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!

Tough Gaussian Integral

  1. Aug 31, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm trying to solve the Gaussian integral:
    [tex]\int_{-∞}^{∞}xe^{-λ(x-a)^2}dx[/tex]
    and
    [tex]\int_{-∞}^{∞}x^2e^{-λ(x-a)^2}dx[/tex]


    2. Relevant equations
    I can't find anything online that gives the Gaussian integral of x times the exponential of -λ(x+(some constant))squared. I was hoping someone here would know. It is the (-a) in the exponential that is throwing me off.

    Thanks!
     
  2. jcsd
  3. Aug 31, 2014 #2

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Well, let's start with something simpler. Do you know how to do the integrals from mini infinity to plus infinity of ## e^{-x^2}, xe^{-x^2}, x^2 e^{-x^2}##? That's the first step. If you know how to do these, it will be easy to show how to the ones you are asking about.
     
  4. Aug 31, 2014 #3

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    How about a change of variables?
     
  5. Aug 31, 2014 #4
    Ya I "know" how do them. It's a QM problem, not a mathematical one, so it is having me look up the integrals. So respectively, the solutions are [itex]\int_{0}^{∞}x^{2n}e^{\frac{-x^2}{a^2}}dx=\sqrt{π}\frac{(2n)!}{n!}(\frac{a}{2})^{2n+1}[/itex]

    In the original problem, when you expand the squared term, you end up with an x and x squared term which is confusing me.
     
  6. Aug 31, 2014 #5

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Then do a change of variable to avoid having an x term in the exponent, as Orodruin suggested.
     
  7. Aug 31, 2014 #6
    Ahh I see. Just to be clear, the correct substitution would be u=x-a, du=dx cause then x=u+a and you end up with:
    [tex]\int_{-∞}^{∞}ue^{-λ(u)^2}+ae^{-λ(u)^2}du[/tex]

    Right? Sorry, running on fumes today :)
     
  8. Aug 31, 2014 #7

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's it! And the same trick will work for the second integral (if you know the integral of ## x e^{-x^2} ## which is trivial, using symmetry.)
     
  9. Aug 31, 2014 #8

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Add a few parentheses so that the expression makes sense, but otherwise yes.
     
  10. Aug 31, 2014 #9
    Ha! I feel like an idiot.

    Thanks a bunch nrqed and Orodruin!
     
  11. Aug 31, 2014 #10

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Don't! I have seen much worse examples among university students ... :rolleyes:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Tough Gaussian Integral
  1. Gaussian Integral (Replies: 1)

Loading...