📂Functions

Various Important Formulas Involving the Gamma Function and Factorials

Formulas

$$\Gamma (\frac{1}{2})=\sqrt{\pi} \tag{a}$$

- Euler’s Reflection Formula: $$\Gamma (p)\Gamma (1-p)=\dfrac{\pi}{\sin(\pi p)} \tag{b}$$

$$\Gamma (n+\frac{1}{2})=\frac{1\cdot 3\cdot \cdot5 \cdots (2n-1)}{2^{n}}\sqrt{\pi}=\frac{(2n-1)!!}{2^n}\sqrt{\pi}=\frac{(2n)!}{4^{n}n!}\sqrt{\pi},\quad n\in \mathbb{N} \tag{c}$$ $!!$ is the Double Factorial.

- Binomial Coefficient: $$\begin{pmatrix} n \\ k \end{pmatrix}=\frac{\Gamma (n+1)}{k! \Gamma (n-k+1)} \tag{d}$$

- Euler-Mascheroni Constant: $$\gamma=-\Gamma^{\prime} (1) \tag{e}$$

- Beta Function:

$$B(p,q)=\frac{\Gamma (p) \Gamma (q)}{\Gamma (p+q)} \tag{f}$$

Proofs

Gamma Function: $$\Gamma (p)=\begin{cases} \displaystyle \int_{0}^\infty x^{p-1}e^{-x}dx & p>0 \\ \frac{1}{p}\Gamma (p+1)& p<0 \end{cases}$$

Recursive Formula of Gamma Function: $$\Gamma (p+1)=p\Gamma (p)$$

$(a)$

By substituting from $(b)$ to $p=\frac{1}{2}$, we can obtain $(a)$, but let’s derive it directly. By definition of the gamma function,

$$\Gamma ({\textstyle \frac{1}{2}})=\int_{0}^{\infty}\frac{1}{\sqrt{x}}e^{-x}dx$$

Substituting with $x=y^{2}$ in the above equation gives $dx=2ydy$, hence

$$\Gamma ({\textstyle\frac{1}{2}}) = \int_{0}^{\infty}\frac{1}{y}e^{-y^{2}}2ydy = 2\int_{0}^{\infty}e^{-y^{2}}dy = \int_{-\infty}^{\infty}e^{-y^{2}}dy$$

The right side is the Gaussian Integral, so

$$\textstyle \Gamma (\frac{1}{2})=\sqrt{\pi}$$

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$(b)$

Not easy. Uses Weierstrass’s infinite product and Euler’s representation of the sinc function.

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$(c)$

By the recursion relation of the gamma function and $(a)$,

$$\begin{align*} \textstyle \Gamma (1+\frac{1}{2}) &= \textstyle \frac{1}{2}\Gamma (\frac{1}{2})=\frac{1}{2}\sqrt{\pi} \\[1em] \textstyle \Gamma (2+\frac{1}{2}) &= \textstyle \frac{3}{2}\Gamma (1+\frac{1}{2})=\frac{3\cdot 1}{2\cdot2}\sqrt{\pi} \\[1em] \textstyle \Gamma (3+\frac{1}{2}) &= \textstyle \frac{5}{2}\Gamma (2+\frac{1}{2})=\frac{5\cdot 3\cdot 1}{2\cdot2 \cdot 2}\sqrt{\pi} \\[1em] \textstyle \Gamma (4+\frac{1}{2}) &= \textstyle \frac{7}{2}\Gamma (3+\frac{1}{2})=\frac{7\cdot 5\cdot 3\cdot 1}{2\cdot 2\cdot2 \cdot 2}\sqrt{\pi} \\ \vdots \\ \textstyle \Gamma (n+\frac{1}{2}) &= \textstyle \frac{(2n-1)(2n-3)\cdots 3\cdot 1}{2^n}\sqrt{\pi} \end{align*}$$

Where

$$\begin{align*} \frac{(2n-1)(2n-3)\cdots 3\cdot 1}{2^n} &=\frac{{\color{blue}2n}(2n-1){\color{blue}(2n-2)}(2n-3){\color{blue}(2n-4)}\cdots {\color{blue}4}\cdot 3\cdot {\color{blue}2}\cdot 1}{2^n {\color{blue}2n(2n-2)(2n-4)\cdots 4\cdot 2}} \\ &=\frac{(2n)!}{2^n 2^n n(n-1)(n-2)\cdots 2\cdot 1} \\ &=\frac{(2n)!}{4^n (n)!} \end{align*}$$

Therefore,

$$\Gamma (n+{\textstyle\frac{1}{2}})=\frac{1\cdot 3\cdot \cdot5 \cdots (2n-1)}{2^{n}}\sqrt{\pi}=\frac{(2n-1)!!}{2^n}\sqrt{\pi}=\frac{(2n)!}{4^{n}n!}\sqrt{\pi}$$

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$(d)$

$$\begin{pmatrix} n \\ k \end{pmatrix} = \frac{ n! }{ k!(n-k)! }=\frac{ \Gamma (n+1) }{ k! \Gamma (n-k+1)! }$$

Using $\Gamma (n+1)=n!$ concludes the proof in one line.

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$(e)$

Utilizes the derivatives of the gamma function and the reciprocals.

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$(f)$

Can be shown as a generalization of the binomial coefficient.

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