Question Number 33695 by math khazana by abdo last updated on 22/Apr/18
$${find}\:{lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{e}^{−\frac{{x}}{{n}}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}. \\ $$
Commented by math khazana by abdo last updated on 29/Apr/18
$${let}\:{put}\:{A}_{{n}} \:\:=\:\int_{\mathrm{0}} ^{{n}} \:\:\:\:\frac{{e}^{−\frac{{x}}{{n}}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${A}_{{n}} \:\:=\:\int_{{R}} \:\:\:\:\frac{{e}^{−\frac{{x}}{{n}}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:\chi_{\left[\mathrm{0},{n}\left[\right.\right.} \left({x}\right){dx}\:\:{but} \\ $$$${f}_{{n}} \left({x}\right)\:=\:\frac{{e}^{−\frac{{x}}{{n}}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:\chi_{\left[\mathrm{0},{n}\left[\right.\right.} \left({x}\right){dx}\:\rightarrow^{{c}.{s}} \:\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{on}\left[\mathrm{0},+\infty\left[\right.\right. \\ $$$$\Rightarrow\:\int_{{R}} {f}_{{n}} \left({x}\right){dx}\:\rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\frac{\pi}{\mathrm{2}}\:. \\ $$