\text{Gamma and Beta functions}

\begin{aligned} & \Gamma(z)&=&\lim _{n \rightarrow \infty} \frac{1 \cdot 2 \cdot 3 \cdots n}{z(z+1)(z+2) \cdots(z+n)} n^z \quad, z \neq 0,-1,-2, \ldots \\ &\Gamma(z+1)&=&\lim _{n \rightarrow \infty} \frac{1 \cdot 2 \cdot 3 \cdots n}{(z+1)(z+2)(z+3) \cdots(z+n+1)} n^{z+1} \\ & &=&\lim _{n \rightarrow \infty} \frac{n z}{z+n+1} \cdot \frac{1 \cdot 2 \cdot 3 \cdots n}{z(z+1) \cdots(z+n)} n^z=z \Gamma(z) \\ & \end{aligned}

\displaystyle \Gamma(1)=\lim _{n \rightarrow \infty} \frac{1 \cdot 2 \cdot 3 \cdots n}{1 \cdot 2 \cdot 3 \cdots n(n+1)} n=1

\begin{aligned} & \Gamma(2)=1 \cdot \Gamma(1)=1 \\ & \Gamma(3)=2 \cdot \Gamma(2)=1 \cdot 2, \Gamma(4)=3 \Gamma(3)=1 \cdot 2 \cdot 3 \end{aligned}

\text{and in general }

\Gamma(n)=1 \cdot 2 \cdot 3 \cdots(n-1)=(n-1)!

\text{We will use three different methods to define the properties of the Gamma function}

1) \text{ Infinite limit (Euler)}

\Gamma (z+1)=z\Gamma(z)

\Gamma (n+1)=n!

2)\text{ Definite integrals (Euler)}

\displaystyle \Gamma(z) \equiv \int^{\infty}_{0} e^{-t} t^{z-1} d t,\hspace{1 cm} \text{Re}(z)>0

\text {a) transform using } t=u^2, d t=2 u d u

\displaystyle \Gamma(z)=\int_0^{\infty} e^{-u^2} u^{2(z-1)} \cdot 2 u d u=2 \int^{\infty} e^{-u^2} u^{2 z-1} d u

\text{b) transform using }\displaystyle t=-\ln u, d t=-\frac{d u}{u}, t=0 \rightarrow u=1\text{ and } t=\infty \rightarrow u=0

\displaystyle \Gamma(z)=\int_1^0 e^{\ln u}\left[\ln \left(\frac{1}{u}\right)\right]^{z-1}\left(-\frac{d u}{u}\right)=\int_0^1\left[\ln \left(\frac{1}{u}\right)\right]^{z-1} d u

\text{Let us consider (a) with }z=\frac{1}{2}

\displaystyle \Gamma \Big(\frac{1}{2}\big)=2 \int_0^{\infty} e^{-u^2} d u=2 \cdot \frac{\sqrt{\pi}}{2}=\sqrt{\pi}

\displaystyle \Gamma(z) \equiv 2\int^{\infty}_{0} e^{-u^2} u^{2z-1} d u,\hspace{1 cm} \text{Re}(z)>0

\displaystyle \Gamma(z) \equiv \int^{1}_{0} \left[\ln \Big(\frac{1}{u}\Big)\right]^{z-1} d u,\hspace{1 cm} \text{Re}(z)>0

\Gamma(\frac{9}{2}), \Gamma(\frac{7}{2}),\Gamma(\frac{5}{2}),\Gamma(-\frac{1}{2}), \Gamma(-\frac{5}{2})

\text{Calculate:}

\text{The definite integrals and the infinite limit definitions are equivalent}

\text { Write } e^{-t} \text { as the limit } \displaystyle \lim _{n \rightarrow \infty}\left(1-\frac{t}{n}\right)^n=e^{-t}, \text{ and consider the function}

\displaystyle F(z, n)=\int_0^n\left(1-\frac{t}{n}\right)^n t^{z-1} d t, \operatorname{Re}(z)>0

\text { Taking the limit } n \rightarrow \infty \text { gives } \Gamma(z) \text { through }

