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Question Number 37309 by math khazana by abdo last updated on 11/Jun/18

calculate ∫_0 ^(+∞) (x^2 /((1+x^2 )^3 )) dx .

$${calculate}\:\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:{dx}\:. \\ $$

Commented by abdo.msup.com last updated on 15/Jun/18

I = ∫_0 ^∞ ((1+x^2 −1)/((1+x^2 )^3 ))dx =∫_0 ^∞ (dx/((1+x^2 )^2 )) −∫_0 ^∞ (dx/((1+x^2 )^3 )) changement x=tanθ give ∫_0 ^∞ (dx/((1+x^2 )^2 )) = ∫_0 ^(π/2) (1/((1+tan^2 θ)^2 ))(1+tan^2 θ )dθ =∫_0 ^(π/2) cos^2 θ dθ =∫_0 ^(π/2) ((1+cos(2θ))/2)dθ =(π/4) ∫_0 ^∞ (dx/((1+x^2 )^3 )) =∫_0 ^(π/2) (1/((1 +tan^2 θ)^3 ))(1+tan^2 θ)dθ =∫_0 ^(π/2) (dθ/((1+tan^2 θ)^2 )) =∫_0 ^(π/2) cos^4 θ dθ =∫_0 ^(π/2) (((1+cos(2θ))/2))^2 dθ =(1/4) ∫_0 ^(π/2) {1+2cos(2θ) +((1+cos(4θ))/2)}dθ =(π/8) +(1/2) ∫_0 ^(π/3) cos(2θ)dθ +(π/(16)) +(1/8) ∫_0 ^(π/2) cos(4θ)dθ =((3π)/(16)) +0 +0=((3π)/(16)) ⇒ I =(π/4) −((3π)/(16)) ⇒ I =(π/(16)) .

$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}+{x}^{\mathrm{2}} \:−\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$${changement}\:{x}={tan}\theta\:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\:\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{cos}^{\mathrm{2}} \theta\:{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}{d}\theta \\ $$$$=\frac{\pi}{\mathrm{4}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{1}}{\left(\mathrm{1}\:+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{3}} }\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{4}} \theta\:{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\right)^{\mathrm{2}} \:{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left\{\mathrm{1}+\mathrm{2}{cos}\left(\mathrm{2}\theta\right)\:+\frac{\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)}{\mathrm{2}}\right\}{d}\theta \\ $$$$=\frac{\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} {cos}\left(\mathrm{2}\theta\right){d}\theta\:+\frac{\pi}{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left(\mathrm{4}\theta\right){d}\theta \\ $$$$=\frac{\mathrm{3}\pi}{\mathrm{16}}\:+\mathrm{0}\:+\mathrm{0}=\frac{\mathrm{3}\pi}{\mathrm{16}}\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{3}\pi}{\mathrm{16}}\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{16}}\:. \\ $$$$ \\ $$

Commented by math khazana by abdo last updated on 15/Jun/18

Residus method 2I = ∫_(−∞) ^(+∞) (x^2 /((1+x^2 )^3 ))dx let the comlex function ϕ(z)= (z^2 /((1+z^2 )^3 )) ϕ(z) = (z^2 /((z−i)^3 (z+i)^3 )) so the poles of ϕ are i and −i (triples) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i)=lim_(z→i) (1/((3−1)!)){ (z−i)^3 ϕ(z)}^((2)) =(1/2)lim_(z→i) { (z^2 /((z+i)^3 ))}^((2)) but { (z^2 /((z+i)^3 ))}^((1)) = ((2z(z+i)^3 −3(z+i)^2 z^2 )/((z+i)^6 )) = ((2z(z+i) −3z^2 )/((z+i)^4 )) = ((−z^2 +2iz)/((z+i)^4 )) ⇒ { (z^2 /((z+i)^3 ))}^((2)) ={ ((−z^2 +2iz)/((z+i)^4 ))}^((1)) =(((−2z+2i)(z+i)^4 −4(z+i)^3 (−z^2 +2iz))/((z+i)^8 )) =(((−2z+2i)(z+i) −4(−z^(2 ) +2iz))/((z+i)^5 )) =((−2z^2 −2iz+2iz −2 +4z^2 −8iz)/((z+i)^5 )) =((2z^2 −8iz −2)/((z+i)^5 )) ⇒ Res(ϕ,i)=(1/2) ((−2+8−2)/((2i)^5 )) = (2/(2^5 i)) = (1/(16i)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (1/(16i)) = (π/8) but I =(1/2) ∫_(−∞) ^(+∞) ϕ(z)dz ⇒ I = (π/(16)) .

