暴力推导 Beta 函数与 Gamma 函数关系式

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$$\mathrm{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.$$
其中
$$\Gamma(x)=\int_0^{+\infty} e^{-t} t^{x-1} \mathrm{d}t,\quad \mathrm{B} (x,y)=\int_0^1 t^{x-1}(1-t)^{y-1} \mathrm{d}t.$$
证明
$$\begin{align} \Gamma(x) \Gamma(y) &= \int_0^{+\infty} e^{-t} t^{x-1} \mathrm{d}t \int_0^{+\infty} e^{-s} s^{y-1} \mathrm{d}s \\ & = \int_0^{+\infty} \int_0^{+\infty} e^{-(s+t)} t^{x-1} s^{y-1} \mathrm{d}s \mathrm{d}t \\ &=4 \int_0^{+\infty} \int_0^{+\infty} e^{-(u^2+v^2)} u^{2x-2} v^{2y-2} \cdot uv \mathrm{d}u \mathrm{d}v \quad (t=u^2,s=v^2)\\ &= 4 \int_0^{+\infty} \int_0^{+\infty} e^{-(u^2+v^2)} u^{2x-1} v^{2y-1} \mathrm{d}u \mathrm{d}v \\ &= \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-(u^2+v^2)} |u|^{2x-1}|v|^{2y-1} \mathrm{d}u \mathrm{d}v \\ &= \int_0^{+\infty} \int_0^{2\pi} r e^{-r^2} r^{2x-1} |\cos \theta|^{2x-1} r^{2y-1} |\sin \theta|^{2y-1} \mathrm{d}r \mathrm{d}\theta \quad (u=r\cos \theta,v=r\sin \theta) \\ &= \int_0^{+\infty} r e^{-r^2} r^{2x+2y-2} \mathrm{d}r \int_0^{2\pi} |\cos \theta|^{2x-1} |\sin \theta|^{2y-1} \mathrm{d} \theta \\ &= \frac{1}{2} \int_0^{+\infty} e^{-r^2} r^{2(x+y-1)} \mathrm{d}r^2 \int_0^{2\pi} |\cos \theta|^{2x-1} |\sin \theta|^{2y-1} \mathrm{d} \theta \\ &= \frac{1}{2} \Gamma(x+y) \int_0^{2\pi} |\cos \theta|^{2x-1} |\sin \theta|^{2y-1} \mathrm{d} \theta \\ &= \Gamma(x+y) \cdot 2\int_0^{\frac{\pi}{2}} \cos^{2x-1} \theta \sin^{2y-1} \theta \mathrm{d} \theta \\ &= \Gamma(x+y) \cdot 2 \int_0^1 t^{x-\frac{1}{2}} (1-t)^{y-\frac{1}{2}} \frac{1}{2} t^{-\frac{1}{2}} (1-t)^{-\frac{1}{2}} \mathrm{d}t \quad (t=\cos^2 \theta,\sin \theta =(1-t)^\frac{1}{2}, \mathrm{d}t = -2 t^{\frac{1}{2}} (1-t)^{\frac{1}{2}} \mathrm{d} \theta) \\ &= \Gamma(x+y) \int_0^1 t^{x-1} (1-t)^{y-1} \mathrm{d}t \\ &= \Gamma(x+y) \mathrm{B} (x,y). \end{align}$$

故

$$\mathrm{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}. \quad \square$$