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Homework Help: Beta Function, Limits Problem

  1. Jul 25, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int_1^{\infty}\frac{dx}{x^2(x-1)^{1/2}}[/tex]

    2. Relevant equations

    [tex]\int_0^1t^{x-1}(1-t)^{y-1}\,dt[/tex]

    [tex]\int_0^\infty\dfrac{t^{x-1}}{(1+t)^{x+y}}\,dt,[/tex]

    3. The attempt at a solution

    Hi all, I have another beta function problem. This time I'm unsure how to deal with the limits, as the book states I have to make the substitution : x = 1/y. Putting in the values of x, gives:

    [tex]x = \infty => y = \frac{1}{\infty}[/tex]

    [tex]x = 1 => y = 1[/tex]

    I'm unsure how to deal with the one over infinity. It doesn't conform to the beta forms I'm aware of ( see relevant equations ). I tried to continue on with my calculation pretending that I had upper limit 1 and lower limit 0. But I got a negative answer as opposed to the positive one I should get. Any help one this question will be appreciated.
     
    Last edited: Jul 25, 2010
  2. jcsd
  3. Jul 25, 2010 #2
    Shouldn't your upper bound be 0, and your lower bound be 1? After all,

    [tex] \frac{1}{\infty} = 0 [/tex]

    If you simply had the bounds mixed up, that will be the culprit of your unwanted negative sign!
     
  4. Jul 25, 2010 #3
    That's what I thought, but can I have an upper bound of 0 and lower bound of 1? I thought b was always supposed to be greater than a:

    [tex]\int_a^bf(x)\,dx[/tex]

    Also, I get the negative sign from differentiating x = 1/y, not from my mixing the bounds around.
     
  5. Jul 25, 2010 #4
    What makes you think that?
     
  6. Jul 25, 2010 #5
    I've never seen an example in which it was less than a.

    EDIT: Also when getting bounds from a graph you always choose b as the largest positive x value and a as the smallest x value.
     
  7. Jul 25, 2010 #6
    I don't think there's any rule to dictate that the upper bound of the integral has to larger than the lower bound.

    [tex]\int_a^bf(x)\,dx = F(b) - F(a) [/tex]

    So,

    [tex]\int_b^af(x)\,dx = F(a) - F(b) [/tex]

    What's wrong with that? I think of it this way, instead of integrating "left to right" we are now integration "right to left".
     
  8. Jul 25, 2010 #7
    Make the substitution

    [tex]
    x = \frac{1}{t}
    [/tex]
     
  9. Jul 25, 2010 #8
    That's already been done. I made the substitution x = 1/y as I stated in my initial post.

    What I'm confused about is the limits of the integration and the fact that I'm getting a negative answer.
     
  10. Jul 25, 2010 #9
    I'm not sure that's correct. The upper limit is b, so really that should be the upper limit, right?
     
  11. Jul 25, 2010 #10
    b and a are just numbers.
     
  12. Jul 25, 2010 #11
    Do you happen to have an example to show me that that has a smaller number on the upper limit than the lower limit?
     
  13. Jul 25, 2010 #12
    Here's an example to show you that when you the bounds of integration are swapped, the answer is simply the negative of each other.

    [tex]\int_0^{\pi}sin(x)\,dx = 2[/tex]

    Swapping the bounds,

    [tex]\int_\pi^0sin(x)\,dx = -2[/tex]
     
  14. Jul 25, 2010 #13
    Show your work. We will tell you where you made a mistake.
     
  15. Jul 25, 2010 #14
    Thanks. I'm not familiar with this sort of work ( swapping the bounds and applying a negative ) I have never done it before, and nothing was stated in the answer section or accompanying question text. So, this is what I get applying your helpful advice :smile::

    [tex]x = \frac{1}{y} => dx = \frac{-1}{y^2}\,dy[/tex]

    [tex]-\int_1^0y^{-\frac{1}{2}}(1-y)^{\frac{1}{2}}\,dy[/tex]

    And swapping the limits around gives :

    [tex]\int_0^1y^{-\frac{1}{2}}(1-y)^{\frac{1}{2}}\,dy[/tex]

    [tex]B\left(\frac{1}{2},\frac{3}{2}\right) = \frac{\sqrt{\pi}\frac{1}{2}\sqrt{\pi}}{1}[/tex]

    [tex]B\left(\frac{1}{2},\frac{3}{2}\right) = \frac{1}{2}\pi[/tex]

    I really wish the text had stated this "bound switch" was needed to achieve the answer. Thanks again, for your help jegues.
     
  16. Jul 25, 2010 #15
    Although the end result is correct, since the Beta function is symmetric with respect to its arguments, you made a mistake in the second step

    I got:

    [tex]-\int_1^0y^{\frac{1}{2}}(1-y)^{-\frac{1}{2}}\,dy[/tex]
     
  17. Jul 25, 2010 #16
    My mistake, I forgot the part I used to get my answer was under 1 as in, a fraction. Thanks.
     
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