Question Number 33344 by prof Abdo imad last updated on 14/Apr/18
$$\left.{prove}\:{that}\:\:\forall\:\alpha\:\in\right]\mathrm{0},+\infty\left[\right. \\ $$$${lim}_{{n}\rightarrow\infty} \:\int_{\mathrm{0}} ^{{n}} \:\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} \:{x}^{\alpha−\mathrm{1}} {dx}\:=\Gamma\left(\alpha\right)\:. \\ $$
Commented by prof Abdo imad last updated on 19/Apr/18
$$\int_{\mathrm{0}} ^{{n}} \:\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {x}^{\alpha−\mathrm{1}} {dx}\:\:=\:\int_{{R}} \:\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {x}^{\alpha−\mathrm{1}} \:\chi_{\left[\mathrm{0},{n}\left[\right.\right.} \left({x}\right){dx} \\ $$$${let}\:{put}\:\:{f}_{{n}} \left({x}\right)\:=\:\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {x}^{\alpha−\mathrm{1}} \:\chi_{\left[\mathrm{0},{n}\left[\right.\right.} \left({x}\right){dx} \\ $$$${f}_{{n}} \left({x}\right)\rightarrow^{{c}.{s}} \:\:{f}\left({x}\right)=\:{e}^{−{x}} \:{x}^{\alpha−\mathrm{1}} \:{if}\:\mathrm{0}\leqslant{x}<{n} \\ $$$${and}\:{f}\left({x}\right)=\mathrm{0}\:{if}\:{x}>{n}\:\:\:{also}\:{we}\:{have} \\ $$$${f}_{{n}} \left({x}\right)\:\leqslant\:{f}\left({x}\right)\:{theorem}\:{of}\:{convergence}\:{dominee} \\ $$$${give}\:\:{lim}_{{n}\rightarrow\infty} \int_{\mathrm{0}} ^{{n}} \:\:{f}_{{n}} \left({x}\right){dx}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{x}^{\alpha−\mathrm{1}} \:{e}^{−{x}} {dx}\:=\Gamma\left(\alpha\right). \\ $$