> [!definition] > > $ > \beta(m, n) = \int_{0}^{1}x^{m - 1}(1 - x)^{n - 1}dx > $ > > The Beta [[Function|Function]] is a function closely related to the [[Gamma Function|gamma function]] and the [[Combination|binomial coefficients]]. > [!theoremb] Connection to the Gamma Function > > $ > \beta(m, n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m + n)} > $ > > *Proof*. Substitute $x = \cos^2\theta$, $dx = -2\sin{\theta}\cos{\theta}d\theta$ into the beta function: > $ > \begin{align*} > \int_{0}^{1}x^{m - 1}(1 - x)^{n - 1}dx > &= -2\int_{\pi / 2}^{0}\cos^{2m - 2}\theta \sin^{2n - 2}\theta \cos\theta \sin\theta d\theta\\ > &= 2\int_{0}^{\pi / 2}\cos^{2m - 1}\theta \sin^{2n - 1}\theta d\theta > \end{align*} > $ > Substitute $x = y^2$, $dx = 2y dy$ into the gamma function: > $ > \begin{align*} > \int_{0}^{\infty}e^{-x}x^{\alpha - 1}dx > &= \int_{0}^{\infty}e^{-y^2}y^{2\alpha - 2} 2ydy \\ > &= 2\int_{0}^{\infty}e^{-y^2}y^{2\alpha - 1}dy > \end{align*} > $ > > Combine the integrals for $\Gamma(m)$ and $\Gamma(n)$: > $ > \begin{align*} > \Gamma(m)\Gamma(n) > &= 2\int_{0}^{\infty}e^{-x^2}x^{2m - 1}dx \cdot 2\int_{0}^{\infty}e^{-y^2}y^{2n - 1}dy \\ > &= 4\int_{x = 0}^{\infty}\int_{y = 0}^{\infty} > e^{-x^2}e^{-y^2}x^{2m - 1}y^{2n - 1}dy\ dx \\ > &= 4\int_{x = 0}^{\infty}\int_{y = 0}^{\infty} > e^{-x^2-y^2}x^{2m - 1}y^{2n - 1}dy\ dx > \end{align*} > $ > ![[beta_polar.png|200]] > > Convert the integral into polar form, with $x = r\cos{\theta}, y = r{\sin{\theta}}$: > $ > \begin{align*} > &4\int_{x = 0}^{\infty}\int_{y = 0}^{\infty} > e^{-x^2-y^2}x^{2m - 1}y^{2n - 1}dy\ dx \\ > &= 4\int_{r = 0}^{\infty}\int_{\theta = 0}^{\pi / 2}e^{-r^2}r^{2m-1}\cos^{2m-1}{\theta}r^{2n-1}\sin^{2n-1}{\theta}\ d\theta\ dr \\ > &= 4\int_{r = 0}^{\infty}e^{-r^2}r^{2m-1}r^{2n-1}\int_{\theta = 0}^{\pi / 2}\cos^{2m-1}{\theta}\sin^{2n-1}{\theta}\ d\theta\ dr > \end{align*} > $ > Plug in previous substitutions: > $ > \begin{align*} > &4\int_{r = 0}^{\infty}e^{-r^2}r^{2m-1}r^{2n-1}\int_{\theta = 0}^{\pi / 2}\cos^{2m-1}{\theta}\sin^{2n-1}{\theta}\ d\theta\ dr \\ > &= 2\int_{r = 0}^{\infty}e^{-r^2}r^{2m-1}r^{2n-1}\cdot 2\int_{\theta = 0}^{\pi / 2}\cos^{2m-1}{\theta}\sin^{2n-1}{\theta}\ d\theta\ dr \\ > &= 2\int_{r = 0}^{\infty}e^{-r^2}r^{2m-1}r^{2n-1}\cdot \beta(m, n)dr\\ > &= 2\beta(m, n)\int_{r = 0}^{\infty}e^{-r^2}r^{2m-1}r^{2n-1}dr\\ > &= 2\beta(m, n)\int_{r = 0}^{\infty}e^{-x}x^{m - 1/2}x^{n - 1/2}dx\\ > &= \beta(m, n) \cdot 2\int_{r = 0}^{\infty}e^{-x}x^{m + n - 1}dx\\ > \Gamma(m)\Gamma(n) &= \beta(m, n) \Gamma(m + n) \\ > \beta(m, n) &= \frac{\Gamma(m)\Gamma(n)}{\Gamma(m + n)} > \end{align*} > $ > [!theorem] Connection to the Binomial Coefficients > > Using the connection to the gamma function: > $ > \begin{align*} > \beta(m + 1, n) &= \frac{\Gamma(m + 1)\Gamma(n)}{\Gamma(m + n + 1)} \\ > &= \frac{m!n!}{n(m + n)!} \\ > &= \frac{1}{n\frac{(m + n)!}{m!n!}} \\ > &= \frac{1}{n{{m + n}\choose{m}}} > \end{align*} > $