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Question Number 59277 by Mr X pcx last updated on 07/May/19
calculate ∫_(−∞) ^(+∞) ((ln(1+x^2 ))/(1+x^2 )) dx
$${calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$
Commented by maxmathsup by imad last updated on 18/May/19
let A =∫_(−∞) ^(+∞) ((ln(1+x^2 ))/(1+x^2 ))dx changement x=tanθ give A =∫_(−(π/2)) ^(π/2) ((ln(1+tan^2 θ))/(1+tan^2 θ))(1+tan^2 θ)dθ =∫_(−(π/2)) ^(π/2) ln((1/(cos^2 θ)))dθ =−2 ∫_(−(π/2)) ^(π/2) ln(cosθ)dθ =−4 ∫_0 ^(π/2) ln(cosθ)dθ let find ∫_0 ^(π/2) ln(cosθ)dθ =I let use his friend J=∫_0 ^(π/2) ln(sinθ)dθ we have J =_(θ=(π/2)−t) −∫_0 ^(π/2) ln(cost)(−dt) =I also I +J =2I =∫_0 ^(π/2) ln(cosθ.sinθ)dθ =∫_0 ^(π/2) ln(((sin(2θ))/2))dθ=∫_0 ^(π/2) ln(sin(2θ))dθ −(π/2)ln(2) ∫_0 ^(π/2) ln(sin(2θ))dθ =_(2θ=t) (1/2)∫_0 ^π ln(sint) dt =(1/2){ ∫_0 ^(π/2) ln(sint)dt +∫_(π/2) ^π ln(sint)dt} =(1/2) I +(1/2) ∫_0 ^(π/2) ln(cosα)dα (t=(π/2)+α) =(1/2)I +(1/2)I =I ⇒ 2I =−(π/2)ln(2) +I ⇒ I =−(π/2)ln(2) ⇒A =−4(−(π/2))ln(2) ⇒ A = 2πln(2) .
$${let}\:{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:{changement}\:{x}={tan}\theta\:{give} \\ $$$${A}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta}\right){d}\theta\:=−\mathrm{2}\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cos}\theta\right){d}\theta \\ $$$$=−\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({cos}\theta\right){d}\theta\:\:\:{let}\:{find}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\theta\right){d}\theta\:={I}\:\:{let}\:{use}\:{his}\:{friend}\:{J}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\theta\right){d}\theta \\ $$$${we}\:{have}\:{J}\:=_{\theta=\frac{\pi}{\mathrm{2}}−{t}} \:\:\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cost}\right)\left(−{dt}\right)\:={I}\:\:{also} \\ $$$${I}\:+{J}\:=\mathrm{2}{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\theta.{sin}\theta\right){d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\right){d}\theta=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({sin}\left(\mathrm{2}\theta\right)\right){d}\theta \\ $$$$−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left(\mathrm{2}\theta\right)\right){d}\theta\:=_{\mathrm{2}\theta={t}} \:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {ln}\left({sint}\right)\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sint}\right){dt}\:+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} {ln}\left({sint}\right){dt}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{I}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\alpha\right){d}\alpha\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({t}=\frac{\pi}{\mathrm{2}}+\alpha\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{I}\:+\frac{\mathrm{1}}{\mathrm{2}}{I}\:={I}\:\Rightarrow\:\mathrm{2}{I}\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:+{I}\:\:\Rightarrow\:{I}\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:\Rightarrow{A}\:=−\mathrm{4}\left(−\frac{\pi}{\mathrm{2}}\right){ln}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\:{A}\:=\:\mathrm{2}\pi{ln}\left(\mathrm{2}\right)\:. \\ $$$$ \\ $$

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