Question Number 56329 by maxmathsup by imad last updated on 14/Mar/19
$$\left.\mathrm{1}\right){calculate}\:{A}_{{n}} =\int_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:{A}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{A}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 17/Mar/19
$$\left.\mathrm{1}\right)\:{A}_{{n}} =_{{x}={tan}\theta} \:\:\:\int_{{arctan}\left(\frac{\mathrm{1}}{{n}}\right)} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\:\int_{{arctan}\left(\frac{\mathrm{1}}{{n}}\right)} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\:\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta}\right){d}\theta\:=−\mathrm{2}\:\int_{{arctan}\left(\frac{\mathrm{1}}{{n}}\right)} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({cos}\theta\right){d}\theta\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =−\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\theta\right){d}\theta\:\:{let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({cos}\theta\right){d}\theta\:{and} \\ $$$${J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({sin}\theta\right){d}\theta\:\:\:{we}\:{have}\:\:{I}\:+{J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:{ln}\left({cos}\theta\:{sin}\theta\right){d}\theta \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\right){d}\theta\:=−\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({sin}\left(\mathrm{2}\theta\right)\right){d}\theta\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({sin}\left(\mathrm{2}\theta\right){d}\theta\:=_{\mathrm{2}\theta\:={t}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({sint}\right)\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right)\:=−\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:\Rightarrow\right. \\ $$$${I}\:+{J}\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\ $$$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({cos}\theta\right){d}\theta\:=_{\theta\:={t}−\frac{\pi}{\mathrm{2}}} \:\:\:\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:{ln}\left({sin}\theta\right){d}\theta\:\:=\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {ln}\left({sin}\theta\right){d}\theta\:+\int_{\mathrm{0}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} {ln}\left({sin}\theta\right){d}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)+\:\int_{\mathrm{0}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} {ln}\left({sin}\theta\right){d}\theta\:\:….{be}\:{continued}…. \\ $$$$ \\ $$