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.Sam.

  • 4 years ago

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  1. .Sam.
    • 4 years ago
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    \[\Gamma(x+1)=x \Gamma(x)~~,~~x>0\]

  2. anonymous
    • 4 years ago
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    gamma got run over by a john deere i think parts does this

  3. lalaly
    • 4 years ago
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    Let u = y^x, dv = e^(-t) dt du = xy^(x-1) dy, v = -e^(-t). Then, Γ(x+1) = ∫(0 to ∞) y^x e^(-t) dt = -y^x e^(-t) {for t = 0 -> ∞} - ∫(0 to ∞) -xy^(x-1) e^(-t) dt = -y^x/e^t {for t = 0 -> ∞} + x ∫(0 to ∞) y^(x-1) e^(-t) dt = 0 + x∫(0 to ∞) y^(x-1) e^(-t) dt, by lohpitals rule = x Γ(x)

  4. .Sam.
    • 4 years ago
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    Nice work lana!!! :D

  5. lalaly
    • 4 years ago
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    :D:D

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