무한서고

Gamma function(감마함수)를 통하여 gamma(n+1)=n!(팩토리얼, factorial) 증명

1. 감마함수 정의

    $\Gamma \left ( n \right ) = \int_{0}^{\infty }e^{-x}\cdot x^{n-1 }\: dx$

2. gamma(n+1) = n! 증명

  2-1) gamma(n+1) 재정의

    $\Gamma \left ( n+1 \right ) = \int_{0}^{\infty }e^{-x}\cdot x^{n }\: dx$

  2-2) gamma(n+1) 부분적분

    부분적분법

    $\int u(x)v'(x) \; dx = u(x)v(x) + \int u'(x)v(x)\: dx$

    부분적분

    $\int_{0}^{\infty }x^{n}e^{-x}\: dx = [-x^{n}e^{-x}]_{0}^{\infty} - \int_{0}^{\infty }nx^{n-1}(-1)e^{-x}\: dx$

    $\int_{0}^{\infty }x^{n}e^{-x}\: dx = \lim_{x\rightarrow \infty}(-x^{n}e^{-x})-(0e^{0}) + n \int_{0}^{\infty }x^{n-1}e^{-x}\: dx$

    $\int_{0}^{\infty }x^{n}e^{-x}\: dx = n \int_{0}^{\infty }x^{n-1}e^{-x}\: dx$

    $\Gamma (n+1) = n \Gamma (n)$

  2-3) gamma(1) 계산

    $\Gamma (1) = \int_{0}^{\infty}e^{-x} \cdot x^{1-1} \: dx$

    $\qquad \, = [-e^{-x}]_{0}^{\infty}$

    $\qquad \, = \lim_{x\rightarrow \infty} (-e^{-x}) - (-e^{0})$

    $\qquad \, = 0 - (-1)$

    $\qquad \, = 1$

  2-4) 순환 반복하므로 gamma(n+1)은 n!

    $\Gamma (n+1) = n \Gamma (n)$

    $\Gamma (n) = (n-1) \Gamma (n-1) = n \times (n-1) \times \Gamma (n-1)$

    $\vdots$

    $\Gamma (2) = 1 \cdot \Gamma (1)$

    $\Gamma (1) = 1$

    $\Gamma (n+1) = n \times (n-1) \times ... \times 2 \times 1$

    $\therefore \Gamma (n+1) = n! = \int_{0}^{\infty}e^{-x}\cdot x^{n} \: dx$