<< calculus

Gamma Function

\[ \newcommand{\d}{\text{d}} \newcommand{\dx}{\d x} \newcommand{\dy}{\d y} \newcommand{\dt}{\d t} \newcommand{\dr}{\d r} \newcommand{\euler}{\text{e}} \]

# Definition

\[ \Gamma(\alpha) = \int_{0}^{\infty} x^{\alpha-1}\euler^{-x} \dx \]

# Properity
  1. \(\Gamma(1) = 1\).
  2. \(\Gamma(\alpha+1) = \alpha\Gamma(\alpha)\)
  3. $B(a, b) = _{0}^{1} t{a-1}(1-t){b-1} = $

Properity 1 and 2 gurantees that for positive integer \(n\), \(\Gamma(n) = n!\).

Furthermore, the value of Gamma function at \(\alpha = \dfrac{1}{2}\) plays important rules in multiple possiblity distribution, especially the normal distribution, so we list it as the fourth property here:

  1. \(\Gamma\left( \dfrac{1}{2} \right) = \sqrt{\pi}\)

Proof.

  1. \(\Gamma(1) = \displaystyle\int_{0}^{\infty} \euler^{-x}\dx = 1\)

  2. \(\Gamma(\alpha+1) = \displaystyle\int_{0}^{\infty} x^{\alpha}\euler^{-x}\dx = -\displaystyle\int_{0}^{\infty} x^{\alpha} \d(\euler^{-x}) = \displaystyle\int_{0}^{\infty} \euler^{-x} \d x^{\alpha} = \alpha\Gamma(\alpha)\)

  3. ==TODO: Finish the proof.==

\[ \Gamma(a)\Gamma(b) = \]

  1. Consider the integral \(I = \displaystyle \int_{0}^{\infty} \euler^{-t} \dt\):

\[ \begin{align} I^2 &= \left(\int_{0}^{\infty} \euler^{-t} \dt \right)^2 \\ &= \int_{0}^{\infty} \euler^{-x^2} \dx \int_{0}^{\infty} \euler^{-y^2} \dy \\ &= \int_{0}^{\infty}\int_{0}^{\infty} \euler^{-(x^2 + y^2)} \dx\dy \\ &= \int_{0}^{ \frac{\pi}{2}}\int_{0}^{\infty} \euler^{-r^2} r ~ \dr\d\theta \\ &= \frac{\pi}{4} \end{align} \]

Hence \(I = \dfrac{\sqrt{\pi}}{2}\). Now we have

\[ \Gamma\left( \frac{1}{2} \right) = \int_{0}^{\infty} x^{-1/2}e^{-x}\dx = \int_{0}^{\infty} t^{-1}e^{-t^2}\d t^2 = 2I = \sqrt{\pi} \]