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Some greek gymnastics

Beta Function

The definition of the beta (B) function is:

B(\alpha, \beta) =\int_0^1 t^{\alpha -1}(1-t)^{\beta -1}\, dt

Gamma Function

The definition of the gamma (\Gamma) function is:

\Gamma (z)= \int_0^\infty e^{-t} t^{z-1}\, dt

Integration by parts yields,

\Gamma (z)= [- e ^{-t} t^{z-1}]_0^\infty - (z-1)\int_0^\infty (-e^{-t}) t^{z-2}\,dt = (z-1) \Gamma (z-1)

So for n \in \mathrm{N} and noting \Gamma (1) =\int_0^\infty e^{-t}\, dt =1,

\Gamma (n) = (n-1) \Gamma (n-1)= (n-1) (n-2) \ldots 1\cdot \Gamma (1)= (n-1)!

Relating Beta and Gamma Functions

Consider,

\Gamma (m+1) \Gamma (n+1) =\int_0^\infty e^{-u} u^{m}\, du \int_0^\infty e^{-v} v^{n}\, dv

Now changing variables: u=x^2, du =2 x\, dx  and v= y^2 , dv = 2 y \,dy and rearranging:

\Gamma (m+1)\Gamma (n+1) =4 \int_0^\infty \int_0^\infty e^{-(x^2+y^2)} x^{2m+1}y^{2n+1}\,dx\,dy

Changing to polar coordinates: x = r \cos \theta and y = r \sin \theta and noting the region is the upper quadrant :

\Gamma(m+1) \Gamma (n+1) = 4\int_0^\infty\int_0^{\pi/2} e^{-r^2} (r^2)^{m+n+1} r\cos ^{2m+1} \theta \sin ^{2n+1} \theta \, dr \, d\theta

Changing q =r^2, dq = 2 r\, dr:

\Gamma (m+1) \Gamma (n+1) =2 \int_0^\infty e^{-q} q^{m+n+1} \, dq \int_0^{\pi/2} \cos ^{2m+1} \theta \sin ^{2n+1} \theta \, d\theta

\Gamma (m+1) \Gamma (n+1) =2 \Gamma ( m+n+2) \int_0^{\pi/2} \cos ^{2m +1} \theta \sin ^{2n +1} \theta\, d\theta

Changing t= \cos ^2 \theta, dt = -2 \sin\theta \cos \theta \, d\theta and noting \cos ^2 \pi/2 =0, \cos ^2 0 = 1:

\Gamma (m+1) \Gamma (n+1) = \Gamma (m+n+2) \int_0^1 t^{m}(1-t)^{n}\,dt

\Gamma (m+1) \Gamma (n+1) = \Gamma (m+ n+ 2) B (m+1, n+1)

Therefore,

B (m+1, n+1) = \frac{\Gamma (m+1) \Gamma (n+1)}{\Gamma (m+n+2)}

or equivalently:

B(\alpha , \beta) =\frac {\Gamma (\alpha ) \Gamma (\beta )}{\Gamma (\alpha + \beta )}

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