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How would you apply the Beta Function to this integral?
How would the beta function be applied to this integral?
The integral is between 1 and infinity of (t^2 - 1)^(-2/3) dt
u = t^2-1
du = 2t dt ==> dt = du/ (2(u+1)^(1/2))
1/2 ∫ u^(-2/3)/(u+1)^(1/2) du from 0 to infinity
we know that B(x,y) = ∫ t^(x-1)/(1+t)^(x+y) dt from 0 to infty
so
I= 1/2 ∫ u^(-2/3)/(u+1)^(1/2) du from 0 to infinity
x-1 = -2/3 ==> x=1/3
y+x = 1/2 ==> y=1/6
I = 1/2 B(1/3,1/6)
=1/2 ( Γ(1/3) Γ(1/6))/ Γ(1/2)
=Γ(1/3) Γ(1/6) / (2√π)
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