Beta and gamma function

[abeyatta]

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.

 

[daon2]

You didn't miss anything. The value you came up with is equal to what you want.

http://www.wolframalpha.com/input/?i=(Γ(1/4)^2)/(6√2π)+=1/4*Beta(1/4+,+3/2)

Perhaps you can use these identities to turn one into the other?

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

and

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

 

[abeyatta]

Thanks

Thanks daon2,

You didn't miss anything. The value you came up with is equal to what you want.

http://www.wolframalpha.com/input/?i=(Γ(1/4)^2)/(6√2π)+=1/4*Beta(1/4+,+3/2)

Perhaps you can use these identities to turn one into the other?

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

and

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