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.Sam.
 4 years ago
Prove below
.Sam.
 4 years ago
Prove below

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.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.0\[\Gamma(x+1)=x \Gamma(x)~~,~~x>0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0gamma got run over by a john deere i think parts does this

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.6Let u = y^x, dv = e^(t) dt du = xy^(x1) 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^(x1) e^(t) dt = y^x/e^t {for t = 0 > ∞} + x ∫(0 to ∞) y^(x1) e^(t) dt = 0 + x∫(0 to ∞) y^(x1) e^(t) dt, by lohpitals rule = x Γ(x)
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