貝它(beta)函數與伽瑪(gamma)函數的關係式證明

$$假設\text{ beta function }B(p,q)= \int_0^1 t^{p-1}(1-t)^{q-1}\,dt ,則B(p,q)={\Gamma(p) \Gamma(q)\over \Gamma(p+q)},\\其中\text{gamma function }\Gamma(x)= \int_0^\infty t^{x-1}e^{-t}\,dt.\\ \bbox[red,2pt]{證明:}\\　取t=x^2,則dt=2xdx \Rightarrow \Gamma(p)=\int_0^\infty t^{p-1} e^{-t}\,dt = \int_0^\infty x^{2p-2} e^{-x^2}\cdot 2x\,dx =2\int_0^\infty x^{2p-1}e^{-x^2}\,dx \\ \Rightarrow \Gamma(q)=2\int_0^\infty y^{2q-1}e^{-y^2}\,dy \Rightarrow \Gamma(p) \Gamma(q)=4\int_0^\infty \int_0^\infty x^{2p-1}y^{2q-1}e^{-(x^2+y^2)}\,dx\,dy\\ 取\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \Gamma(p) \Gamma(q)= 4\int_0^{\pi/2} \int_0^\infty r^{2p+2q-1} e^{-r^2} \cos^{2p-1}\theta \sin^{2q-1}\,drd\theta \\ =\left( 2\int_0^\infty r^{2(p+q)-1}e^{-r^2}\,dr\right)\left( 2\int_0^{ \pi/2}\cos^{2p-1}\theta \sin^{2q-1}\, d\theta\right) =\Gamma(p+q)\left( 2\int_0^{ \pi/2}\cos^{2p-1}\theta \sin^{2q-1}\, d\theta\right)\\ 取t=\sin^2\theta \Rightarrow dt=2\sin \theta\cos\theta d\theta\Rightarrow   \cases{\sin \theta=\sqrt t\\ \cos \theta=\sqrt{1-t}}\\ \Rightarrow 2\int_0^{ \pi/2}\cos^{2p-1}\theta \sin^{2q-1}\, d\theta=2\int_0^1 (1-t)^{p-1/2} t^{q-1/2}\cdot {dt\over 2t^{1/2}(1-t)^{1/2}}\\ =\int_0^1 (1-t)^{p-1}t^{q-1}\,dt =B(p,q)\\因此\Gamma(p)\Gamma(q)=\Gamma(p+q)B(p,q)\Rightarrow B(p,q)={\Gamma(p)\Gamma(q) \over \Gamma(p+q)},\bbox[red, 2pt]{故得證}$$