Factorial expression

Γ(z)Γ(1-z)=π/sin(πz)Part 1: Γ(z)= 2∫[0~∞] x^(2z-1)exp(-x^2) dx
Γ(z)
= ∫[0~∞] t^(z-1)exp(-t) dx (Let t=x^2)
= 2∫[0~∞] x^(2z-1)exp(-x^2) dx Part 2: Γ(z)Γ(1-z)= 2∫[0~π/2] tan^(1-2z) dθ
Γ(z)Γ(1-z)
={2∫[0~∞] x^(2z-1)exp(-x^2) dx}{2∫[0~∞] y^(1-2z)exp(-y^2) dy}
=4∫[0~∞]∫[0~∞] x^(2z-1)y^(1-2z)exp[-(x^2+y^2)] dxdy (Let x=rcosθ and y=rsinθ)
=4∫[0~π/2]∫[0~∞] tan^(1-2z)θ rexp(-r^2) drdθ
={2∫[0~∞] rexp(-r^2) dr}{ 2 ∫[0~π/2] tan^(1-2z)θ dθ}
=2 ∫[0~π/2] tan^(1-2z)θ dθPart 3: Γ(z)Γ(1-z)= ∫[0~∞] x^(z-1)/(1+x) dx
Sub. x=tan^2θ dx=2(sec^2θ)(tanθ)dθ
Γ(z)Γ(1-z)
=2 ∫[0~π/2] tan^(1-2z)θ dθ
=2∫[0~∞] x^(-z)/2sec^2θ dx
=∫[0~∞] x^(-z)/(1+x) dxPart 4: ∫[0~∞] x^(p-1)/(1+x) dx = π/sin(πp)This can be proved by Residue theory. See Schaum's Outline of Complex Variables.Part 5: Γ(z)Γ(1-z)=π/sin(πz)Sub. z=1-p, we have Γ(z)Γ(1-z)
=π/sin[π(1-z)]
=π/(sinπcosπz-cosπsinπz)
=π/sinπz

To:M大:關鍵點part4是本題的重點所在吔,怎可省略?
To A大:給我mail,再寄作法給"妳"