$${Residus}\:{method} \\ $$$$\mathrm{2}{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx}\:\:{let}\:{the}\:{comlex}\:{function} \\ $$$$\varphi\left({z}\right)=\:\frac{{z}^{\mathrm{2}} }{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} }{\left({z}−{i}\right)^{\mathrm{3}} \left({z}+{i}\right)^{\mathrm{3}} }\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i} \\ $$$$\left({triples}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\:\left({z}−{i}\right)^{\mathrm{3}} \varphi\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}{lim}_{{z}\rightarrow{i}} \:\left\{\:\frac{{z}^{\mathrm{2}} }{\left({z}+{i}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \:{but} \\ $$$$\left\{\:\frac{{z}^{\mathrm{2}} }{\left({z}+{i}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{1}\right)} \:=\:\frac{\mathrm{2}{z}\left({z}+{i}\right)^{\mathrm{3}} \:−\mathrm{3}\left({z}+{i}\right)^{\mathrm{2}} \:{z}^{\mathrm{2}} }{\left({z}+{i}\right)^{\mathrm{6}} } \\ $$$$=\:\frac{\mathrm{2}{z}\left({z}+{i}\right)\:−\mathrm{3}{z}^{\mathrm{2}} }{\left({z}+{i}\right)^{\mathrm{4}} }\:=\:\frac{−{z}^{\mathrm{2}} \:+\mathrm{2}{iz}}{\left({z}+{i}\right)^{\mathrm{4}} }\:\Rightarrow \\ $$$$\left\{\:\frac{{z}^{\mathrm{2}} }{\left({z}+{i}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \:=\left\{\:\frac{−{z}^{\mathrm{2}} \:+\mathrm{2}{iz}}{\left({z}+{i}\right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\left(−\mathrm{2}{z}+\mathrm{2}{i}\right)\left({z}+{i}\right)^{\mathrm{4}} \:−\mathrm{4}\left({z}+{i}\right)^{\mathrm{3}} \left(−{z}^{\mathrm{2}} \:+\mathrm{2}{iz}\right)}{\left({z}+{i}\right)^{\mathrm{8}} } \\ $$$$=\frac{\left(−\mathrm{2}{z}+\mathrm{2}{i}\right)\left({z}+{i}\right)\:−\mathrm{4}\left(−{z}^{\mathrm{2}\:} +\mathrm{2}{iz}\right)}{\left({z}+{i}\right)^{\mathrm{5}} } \\ $$$$=\frac{−\mathrm{2}{z}^{\mathrm{2}} \:−\mathrm{2}{iz}+\mathrm{2}{iz}\:−\mathrm{2}\:+\mathrm{4}{z}^{\mathrm{2}} \:−\mathrm{8}{iz}}{\left({z}+{i}\right)^{\mathrm{5}} } \\ $$$$=\frac{\mathrm{2}{z}^{\mathrm{2}} \:−\mathrm{8}{iz}\:−\mathrm{2}}{\left({z}+{i}\right)^{\mathrm{5}} }\:\Rightarrow \\ $$$${Res}\left(\varphi,{i}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{−\mathrm{2}+\mathrm{8}−\mathrm{2}}{\left(\mathrm{2}{i}\right)^{\mathrm{5}} }\:=\:\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{5}} {i}}\:=\:\frac{\mathrm{1}}{\mathrm{16}{i}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{1}}{\mathrm{16}{i}}\:=\:\frac{\pi}{\mathrm{8}}\:\:{but} \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:\Rightarrow \\ $$$${I}\:=\:\frac{\pi}{\mathrm{16}}\:. \\ $$$$ \\ $$$$ \\ $$

Answered by ajfour last updated on 11/Jun/18

let x=tan θ I=∫_0 ^( π/2) tan^2 θ cos^4 θdθ =(1/8)∫_0 ^( π/2) (1+cos 4θ)dθ ⇒ I=(π/(16)) .

$${let}\:{x}=\mathrm{tan}\:\theta \\ $$$${I}=\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta\:\mathrm{cos}\:^{\mathrm{4}} \theta{d}\theta \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\mathrm{4}\theta\right){d}\theta \\ $$$$\Rightarrow\:\:{I}=\frac{\pi}{\mathrm{16}}\:. \\ $$

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