\displaystyle \lim _{n \rightarrow \infty} F(z, n)=F(z, \infty)=\int_0^{\infty} e^{-t} t^{z-1} d t \equiv \Gamma(z)

\text{Substitute }u=\frac{t}{n}\text{ in }F(z, n)\text{ with }t=n u, d t=n d u, t=0 \rightarrow u=0, t=n \rightarrow u=1

\displaystyle F(z, n)=\int_0^1(1-u)^n(n u)^{z-1} \cdot n d u=n^z \int_0^1(1-u)^n u^{z-1} d u

\text { Integrate by parts: } \displaystyle \frac{F(z, n)}{n^z}=\left[(1-u)^n \frac{u^z}{z}\right]_0^1+\frac{n}{z} \int_0(1-u)^{n-1} u^z d u=

\displaystyle =\cdots=\frac{n(n-1) \cdots \cdot 1}{z(z+1) \cdots(z+n-1)} \int_0^z u^{z+n-1} d u

\displaystyle \Rightarrow F(z, n)=\frac{1 \cdot 2 \cdot 3 \cdots n}{z(z+1) \cdots(z+n-1)(z+n)} n^z \quad \text { and } \lim _{n \rightarrow \infty} F(z, n)=F(z, \infty) \equiv \Gamma(z)

\text { 3) Infinite product (Weierstrass) }

\displaystyle \frac{1}{\Gamma(z)} \equiv z e^{\gamma z} \prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right) e^{-z / n}

\gamma=0.5772\dots \text{is the Euler constant}

\text { Euler's reflection formula: }\\

\begin{aligned} \Gamma(z+1) \Gamma(1-z) & =\int_0^{\infty} s^z e^{-s} d s \int_0^{\infty} t^{-z} e^{-t} d t \\ & =\int_0^{\infty} v^z \frac{d v}{(v+1)^2} \int_0^{\infty} e^{-u} u d u=\frac{\pi z}{\sin \pi z} \end{aligned}

\text {We can start from the product of Euler integrals, }

\text{transforming from the variables $s, t$ to $u=s+t, v=s / t$, as suggested by combining the}\\ \text{exponentials and the powers in the integrands. The Jacobian is}

J=-\left|\begin{array}{cc} 1 & 1 \\ \frac{1}{t} & -\frac{s}{t^2} \end{array}\right|=\frac{s+t}{t^2}=\frac{(v+1)^2}{u},

\text{where $(v+1) t=u$. The integral $\int_0^{\infty} e^{-u} u d u=1$, while that over $v$ may }\\ \text{be derived by contour integration, giving $\frac{\pi z}{\sin \pi z}$.}

\text{We can apply this formula for }z=\frac{1}{2}

\displaystyle \Gamma(z) \Gamma(1-z)=\frac{\pi}{\sin \pi z} \Rightarrow \Gamma^2\Bigg(\frac{1}{2}\Bigg)=\pi\Rightarrow\Gamma\Bigg(\frac{1}{2}\Bigg)=\sqrt{\pi}

\text {Legendre's duplication formula:}

\text{Similarly we can establish}

\text{Srtirling formula:}

\displaystyle n ! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n

\text{The factorial of a number can be approximated for large }n\text{ using the Stirling formula}

\text{The }\Gamma \text{ function for real values of } z.\\ z=-1,-2,\dots \text{ are poles of the function}

\text{Poles of the Gamma function:}

\text{From the Weistrass formula, we see that the poles of the }\Gamma \text{ function are }z=-n, n=1,2,\dots

\displaystyle \frac{1}{\Gamma(z)} \equiv z e^{\gamma z} \prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right) e^{-z / n}

\text{Notations}

\begin{aligned} 1 \cdot 3 \cdot 5 \cdots(2 n+1) & =(2 n+1)!! \\ 2 \cdot 4 \cdot 6 \cdots(2 n) & =(2 n)!!. \end{aligned} \\ \\

\text{We also define $(-1)!!=1$, a special case.}

(2 n)!!=2^n n!\quad \text { and } \quad(2 n+1)!!=\frac{(2 n+1)!}{2^n n!}

\text{Clearly, these are related to the regular factorial functions by}\\

\text{In a Maxwellian distribution the fraction of particles with speed between } v\text{ and }v+d v \text{ is}

