Factorial expression

Show that for 0 < a < 1

a! (-a)! = pia / sin(pia)

2 個解答

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  • 1 十年前
    最愛解答

    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"

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