# 17 Gamma Function

The Gamma Function is a function used frequently in statistical theory. It has properties that permit simplified calculations and is used in defining many probability distributions, particularly within the Exponential family of distributions.

## 17.1 Definition

$\Gamma(x) = \int\limits_0^{\infty}t^{x-1}e^{-t}dt$

Note that the definition defines a function of $$x$$ that is integrated over $$t$$. Thus, for each value of $$x$$, a curve is defined, and the Gamma function calculates the area under the curve defined by $$x$$.

## 17.2 Theorems for the Gamma Function

### 17.2.1 Lemma

$\left[ -t^{x-1} e^{-t} \right]_{t = 0}^{t = \infty} = 0$

Proof:

\begin{aligned} \left[ -t^{x-1} e^{-t} \right]_{t = 0}^{t = \infty} &= \lim_{t\to\infty}\big(-t^{x - 1} e^{-t}\big) - 0^{x - 1} e^{-0} \\ &= \lim_{t\to\infty}\big(-t^{x - 1} e^{-t}\big) - 0 \\ &= \lim_{t\to\infty}\big(-t^{x - 1} e^{-t}\big) \\ &= - \lim_{t\to\infty}\frac{t^{x - 1}}{e^t} \\ &= - \lim_{t\to\infty} \left\{exp\left[ (x - 1) \ln t - t\right] \right\} \\ &= - \lim_{t\to\infty} \left\{ exp \left[(x - 1) \cdot t \cdot \Big( \frac{\ln t}{t} - 1\Big) \right]\right\} \\ ^{[1]} &= \lim_{t\to\infty} \left\{ exp \left[(x - 1) \cdot t \cdot \Big( 0 - 1\Big) \right]\right\} \\ &= \lim_{t\to\infty} \left\{ exp \left[- (x - 1) \cdot t \right]\right\} \\ &= \lim_{t\to\infty} \frac{1}{e^{(x - 1) \cdot t}} \\ &= 0 \end{aligned}

1. L’H^{o}pital’s Rule: $$\lim_{x\to u}\frac{f(x)}{g(x)} = \lim_{x\to u} \frac{f'(x)}{g'(x)}$$. This implies $$\lim_{x\to \infty} \frac{\ln x}{x} = \lim_{x\to \infty} \frac{\frac{1}{x}}{1} = \lim_{x \to \infty} \frac{1}{x} = 0$$

### 17.2.2 Theorem: The Reduction Relation

$$\Gamma(x) = (x-1) \cdot \Gamma(x)$$

Proof:

The proof relies on integration by parts. Let:

\begin{aligned} u &= t^{x-1} \\ du &= (x - 1) \cdot t^{(x-2)} \\\\ v &= -e^{-t} \\ dv &= e^{-t} dt \end{aligned}

Integration by parts yields:

\begin{aligned} \Gamma(x) &= \int\limits_0^{\infty}t^{x-1}e^{-t}dt \\ &= u \cdot v - \int\limits_0^\infty v \cdot du \\ &= \Big[t^{x-1} \cdot - e^{-t}\Big]_{t=0}^{t=\infty} - \int\limits_0^\infty-e^{-t} \cdot (x-1)\cdot t^{(x-2)} dt \\ &= -\Big[t^{x-1} \cdot e^{-t}\Big]_{t=0}^{t=\infty} - (- (x - 1)) \int\limits_0^\infty e^{-t} \cdot t^{(x-2)} dt \\ ^{[1]} &= -0 + (x - 1) \int\limits_0^\infty e^{-t} \cdot t^{(x-2)} dt \\ &= (x - 1) \int\limits_0^\infty e^{-t} \cdot t^{((x-1)-1)} dt \\ &= (x - 1) \int\limits_0^\infty t^{((x-1)-1)} \cdot e^{-t} dt \\ &= (x - 1) \cdot \Gamma(x - 1) \end{aligned}

