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Beta Function Problem

  1. Jul 22, 2010 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    [tex]B(p,q) = \int_0^{\infty}\frac{y^{p-1}}{(1+y)^{p+q}}\hspace{1mm}dy[/tex]

    3. The attempt at a solution

    I am completely stuck on this one, just a total mental block. The answer in the book says:

    "First put u = x + 2. Then put u = 4t. Answer = [tex]2\pi[/tex]"

    I'll take you through what I done, but I got nowhere near the Beta function form I require:

    [tex]u = x + 2\hspace{1mm}=>\hspace{1mm}4 - u = 2 - x[/tex]

    Also: du = dx. So:


    So now I thought to do the second part ( set u = 4t ):

    [tex]u = 4t => du = 4 dt[/tex]


    And I take out four as a common factor:


    I don't have a clue on how to proceed, it is nowhere near the equation I need, I'm not sure how to get rid of the square root, or even "turn it upside down", but even then that doesn't work because I have 1 - t and I require something like 1 + t. I would really appreciate any help on this, thank you for your time.
  2. jcsd
  3. Jul 22, 2010 #2
    [tex]\int_0^1 \left(\frac{1-t}{t}\right)^{1/2}dt=\int_0^1 \left(\frac{1}{t}-1\right)^{1/2}dt[/tex]

    Now, you want the limits to go from zero to infinity. So looking at the integrand, what would I have to let say v=f(t) be so that when I plug in t=0 I get infinity but when I plug in t=1 I get zero?

    Also, look at Wikipedia on the beta function to get familiar with it.
  4. Jul 22, 2010 #3
    I think you mixed the two boundaries of the integral up.

    When you plug in t = 0 into v you want 0, and similarly when you plug in t = 1 into v you want infinity.

    Because the resulting boundaries you want are from 0 to infinity, right?
  5. Jul 22, 2010 #4
    Just let:

    [tex]v=1/t-1[/tex] and work it through. Then switch the limits of integration.
  6. Jul 22, 2010 #5
    Your solution doesn't make sense. It doesn't give me the limits I need and it isn't in the form I need it either.
  7. Jul 22, 2010 #6
    [STRIKE]Wouldn't you be factoring out 4/4 = 1? So wouldn't the factor outside still be 1? Or am I just being silly here...Anyway, I hope this helps: [/STRIKE]
    EDIT: yep...I was just being silly. I realized where your 4 came from finally haha.


    = [tex]4\int_{0}^{1}\left(t^{-1/2}(1-t)^{1/2}\right)} \ dt[/tex]

    = [tex]4\int_{0}^{1}\left(t^{1/2 \ - \ 1}(1-t)^{3/2 \ - \ 1}\right)} \ dt[/tex]

    = [tex]4B(1/2, \ 3/2)[/tex]

    = [tex]8\int_{0}^{\pi / 2}\left(sin(\theta)^{2(1/2) \ - \ 1}cos(\theta)^{2(3/2) \ - \ 1}\right)} \ d\theta[/tex]

    = [tex]8\int_{0}^{\pi / 2}\left(sin(\theta)^{0}cos(\theta)^{2}\right)} \ d\theta[/tex]

    = [tex]8[.5(\theta + sin(\theta)cos(\theta))|_{0}^{\pi/2}[/tex]

    = [tex] 4(\pi /2) [/tex]

    = [tex] 2\pi. [/tex]
  8. Jul 22, 2010 #7
  9. Jul 22, 2010 #8
    Thanks Raskolnikov, I've gotten so frustrated with that problem over the last day. I seemed to be too obsessed with getting the problem in the form:

    [tex]B(p,q) = \int_0^{\infty}\frac{y^{p-1}}{(1+y)^{p+q}}\hspace{1mm}dy[/tex]

    That I took no notice of it being any in other form. ( Actually I'm a little confused why it is in a different form to the one I described initially, it was in a section dedicated to those types of Beta functions). Anyway, thanks everyone for your help :)
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