Claim : Γ(α)
——— = ∫(x^α-1)e^-λx dx for x from 0 to ∞
λ^α
pf : ∵ Γ(α) = ∫(t^α-1)e^-t dt for t from 0 to ∞
Let t = λx
∴ dt = λdx
Γ(α) = ∫(λ^α-1)(x^α-1)(e^-λx)λ dx
= (λ^α) ∫(x^α-1)e^-λx dx
Hence Γ(α)
——— = ∫(x^α-1)e^-λx dx
λ^α
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∴ ∫x^(3/2)exp(-ax)dx = Γ(5/2) / a^5/2
= 3π^1/2 / 4a^5/2
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