Integrating a Square Root Function: Solution

Ikastun
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Homework Statement



∫(0,1) √x/√[3]1-x

Homework Equations



\Gammap\Gammaq/\Gammap+q

The Attempt at a Solution



p-1=1/2 →p=3/2
q-1=-1/3 →q=2/3

β(3/2,2/3)=\Gamma(3/2) \Gamma(2/3)/\Gamma(13/6)

\Gamma3/2=1/2\Gamma(1/2)=√π/2
\Gamma2/3=-1/3
\Gamma13/6=7/6 1/6=7/36

β(3/2,2/3)=-6√π/7
 
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Ikastun said:
\Gamma2/3=-1/3
\Gamma13/6=7/6 1/6=7/36
That looks wrong.

Can you explain what you want to calculate, how you attempt to do this and where your problem is?
 
Hello and thank you for answering.

My problem begins with the part you quote. I don't know how to properly use the recursive formula in those cases.
Regarding my attempt to calculate the integral, what I wrote above is everything.
 
##\Gamma(\frac{13}{6})=\frac{7}{6}\Gamma(\frac{7}{6}) =\frac{7}{36}\Gamma(\frac{1}{6})##
For some values, an analytic expression is known, in general this doesn't work and you have to live with the expressions (or find a numerical approximation).

Regarding my attempt to calculate the integral, what I wrote above is everything.
There is some connections between the formulas that you could explain.
 
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