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Question Number 163400 by mnjuly1970 last updated on 06/Jan/22

prove Ω= ∫_0 ^( ∞) cot^( −1) (1+x^( 2) )=((√((1/( (√2)))−(1/2))) ) π

$$ \\ $$$$\:\:\:\:\:{prove}\: \\ $$$$\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} {cot}^{\:−\mathrm{1}} \left(\mathrm{1}+{x}^{\:\mathrm{2}} \right)=\left(\sqrt{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}}\:\right)\:\:\pi \\ $$$$ \\ $$

Answered by Ar Brandon last updated on 06/Jan/22

Ω=∫cot^(−1) (1+x^2 )=xcot^(−1) (1+x^2 )+∫((2x^2 )/(x^4 +2x^2 +1))dx =xcot^(−1) (1+x^2 )+∫(((x^2 +1)+(x^2 −1))/(x^4 +2x^2 +1))dx =xcot^(−1) (1+x^2 )+∫((1+(1/x^2 ))/((x−(1/x))^2 +4))dx+∫((1−(1/x^2 ))/((x+(1/x))^2 ))dx =xcot^(−1) (1+x^2 )+(1/2)tan^(−1) (((x^2 −1)/(2x)))−(x/(x^2 +1))+C

$$\Omega=\int\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)={x}\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)+\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\:\:\:\:={x}\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)+\int\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\:\:\:={x}\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)+\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{4}}{dx}+\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} }{dx} \\ $$$$\:\:\:={x}\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}}\right)−\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}+{C} \\ $$

Answered by Kamel last updated on 07/Jan/22

prove Ω= ∫_0 ^( ∞) cot^( −1) (1+x^( 2) )=((√((1/( (√2)))−(1/2))) ) π =2∫_0 ^(+∞) ((x^2 dx)/(1+(1+x^2 )^2 ))=^(t=x^2 ) ∫_0 ^(+∞) (((√t)dt)/(1+(1+t)^2 )) =((√π)/2)∫_0 ^(+∞) ((sin(t)e^(−t) )/t^(3/2) )dt=((√π)/2)∫_1 ^(+∞) ∫_0 ^(+∞) ((sin(t))/( (√t)))e^(−at) dtda =(√π)∫_1 ^(+∞) ∫_0 ^(+∞) sin(u^2 )e^(−au^2 ) duda =(π/(4i))∫_1 ^(+∞) ((1/( (√(a−i))))−(1/( (√(a+i)))))da=(π/(2i))((√(1+i))−(√(1−i)))=π(√(√2))sin((π/8)) sin((π/8))=(√((1−cos((π/4)))/2))=(√(((√2)−1)/(2(√2))))=(1/( (√(√2))))(√((1/( (√2)))−(1/2))) ∴ Ω=π(√((1/( (√2)))−(1/2)))

$$\:\:\:\:\:{prove}\: \\ $$$$\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} {cot}^{\:−\mathrm{1}} \left(\mathrm{1}+{x}^{\:\mathrm{2}} \right)=\left(\sqrt{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}}\:\right)\:\:\pi \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{{x}^{\mathrm{2}} {dx}}{\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\overset{{t}={x}^{\mathrm{2}} } {=}\int_{\mathrm{0}} ^{+\infty} \frac{\sqrt{{t}}{dt}}{\mathrm{1}+\left(\mathrm{1}+{t}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{+\infty} \frac{{sin}\left({t}\right){e}^{−{t}} }{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{dt}=\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{1}} ^{+\infty} \int_{\mathrm{0}} ^{+\infty} \frac{{sin}\left({t}\right)}{\:\sqrt{{t}}}{e}^{−{at}} {dtda} \\ $$$$\:\:\:\:\:\:\:\:=\sqrt{\pi}\int_{\mathrm{1}} ^{+\infty} \int_{\mathrm{0}} ^{+\infty} {sin}\left({u}^{\mathrm{2}} \right){e}^{−{au}^{\mathrm{2}} } {duda} \\ $$$$\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{4}{i}}\int_{\mathrm{1}} ^{+\infty} \left(\frac{\mathrm{1}}{\:\sqrt{{a}−{i}}}−\frac{\mathrm{1}}{\:\sqrt{{a}+{i}}}\right){da}=\frac{\pi}{\mathrm{2}{i}}\left(\sqrt{\mathrm{1}+{i}}−\sqrt{\mathrm{1}−{i}}\right)=\pi\sqrt{\sqrt{\mathrm{2}}}{sin}\left(\frac{\pi}{\mathrm{8}}\right) \\ $$$${sin}\left(\frac{\pi}{\mathrm{8}}\right)=\sqrt{\frac{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{2}}}=\sqrt{\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}}=\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{2}}}}\sqrt{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\therefore\:\:\Omega=\pi\sqrt{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$

Commented by mnjuly1970 last updated on 07/Jan/22

bravo...

$${bravo}... \\ $$

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