Question Number 54821 by Abdo msup. last updated on 12/Feb/19
$${find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{{n}} \:\:\:\frac{{arctan}\left({nx}\right)}{{n}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 13/Feb/19
$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{{n}} \:\:\frac{{arctan}\left({nx}\right)}{{n}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }{dx}\:\Rightarrow{A}_{{n}} =_{{x}={nt}} \:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({n}^{\mathrm{2}} {t}\right)}{{n}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:{ndt} \\ $$$$=\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{arctan}\left({n}^{\mathrm{2}} {t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:\:{but}\:\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({n}^{\mathrm{2}} {t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\frac{\pi}{\mathrm{2}}\:\frac{\pi}{\mathrm{4}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\:{and}\:{lim}_{{n}\rightarrow+\infty} \:\frac{\mathrm{1}}{{n}}\:=\mathrm{0}\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\mathrm{0}\:. \\ $$