14.2: Definition and properties of the Gamma function

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Definition: Gamma Function

The Gamma function is defined by the integral formula

\[\Gamma (z) = \int_{0}^{\infty} t^{z - 1} e^{-t} \ dt \nonumber \]

The integral converges absolutely for \(\text{Re} (z) > 0\).

Properties

\(\Gamma (z)\) is defined and analytic in the region \(\text{Re} (z) > 0\).
\(\Gamma (n + 1) = n!\), for integer \(n \ge 0\).
\(\Gamma (z + 1) = z \Gamma (z)\) (function equation)
This property and Property 2 characterize the factorial function. Thus, \(\Gamma (z)\) generalizes \(n!\) to complex numbers \(z\). Some authors will write \(\Gamma (z + 1) = z!\).
\(\Gamma (z)\) can be analytically continued to be meromorphic on the entire plane with simple poles at 0, −1, −2 .... The residues are
\[\text{Res} (\Gamma, -m) = \dfrac{(-1)^m}{m!} \nonumber \]
\(\Gamma (z) = [ze^{\gamma z} \prod_{1}^{\infty} (1 + \dfrac{z}{n}) e^{-z/n}]^{-1}\), where \(\gamma\) is Euler's constant
\[\gamma = \lim_{n \to \infty} 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdot\cdot\cdot \dfrac{1}{n} - \log (n) \approx 0.577 \nonumber \]
This property uses an infinite product. Unfortunately we won’t have time, but infinite products represent an entire topic on their own. Note that the infinite product makes the positions of the poles of \(\Gamma\) clear.
\(\Gamma (z) \Gamma (1 - z) = \dfrac{\pi}{\sin (\pi z)}\)
With Property 5 this gives a product formula for \(\sin (\pi z)\).
\(\Gamma (z + 1) \approx \sqrt{2\pi} z^{z + 1/2} e^{-z}\) for \(|z|\) large, \(\text{Re} (z) > 0\).
In particular, \(n! \approx \sqrt{2 \pi} n^{n + 1/2} e^{-n}\). (Stirling's formula)
\(2^{2z - 1} \Gamma (z) \Gamma (z + 1/2) = \sqrt{\pi} \Gamma (2z)\) (Legendre duplication formula)

Note

These are just some of the many properties of \(\Gamma (z)\). As is often the case, we could have chosen to define \(\Gamma (z)\) in terms of some of its properties and derived Equation 14.3.1 as a theorem.

We will prove (some of) these properties below.

Example \(\PageIndex{1}\)

Use the properties of \(\Gamma\) to show that \(\Gamma (1/2) = \sqrt{\pi}\) and \(\Gamma (3/2) = \sqrt{\pi}/2\).

Solution

From Property 2 we have \(\Gamma (1) = 0! = 1\). The Legendre duplication formula with \(z = 1/2\) then shows

\[2^0 \Gamma \left(\dfrac{1}{2}\right) \Gamma (1) = \sqrt{\pi} \Gamma (1) \Rightarrow \Gamma \left(\dfrac{1}{2}\right) = \sqrt{\pi}. \nonumber \]

Now, using the functional equation Property 3 we get

\[\Gamma \left(\dfrac{3}{2}\right) = \Gamma \left(\dfrac{1}{2} + 1\right) = \dfrac{1}{2} \Gamma \left(\dfrac{1}{2}\right) = \dfrac{\sqrt{\pi}}{2}. \nonumber \]