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Beta function

2024/01/02に公開

The definition of the beta function is

B(p,q)=01xp1(1x)q1dx\begin{align*} \Beta(p, q) = \int_{0}^{1} x^{p-1} (1-x)^{q-1} dx \end{align*}

where p>0,q>0p>0, q>0.


A characteristics of the beta function is

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

Let's prove this characteristics.

Γ(p)Γ(q)=0xp1exdx0yq1eydy=00xp1yq1e(x+y)dxdy\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=u2x = u^2, y=v2y=v^2 , dx=2ududx = 2u du , dy=2vdvdy = 2v dv , u:0u: 0 \rightarrow \infty , v:0v: 0 \rightarrow \infty.

=00u2(p1)v2(q1)e(u2+v2)2udu2vdv=400u2p1v2q1e(u2+v2)dudv\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=rcos(θ)u = r \cos(\theta) , v=rsin(θ)v = r \sin(\theta) , r:0r: 0 \rightarrow \infty , θ:02π\theta: 0 \rightarrow 2 \pi.

The Jacorbian is

du/drdu/dθdv/drdv/dθ=cos(θ)rsin(θ)sin(θ)rcos(θ)=rcos2(θ)+rsin2(θ)=r\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*}
=4002π(rcos(θ))2p1(rsin(θ))2q1e((rcos(θ))2+(rsin(θ))2)rdrdθ=4002πr2(p+q)1cos2p1(θ)sin2q1(θ)er2drdθ=40r2(p+q)1er2dr02πcos2p1(θ)sin2q1(θ)dθ\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=r2t = r^2 , x=cos2(θ)x = \cos^2(\theta) , t:0t: 0 \rightarrow \infty , x:10x: 1 \rightarrow 0, dt=2rdrdt = 2 r dr , dx=2cos(θ)sin(θ)dθdx = - 2 \cos(\theta) \sin(\theta) d\theta ,

=0r2(p+q)2er22rdr02πcos2p2(θ)sin2q2(θ)2cos(θ)sin(θ)dθ=0t(p+q)1etdt10xp1(1x)q1dx=0t(p+q)1etdt01xp1(1x)q1dx=Γ(p+q)B(p,q)\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,

Γ(p)Γ(q)=Γ(p+q)B(p,q)    B(p,q)=Γ(p)Γ(q)Γ(p+q)\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*}

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