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Question Number 33344 by prof Abdo imad last updated on 14/Apr/18
prove that ∀ α ∈]0,+∞[ lim_(n→∞) ∫_0 ^n (1−(x/n))^n x^(α−1) dx =Γ(α) .
provethat∀α∈]0,+∞[limn→∞∫0n(1−xn)nxα−1dx=Γ(α).
Commented by prof Abdo imad last updated on 19/Apr/18
∫_0 ^n (1−(x/n))^n x^(α−1) dx = ∫_R (1−(x/n))^n x^(α−1) χ_([0,n[) (x)dx let put f_n (x) = (1−(x/n))^n x^(α−1) χ_([0,n[) (x)dx f_n (x)→^(c.s) f(x)= e^(−x) x^(α−1) if 0≤x<n and f(x)=0 if x>n also we have f_n (x) ≤ f(x) theorem of convergence dominee give lim_(n→∞) ∫_0 ^n f_n (x)dx = ∫_0 ^∞ x^(α−1) e^(−x) dx =Γ(α).
∫0n(1−xn)nxα−1dx=∫R(1−xn)nxα−1χ[0,n[(x)dxletputfn(x)=(1−xn)nxα−1χ[0,n[(x)dxfn(x)→c.sf(x)=e−xxα−1if0⩽x<nandf(x)=0ifx>nalsowehavefn(x)⩽f(x)theoremofconvergencedomineegivelimn→∞∫0nfn(x)dx=∫0∞xα−1e−xdx=Γ(α).

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