moderator: Oct 12, 2012
  7. Oct 12, 2012 #6
    I see what you're talking about. Thanks. I messed up on my math anyway : D
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook