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:
$$A\left( \alpha \right) = \alpha \int\limits_0^1 {{{\left( {1 - u} \right)}^\alpha }{u^{\alpha - 1}}du} = \frac{{\alpha \Gamma \left( {\alpha + 1} \right)\Gamma \left( \alpha \right)}}{{\Gamma \left( {2\alpha + 1} \right)}} = \frac{{\Gamma {{\left( {\alpha + 1} \right)}^2}}}{{\Gamma \left( {2\alpha + 1} \right)}}$$
I wouldn't call it a stutter. It strikes me as closer to someone very famous trying to seem folksy; not quite the same as saying "um," more "I'm as uncertain as you."
Dirichlet wrote a generalization of this in Über eine neue Methode zur Bestimmung vielfacher Integrale. Collected Works, volume 1, page 389.
For integrating the constant 1 on $$ x \geq 0, \; y \geq 0, \; x^{1/\alpha} + y^{1/\alpha} \leq 1, $$ the result is $$ \frac{\alpha^2 \Gamma(\alpha)^2}{\Gamma(1 + 2 \alpha)} = \frac{ \Gamma(1 + \alpha)^2}{\Gamma(1 + 2 \alpha)}, $$ multiply by 4 to get the whole thing, $$ \frac{ 4 \; \Gamma(1 + \alpha)^2}{\Gamma(1 + 2 \alpha)}. $$ Sample points, $\alpha = 1/2$ is the circle, $\Gamma(3/2) = (1/2) \sqrt \pi $ and $\Gamma(2) = 1,$ so we get $\pi.$ With $\alpha = 1,$ we have a tilted square, $\Gamma(3) = 2,$ area is indeed $2.$ As $\alpha \rightarrow \infty,$ area goes to $0,$ it is not necessary to quote Stirling's to believe that $(\alpha!)^2 / (2 \alpha)! \rightarrow 0.$ Finally, as $\alpha \rightarrow 0,$ we approach the entire square, and the area approaches $4.$
I understand part of the Dirichlet technique is in Whittaker and Watson.