Paper 4, Section I, D

Further Complex Methods | Part II, 2009

Show that

Γ(α)Γ(β)=Γ(α+β)01tα1(1t)β1dt,Reα>0,Reβ>0\Gamma(\alpha) \Gamma(\beta)=\Gamma(\alpha+\beta) \int_{0}^{1} t^{\alpha-1}(1-t)^{\beta-1} d t, \quad \operatorname{Re} \alpha>0, \quad \operatorname{Re} \beta>0

where Γ(z)\Gamma(z) denotes the Gamma function

Γ(z)=0xz1exdx,Rez>0\Gamma(z)=\int_{0}^{\infty} x^{z-1} e^{-x} d x, \quad \operatorname{Re} z>0

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