Question Number 117223 by mnjuly1970 last updated on 10/Oct/20
![... nice mathematics... proof :: Ω = (1/π)∫_(0 ) ^( ∞) [tan^(−1) ((1/x))]^2 dx=ln(2) ... m.n.1970...](https://www.tinkutara.com/question/Q117223.png)
$$\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{mathematics}… \\ $$$$\:\:{proof}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega\:=\:\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}\:} ^{\:\infty} \left[{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)\right]^{\mathrm{2}} {dx}={ln}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:…\:\:{m}.{n}.\mathrm{1970}… \\ $$
Answered by mathmax by abdo last updated on 10/Oct/20
![=(1/π)∫_0 ^∞ arctan^2 ((1/x))dx by parts πI =[x arctan^2 ((1/x))]_0 ^∞ −∫_0 ^∞ x.2arctan((1/x)).×((−(1/x^2 ))/(1+(1/x^2 )))dx =2 ∫_0 ^∞ x arctan((1/x))(dx/(1+x^2 )) =∫_0 ^∞ ((2x)/(1+x^2 )) arctan((1/x))dx =[ln(1+x^2 )arctan((1/x))]_0 ^∞ −∫_0 ^∞ ln(1+x^2 ).((−(1/x^2 ))/(1+(1/x^2 )))dx =∫_0 ^∞ ((ln(1+x^2 ))/(1+x^2 ))dx =_(x=tanθ) ∫_0 ^(π/2) ((ln(1+tan^2 θ))/(1+tan^2 θ))(1+tan^2 θ)dθ =∫_0 ^(π/2) ln((1/(cos^2 θ)))dθ =−2 ∫_0 ^(π/2) ln(cosθ)dθ =−2(−(π/2)ln2) =πln(2) ⇒πI =πln(2) ⇒I =ln(2)](https://www.tinkutara.com/question/Q117232.png)
$$\:=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \:\mathrm{arctan}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts}\: \\ $$$$\pi\mathrm{I}\:=\left[\mathrm{x}\:\mathrm{arctan}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)\right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}.\mathrm{2arctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right).×\frac{−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}\:\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{dx} \\ $$$$=\left[\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right).\frac{−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\:=_{\mathrm{x}=\mathrm{tan}\theta} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)\mathrm{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta}\right)\mathrm{d}\theta\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cos}\theta\right)\mathrm{d}\theta\:=−\mathrm{2}\left(−\frac{\pi}{\mathrm{2}}\mathrm{ln2}\right)\:=\pi\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\pi\mathrm{I}\:=\pi\mathrm{ln}\left(\mathrm{2}\right)\:\Rightarrow\mathrm{I}\:=\mathrm{ln}\left(\mathrm{2}\right) \\ $$
Commented by mnjuly1970 last updated on 10/Oct/20
$${thank}\:{you}\:{very}\:{much}\:{mr} \\ $$$${max}\:.{nice}\:{solution}\:{as} \\ $$$${always}.. \\ $$
Commented by mathmax by abdo last updated on 10/Oct/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$