I'm watching this video, where D. Knuth explains the connection of $\pi$ and factorials, and other matters (it is very interesting). Almost at the end of the talk he says the area of the superellipse
$x^{\frac{1}{\alpha}}+y^{\frac{1}{\alpha}}=1$
is given by
$A(\alpha) = \frac{2 \alpha \cdot\Gamma{(\alpha)}^2}{\Gamma{(2 \alpha)}}$ whic would be
$A(\alpha) = 2 \alpha B(\alpha,\alpha) = 2 \alpha\int_0^1(1-u)^{\alpha-1}u^{\alpha-1}du $
I was trying to check this so I put
$A\left( \alpha \right) = \int\limits_0^1 {{{\left( {1 - {x^{1/\alpha }}} \right)}^\alpha }dx} $
Now let $x = {u^\alpha }$
$A\left( \alpha \right) = \alpha \int\limits_0^1 {{{\left( {1 - u} \right)}^\alpha }{u^{\alpha - 1}}du} $
What's going on?
The $2$ in Knuth's formula probably comes from the fact he considers the full figure and not only a fourth, as I am, but I don't know what I'm doing wrong here. If you want to check, it is at $1:21:00$ aproximately.
PS: Just as a curiosity, does Knuth have a stutter or is it he is just thinking about too many things in too little time?
So it was just OK:
$A\left( \alpha \right) = \alpha \int\limits_0^1 {{{\left( {1 - u} \right)}^\alpha }{u^{\alpha - 1}}du} = \frac{{\alpha \Gamma \left( {\alpha + 1} \right)\Gamma \left( \alpha \right)}}{{\Gamma \left( {2\alpha + 1} \right)}} = \frac{{\Gamma {{\left( {\alpha + 1} \right)}^2}}}{{\Gamma \left( {2\alpha + 1} \right)}}$