Beta function

The definition of the beta function is

$$B(p,q)=∫01xp−1(1−x)q−1dx\begin{align*} \Beta(p, q) = \int_{0}^{1} x^{p-1} (1-x)^{q-1} dx \end{align*}$$

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

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A characteristics of the beta function is

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

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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,

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