伽马函数（欧拉第二积分/Gamma Function）详解

一、前言

在本文中，「荒原之梦考研数学」将为同学们详细讲解考研高等数学以及概率论和数理统计课程中常用的伽马函数。

二、正文

什么是伽马函数

伽马函数的标准定义式：

$$
\textcolor{springgreen}{
\Gamma(\alpha) = \int_{0}^{ +\infty } x^{\alpha-1} \mathrm{e}^{-x} \mathrm{~d} x
} \tag{1}
$$

若令 $x = t^{2}$, 则根据上面伽马函数的标准定义式，我们可以得到伽马函数的第二种定义式：

$$
\textcolor{orange}{
\Gamma(\alpha) = 2 \int_{0}^{+ \infty} t^{2 \alpha – 1} \mathrm{e}^{-t^{2}} \mathrm{~d} t
} \tag{2}
$$

伽马函数的计算方法/性质

$$
\begin{aligned}
(3) \quad & \alpha > 0 \Rightarrow \Gamma(\alpha+1) = \alpha \Gamma(\alpha) \\ \\
(4) \quad & \alpha < 1 \Rightarrow \Gamma (\alpha) \Gamma( 1-\alpha ) = \frac{\pi}{\sin \alpha \pi} \\ \\
(5) \quad & n \in N^{+} \Rightarrow \begin{cases}
\Gamma(n + 1 ) = n! \\
\Gamma(n) = (n-1)!
\end{cases}
\end{aligned}
$$

关于上面 $(4)$ 式的证明：

$$
\begin{aligned}
\Gamma(\alpha+1) & = \int_0^{+\infty} x^\alpha e^{-x} \mathrm{~d} x \\ \\
& = – \int_0^{+\infty} x^\alpha \mathrm{~d} \left(e^{-x}\right) \\ \\
& = -\left.e^{-x} x^\alpha\right|_0 ^{+\infty} + \alpha \textcolor{blue}{ \int_0^{+\infty} e^{-x} \cdot x^{\alpha-1} \mathrm{~d} x } \\ \\
& =\alpha \textcolor{blue}{ \Gamma(\alpha) }
\end{aligned}
$$

zhaokaifeng.com

关于上面 $(5)$ 式的证明（根据 $(4)$ 式的结论）：

$$
\begin{aligned}
\Gamma(n) \\
& = (n-1) \Gamma(n-1) \\
& = \cdots \\
& = (n-1) \cdot (n-2) \cdots 2 \cdot \textcolor{pink}{ \Gamma(1) } \\
& = (n-1) \cdot (n-2) \cdots 2 \cdot \textcolor{pink}{ 1 } \\
& = ( n-1 )!
\end{aligned}
$$

zhaokaifeng.com

伽马函数的特殊值

$$
\begin{aligned}
\textcolor{springgreen}{\Gamma(1)} = & \int_{0}^{+ \infty} \mathrm{e}^{-x} \mathrm{~d} x = \textcolor{springgreen}{1} \\ \\
\textcolor{springgreen}{\Gamma(2)} = & \int_{0}^{ + \infty } x \mathrm{e}^{-x} \mathrm{~d} x = \textcolor{springgreen}{1} \\ \\
\textcolor{springgreen}{\Gamma \left( \frac{1}{2} \right)} = & \int_{0}^{ + \infty } x^{-\frac{1}{2}} \mathrm{e}^{-x} \mathrm{~d} x = \textcolor{springgreen}{\sqrt{\pi}} \\ \\
\textcolor{springgreen}{\Gamma \left( \frac{3}{2} \right)} = & \int_{0}^{+ \infty} x^{\frac{1}{2}} \mathrm{e}^{-x} \mathrm{~d} x = \textcolor{springgreen}{ \frac{\sqrt{\pi}}{2} }
\end{aligned}
$$

例题

$$
\begin{aligned}
(Ⅰ) \quad & \int_{0}^{ +\infty } \textcolor{springgreen}{x}^{\textcolor{orange}{5}} \mathrm{e}^{ -\textcolor{springgreen}{x} } \mathrm{ ~d} \textcolor{springgreen}{x} = \Gamma(\textcolor{orange}{5} + 1) \\ \\
(Ⅱ) \quad & \int_{0}^{ + \infty} \sqrt{x} \mathrm{e}^{- x} \mathrm {~d} x = \int_{0}^{ + \infty} \textcolor{springgreen}{x}^{\textcolor{orange}{\frac{1}{2}}} \mathrm{e}^{- \textcolor{springgreen}{x}} \mathrm {~d} \textcolor{springgreen}{x} = \Gamma \left( \textcolor{orange}{\frac{1}{2}} + 1 \right) \\ \\
(Ⅲ) \quad & \int_{0}^{ + \infty } x^{3} \mathrm{e}^{-2x} \mathrm {~d} x \\
& = \frac{1}{16} \int_{0}^{+ \infty}(\textcolor{springgreen}{2x})^{\textcolor{orange}{3}} \mathrm{e}^{- \textcolor{springgreen}{2x} } \mathrm{~d}(\textcolor{springgreen}{2x}) \\
& = \frac{1}{16} \Gamma(\textcolor{orange}{3} + 1) \\
& = \frac{3!}{16} = \frac{3}{8} \\ \\
(Ⅳ) \quad & \int_{0}^{ + \infty} x^{4} \mathrm{e}^{−x^{2}} \mathrm{~d} x \\
& = \frac{1}{2} \int_{0}^{ + \infty} \left( \textcolor{springgreen}{x^{2}} \right)^{\textcolor{orange}{\frac{3}{2}}} \mathrm{e}^{− \textcolor{springgreen}{x^{2}}} \mathrm{~d} (\textcolor{springgreen}{x^{2}}) \\
& = \frac{1}{2} \Gamma \left(\textcolor{orange}{\frac{3}{2}} + 1 \right) \\
& = \frac{1}{2} \textcolor{orange}{\frac{3}{2}} \Gamma \left( \textcolor{magenta}{\frac{1}{2}} + 1 \right) \\
& = \frac{1}{2} \textcolor{orange}{\frac{3}{2}} \textcolor{magenta}{\frac{1}{2}} \textcolor{tan}{ \Gamma \left( \frac{1}{2} \right) } \\
& = \frac{1}{2} \frac{3}{2} \frac{1}{2} \textcolor{tan}{ \sqrt{\pi} } \\
& = \frac{3}{8} \sqrt{\pi}
\end{aligned}
$$

---