\displaystyle \frac{d N}{N}=4 \pi\left(\frac{m}{2 \pi k T}\right)^{3 / 2} \exp \left(-\frac{m v^2}{2 k T}\right) v^2 d v

\displaystyle \left\langle v^n\right\rangle=\left(\frac{2 k T}{m}\right)^{n / 2} \frac{\Gamma\left(\frac{n+3}{2}\right)}{\Gamma(3 / 2)}

N \text { being the total number of particles. The average or expectation value of } v^n \text { is defined }

\text { as }\displaystyle \left\langle v^n\right\rangle=N^{-1} \int v^n d N \text {. Show that }

\text{Problem:}

\text{In a Maxwellian distribution the fraction of particles with speed between } v\text{ and }v+d v \text{ is}

\displaystyle \frac{d N}{N}=4 \pi\left(\frac{m}{2 \pi k T}\right)^{3 / 2} \exp \left(-\frac{m v^2}{2 k T}\right) v^2 d v

\displaystyle \left\langle v^n\right\rangle=\left(\frac{2 k T}{m}\right)^{n / 2} \frac{\Gamma\left(\frac{n+3}{2}\right)}{\Gamma(3 / 2)}

\text { as }\displaystyle \left\langle v^n\right\rangle=N^{-1} \int v^n d N \text {. Show that }

\text{Problem:}

\begin{aligned} d v & =\left(\frac{2 k T}{m}\right)^{\frac{1}{2}} \frac{d u}{2 u^{\frac{1}{2}}} \\ \left\langle v^n\right\rangle & =\int_0^{\infty} v^n \frac{d N}{N}=4 \pi\left(\frac{m}{2 k T}\right)^{3 / 2} \cdot \int_0^{+\infty} v^{2+n} e^{-\frac{m v^2}{2 k T}} d v \\ & =4 \pi\left(\frac{m}{ 2\pi k T}\right)^{\frac{3}{2}} \cdot\left(\frac{2 k T}{m}\right)^{\frac{n+2}{2}} \int_0^{+\infty} u^{\frac{n+2}{2}} e^{-u} \frac{d u}{2 u^{\frac{1}{2}}} \cdot\left(\frac{2 k T}{m}\right) \\ & =\frac{2}{\pi^{1 / 2}} \cdot\left(\frac{2 k T}{m}\right)^{\frac{n}{2}} \Gamma\left(\frac{n+3}{2}\right) \end{aligned}

\displaystyle \text{let us put }u=\frac{m v^2}{2 k T} \Rightarrow v=\sqrt{\frac{2 k T}{m}} u^{\frac{1}{2}}

\displaystyle v=\left(\frac{2 k T}{m}\right)^{\frac{1}{2}} u^{\frac{1}{2}}, v^n=\left(\frac{2 k T}{m}\right)^{\frac{n}{2}} \cdot u^{\frac{n}{2}}

\displaystyle \Gamma\Big(\frac{3}{2}\Big)=\frac{\sqrt{\pi}}{2}

\text{Beta function}

\begin{array}{ll} B(x, y)=\displaystyle \int_0^1 t^{x-1}(1-t)^{y-1} d t & \text { for } x, y>0 \\ B(x, y)=2 \displaystyle \int_0^{\pi / 2}(\sin \theta)^{2 x-1}(\cos \theta)^{2 y-1} d \theta \quad\left(\text { Put } t=\sin ^2 \theta\right) \\ B(x, y)=\displaystyle \int_0^{\infty} \frac{u^{x+1}}{(1+u)^{x+y}} d u \quad\left(\text { put } t=\frac{1}{1+u}\right) \end{array}

\text{The Beta function can be defined as follows:}

\text{Show that:}

B(m, n)=B(n, m)

