暴力推导 Beta 函数与 Gamma 函数关系式

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

其中

Γ(x)=∫+∞0e−ttx−1dt,B(x,y)=∫10tx−1(1−t)y−1dt.$$\mathrm{\Gamma }(x)={\int }_0^{+\mathrm{\infty }}{e}^{-t}{t}^{x-1}\mathrm{d}t,\mathrm{B}(x,y)={\int }_0^1{t}^{x-1}(1-t{)}^{y-1}\mathrm{d}t.$$

证明

Γ(x)Γ(y)=∫+∞0e−ttx−1dt∫+∞0e−ssy−1ds=∫+∞0∫+∞0e−(s+t)tx−1sy−1dsdt=4∫+∞0∫+∞0e−(u2+v2)u2x−2v2y−2⋅uvdudv(t=u2,s=v2)=4∫+∞0∫+∞0e−(u2+v2)u2x−1v2y−1dudv=∫+∞−∞∫+∞−∞e−(u2+v2)|u|2x−1|v|2y−1dudv=∫+∞0∫2π0re−r2r2x−1|cosθ|2x−1r2y−1|sinθ|2y−1drdθ(u=rcosθ,v=rsinθ)=∫+∞0re−r2r2x+2y−2dr∫2π0|cosθ|2x−1|sinθ|2y−1dθ=12∫+∞0e−r2r2(x+y−1)dr2∫2π0|cosθ|2x−1|sinθ|2y−1dθ=12Γ(x+y)∫2π0|cosθ|2x−1|sinθ|2y−1dθ=Γ(x+y)⋅2∫π20cos2x−1θsin2y−1θdθ=Γ(x+y)⋅2∫10tx−12(1−t)y−1212t−12(1−t)−12dt(t=cos2θ,sinθ=(1−t)12,dt=−2t12(1−t)12dθ)=Γ(x+y)∫10tx−1(1−t)y−1dt=Γ(x+y)B(x,y).$$\begin{array}{rl}\mathrm{\Gamma }(x)\mathrm{\Gamma }(y)& ={\int }_0^{+\mathrm{\infty }}{e}^{-t}{t}^{x-1}\mathrm{d}t{\int }_0^{+\mathrm{\infty }}{e}^{-s}{s}^{y-1}\mathrm{d}s\\ & ={\int }_0^{+\mathrm{\infty }}{\int }_0^{+\mathrm{\infty }}{e}^{-(s+t)}{t}^{x-1}{s}^{y-1}\mathrm{d}s\mathrm{d}t\\ & =4{\int }_0^{+\mathrm{\infty }}{\int }_0^{+\mathrm{\infty }}{e}^{-(u^2+v^2)}{u}^{2x-2}{v}^{2y-2}\cdot uv\mathrm{d}u\mathrm{d}v(t=u^2,s=v^2)\\ & =4{\int }_0^{+\mathrm{\infty }}{\int }_0^{+\mathrm{\infty }}{e}^{-(u^2+v^2)}{u}^{2x-1}{v}^{2y-1}\mathrm{d}u\mathrm{d}v\\ & ={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-(u^2+v^2)}|u{|}^{2x-1}|v{|}^{2y-1}\mathrm{d}u\mathrm{d}v\\ & ={\int }_0^{+\mathrm{\infty }}{\int }_0^{2\pi }r{e}^{-r^2}{r}^{2x-1}|\cos \theta {|}^{2x-1}{r}^{2y-1}|\sin \theta {|}^{2y-1}\mathrm{d}r\mathrm{d}\theta (u=r\cos \theta ,v=r\sin \theta )\\ & ={\int }_0^{+\mathrm{\infty }}r{e}^{-r^2}{r}^{2x+2y-2}\mathrm{d}r{\int }_0^{2\pi }|\cos \theta {|}^{2x-1}|\sin \theta {|}^{2y-1}\mathrm{d}\theta \\ & =\frac{1}{2}{\int }_0^{+\mathrm{\infty }}{e}^{-r^2}{r}^{2(x+y-1)}\mathrm{d}{r}^2{\int }_0^{2\pi }|\cos \theta {|}^{2x-1}|\sin \theta {|}^{2y-1}\mathrm{d}\theta \\ & =\frac{1}{2}\mathrm{\Gamma }(x+y){\int }_0^{2\pi }|\cos \theta {|}^{2x-1}|\sin \theta {|}^{2y-1}\mathrm{d}\theta \\ & =\mathrm{\Gamma }(x+y)\cdot 2{\int }_0^{\frac{\pi }{2}}{\cos }^{2x-1}\theta {\sin }^{2y-1}\theta \mathrm{d}\theta \\ & =\mathrm{\Gamma }(x+y)\cdot 2{\int }_0^1{t}^{x-\frac{1}{2}}(1-t{)}^{y-\frac{1}{2}}\frac{1}{2}{t}^{-\frac{1}{2}}(1-t{)}^{-\frac{1}{2}}\mathrm{d}t(t={\cos }^2\theta ,\sin \theta =(1-t{)}^{\frac{1}{2}},\mathrm{d}t=-2t^{\frac{1}{2}}(1-t{)}^{\frac{1}{2}}\mathrm{d}\theta )\\ & =\mathrm{\Gamma }(x+y){\int }_0^1{t}^{x-1}(1-t{)}^{y-1}\mathrm{d}t\\ & =\mathrm{\Gamma }(x+y)\mathrm{B}(x,y).\end{array}$$

故

B(x,y)=Γ(x)Γ(y)Γ(x+y).□$$\mathrm{B}(x,y)=\frac{\mathrm{\Gamma }(x)\mathrm{\Gamma }(y)}{\mathrm{\Gamma }(x+y)}.◻$$