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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) .

$$\left.{prove}\:{that}\:\:\forall\:\alpha\:\in\right]\mathrm{0},+\infty\left[\right. \\ $$$${lim}_{{n}\rightarrow\infty} \:\int_{\mathrm{0}} ^{{n}} \:\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} \:{x}^{\alpha−\mathrm{1}} {dx}\:=\Gamma\left(\alpha\right)\:. \\ $$

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).

$$\int_{\mathrm{0}} ^{{n}} \:\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {x}^{\alpha−\mathrm{1}} {dx}\:\:=\:\int_{{R}} \:\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {x}^{\alpha−\mathrm{1}} \:\chi_{\left[\mathrm{0},{n}\left[\right.\right.} \left({x}\right){dx} \\ $$$${let}\:{put}\:\:{f}_{{n}} \left({x}\right)\:=\:\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {x}^{\alpha−\mathrm{1}} \:\chi_{\left[\mathrm{0},{n}\left[\right.\right.} \left({x}\right){dx} \\ $$$${f}_{{n}} \left({x}\right)\rightarrow^{{c}.{s}} \:\:{f}\left({x}\right)=\:{e}^{−{x}} \:{x}^{\alpha−\mathrm{1}} \:{if}\:\mathrm{0}\leqslant{x}<{n} \\ $$$${and}\:{f}\left({x}\right)=\mathrm{0}\:{if}\:{x}>{n}\:\:\:{also}\:{we}\:{have} \\ $$$${f}_{{n}} \left({x}\right)\:\leqslant\:{f}\left({x}\right)\:{theorem}\:{of}\:{convergence}\:{dominee} \\ $$$${give}\:\:{lim}_{{n}\rightarrow\infty} \int_{\mathrm{0}} ^{{n}} \:\:{f}_{{n}} \left({x}\right){dx}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{x}^{\alpha−\mathrm{1}} \:{e}^{−{x}} {dx}\:=\Gamma\left(\alpha\right). \\ $$

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