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# Beta function

2024/01/02に公開

The definition of the beta function is

\begin{align*} \Beta(p, q) = \int_{0}^{1} x^{p-1} (1-x)^{q-1} dx \end{align*}

where p>0, q>0.

A characteristics of the beta function is

\begin{align*} \Beta(p, q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)} \end{align*}

Let's prove this characteristics.

\begin{align*} \Gamma(p) \Gamma(q) &= \int_{0}^{\infty} x^{p-1} e^{-x} dx \int_{0}^{\infty} y^{q-1} e^{-y} dy \\ &=\int_{0}^{\infty} \int_{0}^{\infty} x^{p-1} y^{q-1} e^{-(x+y)} dx dy \\ \end{align*}

Let x = u^2, y=v^2 , dx = 2u du , dy = 2v dv , u: 0 \rightarrow \infty , v: 0 \rightarrow \infty.

\begin{align*} &= \int_{0}^{\infty} \int_{0}^{\infty} u^{2(p-1)} v^{2(q-1)} e^{-(u^2+v^2)} 2u du 2v dv \\ &= 4 \int_{0}^{\infty} \int_{0}^{\infty} u^{2p-1} v^{2q-1} e^{-(u^2+v^2)} du dv \\ \end{align*}

Let u = r \cos(\theta) , v = r \sin(\theta) , r: 0 \rightarrow \infty , \theta: 0 \rightarrow 2 \pi.

The Jacorbian is

\begin{align*} \begin{vmatrix} du/dr & du/d\theta \\ dv/dr & dv/d\theta \\ \end{vmatrix} &= \begin{vmatrix} \cos(\theta) & -r \sin(\theta) \\ \sin(\theta) & r \cos(\theta) \\ \end{vmatrix} \\ &= r \cos^2(\theta) + r \sin^2(\theta) \\ &= r \end{align*}
\begin{align*} &= 4 \int_{0}^{\infty} \int_{0}^{2 \pi} (r \cos(\theta))^{2p-1} (r \sin(\theta))^{2q-1} e^{-((r \cos(\theta))^2+(r \sin(\theta))^2)} r dr d\theta \\ &= 4 \int_{0}^{\infty} \int_{0}^{2 \pi} r^{2(p+q)-1} \cos^{2p-1}(\theta) \sin^{2q-1}(\theta) e^{-r^2} dr d\theta \\ &= 4 \int_{0}^{\infty} r^{2(p+q)-1} e^{-r^2} dr \int_{0}^{2 \pi} \cos^{2p-1}(\theta) \sin^{2q-1}(\theta) d\theta \\ \end{align*}

Let t = r^2 , x = \cos^2(\theta) , t: 0 \rightarrow \infty , x: 1 \rightarrow 0, dt = 2 r dr , dx = - 2 \cos(\theta) \sin(\theta) d\theta ,

\begin{align*} &= - \int_{0}^{\infty} r^{2(p+q)-2} e^{-r^2} 2 r dr \int_{0}^{2 \pi} \cos^{2p-2}(\theta) \sin^{2q-2}(\theta) - 2 \cos(\theta) \sin(\theta) d\theta \\ &= - \int_{0}^{\infty} t^{(p+q)-1} e^{-t} dt \int_{1}^{0} x^{p-1} (1-x)^{q-1} dx \\ &= \int_{0}^{\infty} t^{(p+q)-1} e^{-t} dt \int_{0}^{1} x^{p-1} (1-x)^{q-1} dx \\ &= \Gamma(p+q) \Beta(p, q) \end{align*}

Here, we have gotten,

\begin{align*} \Gamma(p) \Gamma(q) = \Gamma(p+q) \Beta(p, q) \\ \iff \Beta(p, q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)} \end{align*}