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Homework Help: Problem involving Gaussian Integrals

  1. Jan 3, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int e^{-\frac{1}{2}(ax^{2}+\frac{b}{x^{2}})} dx[/tex]

    from [tex]-\infty[/tex] to [tex]\infty[/tex]




    2. Relevant equations
    it says to solve this, given that, again from [tex]-\infty[/tex]to [tex]\infty[/tex]:


    [tex]\int e^{-\frac{1}{2}x^{2}}}dx[/tex]

    = [tex]\sqrt{}2\pi[/tex]

    and suggests that the substitution

    [tex]u = x\sqrt{a}- \frac{\sqrt{b}}{x}[/tex]

    may be helpful taking into account the limits carefully

    3. The attempt at a solution

    i did all this and all i ended up with was another integral which was no more helpful than the one i had in the beginning, and didnt resemble the stated gaussian integral. i am also unsure as to what to make of the "taking into account the limits carefully" im not sure if there is something that im missing out which is blindingly obvious, but any help would be greatly appreciated!
     
  2. jcsd
  3. Jan 4, 2010 #2
    A good place to start is to sketch the function you are trying to substitute and work out whether it makes for a reasonable substitution (that way you should be able to see the limits and all that straight off your sketch). Also find the inverse function, u as a function of x.

    What is dx in terms of du?

    What is the integral you get in the substitution?

    Show us where you got stuck, and hopefully we can get you unstuck!
     
  4. Jan 4, 2010 #3
    well i get

    [tex]\frac{2du}{\sqrt{a}+\frac{\sqrt{b}}{x}} = dx[/tex]

    leading to

    and then

    [tex]-(\sqrt{ab}+2u^{2}) = -\frac{1}{2}(ax^{2}+\frac{b}{x^{2}})[/tex]

    [tex]x=\frac{u}{\sqrt{a}}\pm \sqrt{\frac{u^{2}}{a}+\sqrt{\frac{b}{a}}}[/tex]
    which i make completely unhelpful

    all in all, this gives me an integral of...

    [tex]2\int\frac{e^{-(\sqrt{ab}+2u^{2})}}{\sqrt{a}+\frac{\sqrt{b}}{x^{2}}} du[/tex]

    with that horrendous expression for [tex]x^{2}[/tex] substituted in.
    also, i'm pretty sure that the limits would remain as [tex]\pm\infty[/tex] but the question looks as though it appears otherwise...?
     
  5. Jan 5, 2010 #4
    Firstly the limits: you're not being careful, you're just guessing. Sketch u as a function of x. (Use a graphing program if you must.)

    Your inverse looks good: which sign should you take and why (hint: think about the limits).

    Your expression looks pretty good; but as you say it's pretty messy. An equivalent thing to do would be to differentiate your inverse expression to find dx in terms of du; that should keep everything on the top line and in terms of u. Is that expression any simpler?

    Keep plugging at it and you'll get there.
     
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