伽瑪函數 Gamma function 的特性

 $$\text{GAMMA 函數: } \Gamma(\alpha)=\int_0^\infty t^{\alpha-1} e^{-t}\,dt$$ $$ 性質1: \Gamma(\alpha+1) =\alpha\Gamma(\alpha)\\ \bbox[cyan, 2pt]{證明}:\\取\cases{u=t^\alpha \\dv = e^{-t}dt} \Rightarrow \cases{du=\alpha t^{\alpha-1}\,dt\\ v=-e^{-t}} \Rightarrow \Gamma(\alpha+1)=\int_0^\infty t^{\alpha} e^{-t}\,dt =\left. \left[-t^\alpha e^{-t} \right] \right|_0^\infty+ \alpha \int_0^\infty t^{\alpha-1}e^{-t}\,dt\\ =0+\alpha \Gamma(\alpha) \Rightarrow \Gamma(\alpha+1)=\alpha \Gamma(\alpha),\bbox[red,2pt]{故得證}$$ $$ 性質2: \Gamma(n)=(n-1)!\\ \bbox[cyan,2pt]{證明}:\\ \Gamma(1)= \int_0^\infty t^0e^{-t}\,dt =\int_0^\infty  e^{-t}\,dt =\left.\left[ -e^{-t} \right]\right|_0^\infty =1\\因此\Gamma(n)=(n-1) \Gamma(n-1) =(n-1)(n-2)\Gamma(n-2)=\cdots= (n-1)(n-2)\cdots \Gamma(1)\\ \quad =(n-1)!,\bbox[red,2pt]{故得證} $$ $$ 性質3: \Gamma({1\over 2}) =\sqrt \pi\\ \bbox[cyan,2pt]{證明}:\\ 取t=u^2 ,則dt=2u\,du \Rightarrow \Gamma({1\over 2}) = \int_0^\infty t^{-1/2} e^{-t}\,dt = \int_0^\infty u^{-1}e^{-u^2}\cdot 2u\,du= 2\int_0^\infty e^{-u^2}\,du\\ 令I={1\over 2}\Gamma({1\over 2}) =\int_0^\infty e^{-x^2}\,dx \Rightarrow I^2= \int_0^\infty \int_0^\infty e^{-(x^2+y^2)}\,dxdy =\int_0^{\pi/2} \int_0^\infty re^{-r^2}\,dr d\theta \\=\int_0^{\pi/2} \left. \left[ -{1\over 2}e^{-r^2} \right] \right|_0^\infty \,d\theta =\int_0^{\pi/2} {1\over 2}\,d\theta ={\pi\over 4} \Rightarrow I={1\over 2} \Gamma({1\over 2})=\sqrt{\pi\over 4} \Rightarrow \Gamma({1\over 2})=\sqrt \pi,\bbox[red,2pt]{故得證}$$