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Question Number 56329 by maxmathsup by imad last updated on 14/Mar/19

1)calculate A_n =∫_(1/n) ^1   ((ln(1+x^2 ))/(1+x^2 ))dx    with n integr and n≥1  2) find lim_(n→+∞)     A_n   3)  study the convergence of Σ A_n

$$\left.\mathrm{1}\right){calculate}\:{A}_{{n}} =\int_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:{A}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{A}_{{n}} \\ $$

Commented by maxmathsup by imad last updated on 17/Mar/19

1) A_n =_(x=tanθ)    ∫_(arctan((1/n))) ^(π/4)  ((ln(1+tan^2 θ))/(1+tan^2 θ)) (1+tan^2 θ)dθ  = ∫_(arctan((1/n))) ^(π/4)  ln ((1/(cos^2 θ)))dθ =−2 ∫_(arctan((1/n))) ^(π/4)  ln(cosθ)dθ ⇒  lim_(n→+∞)  A_n =−2 ∫_0 ^(π/4) ln(cosθ)dθ  let I =∫_0 ^(π/4)  ln(cosθ)dθ and  J =∫_0 ^(π/4)  ln(sinθ)dθ   we have  I +J =∫_0 ^(π/4)   ln(cosθ sinθ)dθ  = ∫_0 ^(π/4) ln(((sin(2θ))/2))dθ =−(π/4)ln(2) +∫_0 ^(π/4)  ln(sin(2θ))dθ but  ∫_0 ^(π/4) ln(sin(2θ)dθ =_(2θ =t)    (1/2)∫_0 ^(π/2)  ln(sint) dt =(1/2)(−(π/2)ln(2)) =−(π/4)ln(2) ⇒  I +J =−(π/2)ln(2)  I =∫_0 ^(π/4)  ln(cosθ)dθ =_(θ =t−(π/2))    ∫_(π/2) ^((3π)/4)  ln(sinθ)dθ  =∫_(π/2) ^0 ln(sinθ)dθ +∫_0 ^((3π)/4) ln(sinθ)dθ  =(π/2)ln(2)+ ∫_0 ^((3π)/4) ln(sinθ)dθ  ....be continued....

$$\left.\mathrm{1}\right)\:{A}_{{n}} =_{{x}={tan}\theta} \:\:\:\int_{{arctan}\left(\frac{\mathrm{1}}{{n}}\right)} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\:\int_{{arctan}\left(\frac{\mathrm{1}}{{n}}\right)} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\:\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta}\right){d}\theta\:=−\mathrm{2}\:\int_{{arctan}\left(\frac{\mathrm{1}}{{n}}\right)} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({cos}\theta\right){d}\theta\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =−\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\theta\right){d}\theta\:\:{let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({cos}\theta\right){d}\theta\:{and} \\ $$$${J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({sin}\theta\right){d}\theta\:\:\:{we}\:{have}\:\:{I}\:+{J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:{ln}\left({cos}\theta\:{sin}\theta\right){d}\theta \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\right){d}\theta\:=−\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({sin}\left(\mathrm{2}\theta\right)\right){d}\theta\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({sin}\left(\mathrm{2}\theta\right){d}\theta\:=_{\mathrm{2}\theta\:={t}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({sint}\right)\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right)\:=−\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:\Rightarrow\right. \\ $$$${I}\:+{J}\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\ $$$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({cos}\theta\right){d}\theta\:=_{\theta\:={t}−\frac{\pi}{\mathrm{2}}} \:\:\:\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:{ln}\left({sin}\theta\right){d}\theta\:\:=\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} {ln}\left({sin}\theta\right){d}\theta\:+\int_{\mathrm{0}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} {ln}\left({sin}\theta\right){d}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)+\:\int_{\mathrm{0}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} {ln}\left({sin}\theta\right){d}\theta\:\:....{be}\:{continued}.... \\ $$$$ \\ $$

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