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The Gamma and Beta functions

  1. The Gamma function: \[\Gamma(s)=\int_{0}^{+\infty}t^{s-1}e^{-t}dt\quad \Bigl(=(s-1)! \quad s\in \mathbb Z^+\Bigr) (0<s<\infty)\] Because \[\begin{align} \Gamma(s+1)&=\int_{0}^{+\infty}t^{s}e^{-t}dt\\ &=-\int_{0}^{+\infty}t^{s}d(e^{-t})\\ &=-\Biggl[t^{s}e^{-t}|_{0}^{\infty}-\int_{0}^{+\infty}st^{s-1}e^{-t}dt\Biggl]\\ &=-\Biggl[0-s\Gamma(s)\Biggl]\\ &=s\Gamma(s) \end{align}\] and \[\Gamma(1)=\int_{0}^{+\infty}t^{1-1}e^{-t}dt=\int_{0}^{+\infty}e^{-t}dt=1\]

The product of two Gamma functions: \[\begin{align} \Gamma(x)\Gamma(y)&=\int_{0}^{+\infty}u^{x-1}e^{-u}du\int_{0}^{+\infty}v^{y-1}e^{-v}dv\\ &=\int_{u=0}^{+\infty}\int_{v=0}^{+\infty}e^{-(u+v)}u^{x-1}v^{y-1}dudv \quad (let\quad u+v=z; \quad u/z=t; \quad v/z=1-t; \quad dudv=zdtdz)\\ &=\int_{z=0}^{+\infty}\int_{t=0}^{t=1}e^{-z}(zt)^{x-1}(z(1-t))^{y-1}zdtdz\\ &=\int_{z=0}^{+\infty}e^{-z}z^{(x+y-1)}dz\int_{t=0}^{t=1}t^{(x-1)}(1-t)^{(y-1)}dt\\ &=\Gamma(x+y)\int_{t=0}^{t=1}t^{(x-1)}(1-t)^{(y-1)}dt \end{align}\]

We define this integral \(\int_{t=0}^{t=1}t^{(x-1)}(1-t)^{(y-1)}dt\) as \(B(x,y),\quad (x>0 ;\quad y>0)\), this is the Beta function. \(B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\) \(\Bigl(=\frac{(x-1)!(y-1)!}{(x+y-1)!}\quad x;y\in \mathbb Z^+ \Bigr)\) this is the complete Beta function. \(B(x;a,b)=\int_{t=0}^{t=x}t^{(a-1)}(1-t)^{(b-1)}dt\) is the incomplete Beta function. And \(I_x(a,b)=\frac{B(x;a,b)}{B(a,b)}\) is the Regularized incomplete Beta function.
\[\begin{align} B(x;a,b)&=\frac{x^a}{a} 2F_1(a,1-b;a+1;x)\\ &=\frac{x^a}{a}\sum_{n=0}^{+\infty}\frac{(a)_n(1-b)_n}{(a+1)_n}\frac{x^n}{n!} \end{align}\]

Let’s calculate the very important \(\Gamma(\frac{1}{2})\): \[\begin{align} \Gamma^2(\frac{1}{2})&=\Bigl(\int_{0}^{+\infty}t^{\frac{1}{2}-1}e^{-t}dt\Bigr)^2\\ &=\Bigl(\int_{0}^{+\infty}u^{-1}e^{-u^2}2udu\Bigr)^2\quad(t=u^2)\\ &=\Bigl(2\int_{0}^{+\infty}e^{-u^2}du\Bigr)^2\\ &=4\int_{0}^{+\infty}e^{-u^2}du\int_{0}^{+\infty}e^{-v^2}dv\\ &=4\int_{0}^{+\infty}\int_{0}^{+\infty}e^{-(u^2+v^2)}dudv\\ &=4\int_{0}^{\pi/2}\int_{0}^{+\infty}e^{-r^2}rdrd\theta\\ &=4\int_{0}^{\pi/2}d\theta\int_{0}^{+\infty}e^{-r^2}rdr\\ &=4\frac{\pi}{2}(-\frac{1}{2})e^{-r^2}\Bigl|_{0}^{+\infty}\\ &=\pi \end{align}\]

Then, \(\Gamma(\frac{1}{2})=\sqrt{\pi}\);
and \(\Gamma(\frac{3}{2})=(\frac{3}{2}-1)\Gamma(\frac{3}{2}-1)=\frac{1}{2}\sqrt{\pi}\)

