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Question Number 134418 by mnjuly1970 last updated on 03/Mar/21

𝛗=∫_0 ^( ∞) (arctan((1/x)))^2 =???

$$ \\ $$ $$\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \left({arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)^{\mathrm{2}} =??? \\ $$

Answered by Ñï= last updated on 03/Mar/21

∫(tan^(−1) (1/x))^2 dx=x(tan^(−1) (1/x))^2 +∫(2tan^(−1) (1/x))(x/(1+x^2 ))dx =x(tan^(−1) (1/x))^2 +ln(1+x^2 )tan^(−1) (1/x)+∫((ln(1+x^2 ))/(1+x^2 ))dx ⇒φ=∫_0 ^∞ ((ln(1+x^2 ))/(1+x^2 ))dx x=tan θ ⇒φ=−2∫_0 ^(π/2) lncosθdθ=−2∙(−(π/2)ln2)=πln2

$$\int\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} {dx}={x}\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\int\left(\mathrm{2}{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right)\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$ $$={x}\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}+\int\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$ $$\Rightarrow\phi=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$ $${x}=\mathrm{tan}\:\theta \\ $$ $$\Rightarrow\phi=−\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {lncos}\theta{d}\theta=−\mathrm{2}\centerdot\left(−\frac{\pi}{\mathrm{2}}{ln}\mathrm{2}\right)=\pi{ln}\mathrm{2} \\ $$

Commented bymnjuly1970 last updated on 03/Mar/21

thank you so much..

$${thank}\:{you}\:{so}\:{much}.. \\ $$

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