\displaystyle B(m, n)=\int_0^1 x^{m-1}(1-x)^{n-1} d x , m, n>0

t=1-x \Rightarrow d t=-d x

x \rightarrow 0, t \rightarrow 1 \text{ and } x \rightarrow 1, t \rightarrow 0

\text{Solution:}

\displaystyle B(m, n)=-\int_1^0 x^{m-1}(1-x)^{n-1} d x , m, n>0

\begin{aligned} B(m, n) & =-\int_1^0(1-t)^{m-1} t^{n-1} d t \\ & =\int_0^1(1-t)^{m-1} t^{n-1} d t \\ & =\int_0^1 t^{n-1}(1-t)^{m-1} d t \\ & =B(n, m) \end{aligned}

\text{Show that:}

\displaystyle B(m, n)=2 \int_0^{\pi / 2} \sin ^{2 m-1}(\theta) \cos ^{2 n-1}(\theta) d \theta=2 \int_0^{\pi / 2} \cos ^{2 m-1}(\theta) \sin ^{2 n-1}(\theta) d \theta

B(m, n)=\int_0^1 t^{m-1}(1-t)^{n-1} d t

\text{we have, }

\text{we put $t=\sin^2\theta$, so $dt=2 \sin \theta \cos\theta$}

B(m,n)=\int_0^{\pi/2} \sin(\theta)^{2m-2} \cos(\theta)^{2n-2} 2\sin\theta \cos \theta d\theta

\begin{aligned} B(m, n) & =-\int_1^0(1-t)^{m-1} t^{n-1} d t \\ & =\int_0^1(1-t)^{m-1} t^{n-1} d t \\ & =\int_0^1 t^{n-1}(1-t)^{m-1} d t \\ & =B(n, m) \end{aligned}

\text{Show that:}

\displaystyle B(m, n)=2 \int_0^{\pi / 2} \sin ^{2 m-1}(\theta) \cos ^{2 n-1}(\theta) d \theta=2 \int_0^{\pi / 2} \cos ^{2 m-1}(\theta) \sin ^{2 n-1}(\theta) d \theta

B(m, n)=\int_0^1 t^{m-1}(1-t)^{n-1} d t

\text{we have, }

\text{we put $t=\sin^2\theta$, so $dt=2 \sin \theta \cos\theta$}

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

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

\begin{gathered} B(m+1, n)=2 \int_0^{\pi / 2} \sin ^{2 m+2-1} \cos ^{2 n-1}(\theta) d \theta \\ =2 \int_0^{\pi / 2} \sin ^{2 m+1} \cos ^{2 n-1} d \theta \\ \end{gathered}

\sin ^{2 m+1}\theta=\sin ^{2 m-1}\theta \sin ^2 \theta \\

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

\text{so}, B(m, n)=B(m+1, n)+B(m, n+1)

\text{show that, }\int_0^1 \sqrt{x}(1-\sqrt{x})^3dx=\frac{1}{30}

\text{Prove that} \int_0^1 \frac{dx}{\sqrt{1-x^4}}=\frac{\sqrt{\pi}}{4} \frac{\Gamma{(1/4)}}{3/4}

\text{Prove that } \Gamma(n)\Gamma(n+\frac{1}{2})=2^{1-2n}\sqrt{\pi}\Gamma(2n)

\text{Prove that }\Gamma(\frac{1}{2})=\sqrt{\pi}

\text{Prove that }\Gamma(-\frac{1}{2})=-2\sqrt{\pi}

\text{Problems}

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

\Gamma(m)=\int_0^{\infty} x^{m-1} e^{-x} d x \text { and } \Gamma(n)=\int_0^{\infty} y^{n-1} e^{-y} d y\\

\text { Now, let us put } x=u v \text { and } y=u(1-v)\\

\Gamma(m) \Gamma(n)=\int_0^{\infty} \int_0^{\infty} e^{-(x+y)} x^{m-1} y^{n-1} d x d y

J=\frac{\partial(x, y)}{\partial(u, v)}=\left|\begin{array}{cc} v & u \\ 1-v & -u \end{array}\right|=-u

\Gamma(m) \Gamma(n)=\int_0^{\infty} \int_0^{1} e^{-u} u^{m+n-2} v^{m-1}(1-v)^{n-1} |J| d u d v

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

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