The \(\Gamma(\frac{m}{2})\) is very important for chi square distribution:
\[\Gamma(\frac{m}{2})= \begin{cases} (\frac{m}{2})!, & \text{if $m$ is even} \\[2ex] (\frac{m}{2}-1)(\frac{m}{2}-2)...\frac{1}{2}\sqrt{\pi}, & \text{if $m$ is odd} \end{cases}\]

  1. If \(p>0\),\(q>0\) and \(\frac{1}{p}+\frac{1}{q}=\frac{p+q}{pq}=1\). Let \(u\ge 0\) and \(v\ge 0\), then \[uv\leq \frac{u^p}{p}+\frac{v^q}{q}=\frac{qu^p+pv^q}{pq}\] \[\Gamma(\frac{x}{p}+\frac{y}{q})\leq\Gamma(x)^{1/p}\Gamma(y)^{1/q}\] which means \(\log\Gamma\) is convex on \((0,\infty)\).

  2. If \(f\) is a positive function on \((0,\infty)\) such that (a) \(f(x+1)=xf(x)\), (b) \(f(1)=1\), (c) \(\log f\) is convex, then \(f(x)=\Gamma(x)\). Put \(\varphi=\log f\) and \(f=e^{\varphi}\). Because \(f(x+1)=xf(x)\) then \[e^{\varphi(x+1)}=xe^{\varphi(x)}\] and \[\varphi(x+1)=\log x+\varphi(x)\quad(0<x<\infty)\] \[\varphi(1)=\log f(1)=\log 1=0\] and \(\varphi\) is convex. Suppose \(0<x<1\) and \(n\) is a positive integer. \[\varphi(n+1)=\varphi(n)+\log n=\varphi(n-1)+\log n+\log (n-1)=\log(n!)\] Consider the difference quotients of \(\varphi\) on the intervals \([n,n+1], [n+1,n+1+x],[n+1,n+2]\), \[\varphi(n+1+x)-\varphi(n+1)=\log((n+x)!)-\log(n!)=\log\frac{(n+x)!}{n!}=\log(n+x)(n+x-1)\cdots(n+1)\\ =\underset{\text{x terms}}{\underbrace{\log(n+x)+\log(n+x-1)+\cdots+\log(n+1)}}\] Then \[\log n\leq\frac{\varphi(n+1+x)-\varphi(n+1)}{x}\leq\log(n+1)\] Because \[\varphi(x+1)=\varphi(x)+\log x\quad(0<x<\infty)\] \[\varphi(x+1+1)=\varphi(x+1)+\log (x+1)=\varphi(x)+\log x+\log (x+1)=\varphi(x)+\log [x(x+1)]\] \[\varphi(x+1+n)=\varphi(x)+\log [x(x+1)\cdots(x+n)]\] Thus \[\varphi(n+1+x)-\varphi(n+1)=\varphi(n+1+x)-\log(n!)\ge x\log n\] \[\varphi(n+1+x)=\varphi(x)+\log [x(x+1)\cdots(x+n)]\ge x\log n+\log(n!)\] \[\varphi(x)+\log [x(x+1)\cdots(x+n)]- x\log n-\log(n!)\ge0\] and \[\varphi(n+1+x)-\varphi(n+1)- x\log n=\varphi(x)+\log [x(x+1)\cdots(x+n)]-\log(n!)- x\log n\leq x\log(n+1)- x\log n=x\log(\frac{n+1}{n})\] Then \[0\leq\varphi(x)+\log [x(x+1)\cdots(x+n)]- x\log n-\log(n!)\le x\log(\frac{n+1}{n})\] The last equation tends to \(0\) as \(n\to \infty\). Hence \[\varphi(x)=x\log n-\log(n!)-\log [x(x+1)\cdots(x+n)]=\log\Biggl[\frac{n!n^x}{x(x+1)\cdots(x+n)}\Biggr]\] and \[f(x)=\Gamma(x)=\lim_{n\to\infty}\frac{n!n^x}{x(x+1)\cdots(x+n)}\] is determined.

  3. The Stirling’s formula: \[\lim_{x\to\infty}\frac{\Gamma(x+1)}{(\frac{x}{e})^x\sqrt{2\pi x}}=1\] \[\begin{align} \Gamma(x+1)&=\int_{0}^{+\infty}t^{x}e^{-t}dt\\ &=\int_{0}^{+\infty}(x(1+u))^{x}e^{-x(1+u)}d(x(1+u))\\ &=x^{x+1}e^{-x}\int_{-1}^{+\infty}(1+u)^{x}e^{-xu}du\\ &=x^{x+1}e^{-x}\int_{-1}^{+\infty}[(1+u)e^{-u}]^xdu,\\ &\{t=x(1+u), 0<x, -1<u<+\infty\} \end{align}\]