Integrating a Square Root Function: Solution

[Ikastun]

Homework Statement

∫(0,1) √x/√[3]1-x

Homework Equations

$\Gamma$p$\Gamma$q/$\Gamma$p+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)

$\Gamma$3/2=1/2$\Gamma$(1/2)=√π/2
$\Gamma$2/3=-1/3
$\Gamma$13/6=7/6 1/6=7/36

β(3/2,2/3)=-6√π/7

 

[mfb]

$\Gamma$2/3=-1/3
$\Gamma$13/6=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?

 

[Ikastun]

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.

 

[mfb]

$\Gamma(\frac{13}{6})=\frac{7}{6}\Gamma(\frac{7}{6}) =\frac{7}{36}\Gamma(\frac{1}{6})$
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.