Question

Prove below

Answer 1

\[\Gamma(x+1)=x \Gamma(x)~~,~~x>0\]

Answer 2

gamma got run over by a john deere
i think parts does this

Answer 3

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)

Answer 4

Nice work lana!!! :D

Answer 5

:D:D