The nth Catalan Number Expressed as a Beta Function

John Cook who blogs here, posted the following expression for the nth Catalan number on one of his twitter accounts

C_{n} = \frac{1}{2\pi} \int\limits_{0}^{4} x^{n} \sqrt{\frac{4-x}{x}} \mathrm{d} x

Let us express this as a Beta function
\int\limits_{0}^{4} x^{n} \sqrt{\frac{4-x}{x}} \mathrm{d} x & = x^{n-\frac{1}{2}} (4-x)^{\frac{1}{2}} \mathrm{d} x \\
& = 4^{n+1} \int\limits_{0}^{1} y^{n-\frac{1}{2}} (1-y)^{\frac{1}{2}} \mathrm{d} y \\
& = 4^{n+1} \, \mathrm{B}\left(n + \frac{1}{2}, \frac{3}{2} \right)
We used the substitution \(y=\frac{x}{4}\).

Now we have
C_{n} = \frac{1}{2\pi} \int\limits_{0}^{4} x^{n} \sqrt{\frac{4-x}{x}} \mathrm{d} x = \frac{2^{2n+1}}{\pi} \mathrm{B}\left(n + \frac{1}{2}, \frac{3}{2} \right)

We will check this by using the following definition of the nth Catalan number
C_{n} = \frac{(2n)!}{(n+1)! \, n!}

Using the following Gamma function expressions
\Gamma(n+1) = n\Gamma(n), \quad
\Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}, \quad
\Gamma\left(n + \frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{4^{n}n!}

We have
\frac{2^{2n+1}}{\pi} \mathrm{B}\left(n + \frac{1}{2}, \frac{3}{2} \right) & = \frac{2^{2n+1}}{\pi} \frac{\Gamma\left(n + \frac{1}{2}\right) \Gamma\left(\frac{3}{2}\right)}{\Gamma(n+2)} \\
& = \frac{2^{2n+1}}{\pi} \frac{(2n)!\sqrt{\pi}}{2^{2n}n!} \frac{\sqrt{\pi}}{2} \frac{1}{(n+1)!} \\
& = \frac{(2n)!}{(n+1)! \, n!}
thus equation \eqref{eq:1608123} is correct.