> [!definition] > > $\Gamma(\alpha) \equiv \int_{0}^{\infty}e^{-x}x^{\alpha - 1}dx \equiv (\alpha - 1)!$ > > The Gamma [[Function|Function]] is the extension of the factorial operation from natural numbers to all complex numbers. > [!theoremb] Derivation > > For all non-negative whole numbers $n$, > $ > n! = \int_{0}^{\infty}e^{-x}x^n dx > $ > > *Proof* by [[Axiom of Induction|induction]]: Observe that when $n = 0$, > $ > \int_{0}^{\infty}e^{-x}dx = 1 = 0! > $ > > Suppose that the result holds for some $n$, then: > $ > \begin{align*} > \int_{0}^{\infty}e^{-x}x^{n + 1}dx &= \paren{-e^{-x}x^{n + 1}}|_{0}^{\infty} + \int_{0}^{\infty}e^{-x}(n + 1)x^{n}dx\\ > &= (n + 1)\int_{0}^{\infty}e^{-x}x^{n}dx \\ > &= (n + 1)n! = (n + 1)! > \end{align*} > $ > > Since $\Gamma(\alpha + 1) = \alpha \Gamma(\alpha)$, and that $\Gamma(1) = 0!$, $\Gamma(\alpha) = (\alpha - 1)!$.