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Question Number 141752 by 0731619 last updated on 23/May/21

Answered by qaz last updated on 23/May/21

Ω=∫_0 ^∞ ((tan^(−1) x)/((x+1)(x^2 +1)))dx =∫_0 ^(π/2) (u/(tan u+1))du =∫_0 ^(π/2) ((ucos u)/(sin u+cos u))du.......................(1) =∫_0 ^(π/2) ((((π/2)−u)sin u)/(sin u+cos u))du =(π/2)∫_0 ^(π/2) ((sin u)/(sin u+cos u))du−∫_0 ^(π/2) ((usin u)/(sin u+cos u))du =(π^2 /8)−∫_0 ^(π/2) ((usin u)/(sin u+cos u))du.................(2) (1)+(2) ⇒2Ω=(π^2 /8)+∫_0 ^(π/2) ((u(cos u−sin u))/(sin u+cos u))du =(π^2 /8)−∫_0 ^(π/2) ln(sin u+cos u)du =(π^2 /8)−∫_0 ^(π/2) ln(cos u(tan u+1))du =(π^2 /8)+(π/2)ln2−∫_0 ^(π/2) ln(1+tan u)du =(π^2 /8)+(π/2)ln2−∫_0 ^∞ ((ln(1+u))/(1+u^2 ))du =(π^2 /8)+(π/2)ln2−2∫_0 ^1 ((ln(1+u))/(1+u^2 ))du+∫_0 ^1 ((lnu)/(1+u^2 ))du =(π^2 /8)+(π/2)ln2−2∙(π/8)ln2−G =(π^2 /8)+(π/4)ln2−G ⇒Ω=(π^2 /(16))+(π/8)ln2−(G/2)

$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} {x}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{u}}{\mathrm{tan}\:{u}+\mathrm{1}}{du} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{u}\mathrm{cos}\:{u}}{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}{du}.......................\left(\mathrm{1}\right) \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\left(\frac{\pi}{\mathrm{2}}−{u}\right)\mathrm{sin}\:{u}}{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}{du} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{sin}\:{u}}{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}{du}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{u}\mathrm{sin}\:{u}}{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}{du} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{u}\mathrm{sin}\:{u}}{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}{du}.................\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{2}\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}+\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{u}\left(\mathrm{cos}\:{u}−\mathrm{sin}\:{u}\right)}{\mathrm{sin}\:{u}+\mathrm{cos}\:{u}}{du} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left(\mathrm{sin}\:{u}+\mathrm{cos}\:{u}\right){du} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left(\mathrm{cos}\:{u}\left(\mathrm{tan}\:{u}+\mathrm{1}\right)\right){du} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}+\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left(\mathrm{1}+\mathrm{tan}\:{u}\right){du} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}+\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}−\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{u}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}+\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{u}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }{du}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnu}}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}+\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}−\mathrm{2}\centerdot\frac{\pi}{\mathrm{8}}{ln}\mathrm{2}−{G} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}+\frac{\pi}{\mathrm{4}}{ln}\mathrm{2}−{G} \\ $$$$\Rightarrow\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{\pi}{\mathrm{8}}{ln}\mathrm{2}−\frac{{G}}{\mathrm{2}} \\ $$

Commented by 0731619 last updated on 23/May/21

tanks

$${tanks} \\ $$

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