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

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 ,

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 $\displaystyle \Gamma(n+1)=n\Gamma(n)).$

Re: Gamma and Beta Functions

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