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Improper integral

  1. Jul 29, 2010 #1
    How do you get this:

    [tex]\int_{-\infty}^{\infty}e^{-(ax^2+b/x^2)}dx = \sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}[/tex]

    I've been trying all the tricks I know, like differentiating under the integral sign and whatnot, but I can't get it. Thanks.
     
  2. jcsd
  3. Jul 29, 2010 #2
    In which course did you run into this monster?
     
  4. Jul 29, 2010 #3
    Not in any course. I came across it in a paper where they calculate the probability that a Brownian motion X(t) crosses the line at+b at some point during the interval (0, T).
     
  5. Jul 29, 2010 #4
    Send a link of the paper?
     
  6. Jul 30, 2010 #5

    lurflurf

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    Homework Helper

    1)Let I be your integral
    2)Find I' by differentiating under the integral sign with respect to sqrt(b)
    3)Effect the change of variable u=sqrt(b/a)/x (break up the interval at 0)
    4)Notice I=-2sqrt(a)I'
    5)Conclude I=I0 exp(-2sqrt(ab))
    6)Compute I0

    sqrt() denotes the square root
    I' is the derivative of I with respect to sqrt(b)
    I0 is your integral when b=0
    See this thread
    Feynman's Calculus
     
  7. Jul 30, 2010 #6
    Thanks, lurflurf! That's exactly what I needed. Well done. That's an excellent thread you linked to, also. Surely, You're Joking is where I first learned about differentiating under the integral sign. I apparently need more practice at applying it, though.

    Sure. Here is the paper where the probability is calculated (near the end) using Laplace transforms:

    http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pjm/1102911625

    Here is a paper that makes use of the result. Scroll down to where it says "Kelly Criterion"--that's it. In the section on probability of reaching a goal within a certain time.

    http://edwardothorp.com/id10.html
     
    Last edited: Jul 30, 2010
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