Here's the question you clicked on:
.Sam.
Prove below
\[\Gamma(x+1)=x \Gamma(x)~~,~~x>0\]
gamma got run over by a john deere i think parts does this
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)