### 17.2.3 Corollary

$\Gamma(x) = \frac{1}{x} \cdot \Gamma(x + 1)$

Proof:

Theorem 17.2.2 establishes

$\Gamma(x) = (x - 1) \cdot \Gamma(x - 1)$

Let $$y = x + 1$$. Then

\begin{aligned} \Gamma(y) &= (y - 1) \cdot \Gamma(y - 1) \\ \Rightarrow \Gamma(x + 1) &= (x + 1 - 1) \cdot \Gamma(x + 1 - 1) \\ &= x \cdot \Gamma(x) \\ \Rightarrow \frac{1}{x} \cdot \Gamma(x + 1) &= \Gamma(x) \\ \Rightarrow \Gamma(x) &= \frac{1}{x} \cdot \Gamma(x + 1) \end{aligned}

This allows the recurrence relation to move toward $$\Gamma(0)$$ for any value of $$x$$. Note, however, that $$\Gamma(0)$$ is undefined. Thus, solutions for the Gamma Function may be determined for positive integers, since $$\Gamma(1)$$ can be solved. On the other hand, $$\Gamma(-1)$$ can not be solved, and the recurrence relation results in a zero denominator. Hence, the Gamma Function is defined for all $$x \in \mathbb{R}$$ so long as $$x \not\in \mathbb{Z^-}$$

### 17.2.4 Theorem:

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

Proof:

\begin{aligned} \Gamma\Big(\frac{1}{2}\Big) &= \int\limits_0^\infty t^{-\frac{1}{2}} e ^{-t} dt \\ &= \frac{2}{2} \int\limits_0^\infty t^{-\frac{1}{2}} e ^{-t} dt \\ &= 2 \int\limits_0^\infty \frac{1}{2} \cdot t ^{-\frac{1}{2}} e^{-(\sqrt{t})^2} dt \\ &= 2 \int\limits_0^\infty \frac{1}{2} \cdot t ^{-\frac{1}{2}} e^{-(\sqrt{t})^2} dt \\ ^{[1]} &= 2 \int\limits_0^\infty \frac{1}{2} \cdot x ^{-1} e^{-x^2} \cdot 2\cdot x \ dx \\ &= 2 \int\limits_0^\infty \frac{2x}{2x} e^{-x^2} dx \\ &= 2 \int\limits_0^\infty e^{-x^2} dx \\ ^{[2]} &= 2 \cdot \frac{\sqrt{\pi}}{2} \\ &= \sqrt{\pi} \end{aligned}

1. $$x = \sqrt{t}$$; $$t = x^2$$; and $$dt = 2x dx$$
2. Theorem 18.1.2

### 17.2.5 Theorem:

Let $$c$$ be a constant such that $$c > 0$$. Then

$\int_0^\infty t^x e^{-ct} \cdot dt = \frac{\Gamma(x+1)}{c^{x+1}}$

Proof:

\begin{aligned} \int\limits_0^\infty t^x e^{-ct} dt \\ ^{[1]} &= \int\limits_0^\infty \big(\frac{y}{c}\big) ^ x e ^{-c \frac{y}{c}} \frac{dy}{c} \\ &= \frac{1}{c^x} \cdot \frac{1}{c} \int\limits_0^\infty y ^ x e ^{-y} dy \\ &= \frac{1}{c^{x + 1}} \int\limits_0^\infty y ^ {(x+1)-1} e ^{-y} dy \\ ^{[2]} &= \frac{1}{c^{x + 1}} \cdot \Gamma(x + 1) \\ &= \frac{\Gamma(x + 1)}{c^{x + 1}} \end{aligned}

1. Let $$y = ct$$. Then $$y = \frac{y}{c}$$ and $$dt = \frac{dy}{c}$$.
2. $$\int\limits_0^\infty y^{(x+1)-1} e^{-y} dy$$ satisfies the form of a Gamma function. 17.1