Связь Гамма и Бета-функций

$$B(x, y) = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)},~ \forall{x, y > 0}$$

Сделаем замену $t = (1 + q)p$, $dt = (1 + q)dp$. $$\Gamma(x + y) = \int_{0}^{\infty} t^{x+y-1}e^{-t}dt = (1 + q)^{x+y} \int_{0}^{\infty} p^{x+y-1}e^{-(1+q)p}dp$$ А для $B(x, y)$ замену $t = \dfrac{q}{1 + q}$, $1 - t = \dfrac{1}{1 + q}$, $dt = \dfrac{dq}{(1 + q)^2}$: $$B(x, y) = \int_{0}^{\infty} \left(\dfrac{q}{1 + q}\right)^{x-1} \left(\dfrac{1}{1 + q}\right)^{y-1} \dfrac{dq}{(1 + q)^2} = \int_{0}^{\infty} \dfrac{q^{x-1}}{(1 + q)^{x+y}}dq$$ $$\begin{aligned} \Gamma(x + y)B(x, y) &= \int_{0}^{\infty} \dfrac{\Gamma(x + y)q^{x-1}}{(1 + q)^{x+y}}dq = \\ &= \int_{0}^{\infty} \left(\int_{0}^{\infty} p^{x+y-1}q^{x-1}e^{-(1+q)p}dp\right)dq = \\ &= \int_{0}^{\infty} \left(\int_{0}^{\infty} p^{x+y-1}q^{x-1}e^{-(1+q)p}dq\right)dp = \\ &= \int_{0}^{\infty} p^{y-1}e^{-p} \left(\int_{0}^{\infty} p^{x-1}q^{x-1}e^{-pq}pdq\right)dp \end{aligned}$$ Делая замену $pq = t$, $pdq = dt$, получаем $$\begin{aligned} \Gamma(x + y)B(x, y) &= \int_{0}^{\infty} p^{y-1}e^{-p} \left(\int_{0}^{\infty} t^{x-1}e^{-t}dt\right)dp = \\ &= \left(\int_{0}^{\infty} p^{y-1}e^{-p}dp\right) \left(\int_{0}^{\infty} t^{x-1}e^{-t}dt\right) = \Gamma(y)\Gamma(x) \end{aligned}$$ $\square$