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Thread: Gamma and Beta Functions

  1. #1
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    Gamma and Beta Functions

    the integral from 0-1 of √(1-x^4) dx = (Γ(1/4)^2)/(6√2π)
    Here is how i tried to tackle it...

    let t= x^4 , then dt/dx = 4x^3. And also x^3 = t ^(3/4).

    Now the integral from 0-1 of (1-t) ^1/2 dt/4x^3

    1/4 of the integral from 0-1 of x^(-3) (1-t)^1/2 dt

    1/4 of the integral from 0-1 of. t^(-3/4) (1-t) ^1/2 dt

    x-1 = -3/4 therefore x= 1/4
    y-1 = 1/2 then. y= 3/2

    which is 1/4 B(1/4 , 3/2) NOT equal to the LHS of the equation.

    i also tried it by substituting x^2 = sin u and end up with a complete different answer of B(1/2, 3/2).

    can some one pls point out what I missed.

    Any help would be much appreciated
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  2. #2
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    Re: Gamma and Beta Functions

    $\displaystyle I=\int_{0}^1 \sqrt{1-x^4}dx$

    Let $\displaystyle t=x^4$ so $\displaystyle dt=4x^3dx$ this is $\displaystyle x=t^{1/4}$ and $\displaystyle dx=\frac{dt}{4x^3}=\frac{dt}{4t^{3/4}}$

    $\displaystyle \int_{0}^1(1-t)^{1/2} t^{-3/4}dt=\int_{0}^1(1-t)^{1/2} t^{-3/4}dt$

    The definition of the beta function reads

    $\displaystyle B(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt$

    so that $\displaystyle y-1=\frac{1}{2}$ and $\displaystyle x-1=\frac{3}{4}$ hence $\displaystyle y=\frac{3}{2}, x=\frac{7}{4}$

    so

    $\displaystyle \int_{0}^1(1-t)^{1/2} t^{-3/4}dt=B(\frac{7}{4},\frac{3}{2})$

    Using that

    $\displaystyle B(x,y)=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$

    my answer is

    $\displaystyle 4I=\frac{\Gamma(\frac{7}{4})\Gamma( \frac{3}{2})}{ \Gamma (\frac{13}{4})}$

    which is not yours either
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  3. #3
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    Re: Gamma and Beta Functions

    Thanks Ruun

    I think we have the same answer if it wasn't for the following

    when you work out for x, you did x-1 = 3/4, but it should have been (-3/4) , which gives x = 1/4

    But either way the answer should be equal to Γ(1/4)^2/(6√2π).....its 6 root 2pie in the denominator ,
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  4. #4
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    Re: Gamma and Beta Functions

    Hi Abeyatta,

    I agree with your answer of $\displaystyle \frac{1}{4}B(1/4,3/2).$

    It can be converted to the given answer by the use of a few identities.

    Start with $\displaystyle B(m,n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}.$

    That gets you a $\displaystyle \Gamma(7/4)$ in the denominator and you deal with that by using the so called duplication formula

    $\displaystyle 2^{2x-1}\Gamma(x)\Gamma(x+\frac{1}{2})=\sqrt{\pi} \, \Gamma(2x)$

    (Substitute x = 5/4).

    Remove the $\displaystyle \Gamma(3/2)$ and $\displaystyle \Gamma(5/2)$ by use of the identities $\displaystyle \Gamma(n+1)=n\Gamma(n)$ and finally $\displaystyle \Gamma(1/2)=\sqrt{\pi}.$

    (The $\displaystyle \Gamma(5/4)$ also requires the use of $\displaystyle \Gamma(n+1)=n\Gamma(n)).$
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  5. #5
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    Re: Gamma and Beta Functions

    Thank you very much Bob, and apologies for late reply.
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