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Question Number 29855 by abdo imad last updated on 13/Feb/18

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

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

Commented by abdo imad last updated on 13/Feb/18

let put I= ∫_0 ^∞ (x^2 /((1+x^2 )(3+x^2 ))) I=(1/2) ∫_(−∞) ^(+∞) (x^2 /((1+x^2 )(3+x^2 ))) let introduce the complex function f(z)= (z^2 /((1+z^2 )(3+z^2 ))) the poles of f are i,−i (√3) i and−(√3) i and ∫_(−∞) ^(+∞) f(z)dz=2iπ( Res(f,i) +Res(f,i(√3))) Res(f,i)=lim_(z→i) (z−i)f(z)= ((−1)/((2i)(3−1)))=((−1)/(4i)) Res(f,i(√3))=lim_(z→i(√3)) (z−i(√3) )f(z)= ((−3)/(2i(√3)(−2)))=(3/(4i(√3))) ∫_(−∞) ^(+∞) f(z)dz=2iπ( ((−1)/(4i)) +(3/(4i(√3)))) =−(π/2) +((3π)/(2(√3))) =(π/2)( (√3) −1) ⇒ I = (π/4)((√3) −1) .

$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{3}+{x}^{\mathrm{2}} \right)} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{3}+{x}^{\mathrm{2}} \right)}\:{let}\:{introduce}\:{the}\:{complex} \\ $$$${function}\:{f}\left({z}\right)=\:\frac{{z}^{\mathrm{2}} }{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)\left(\mathrm{3}+{z}^{\mathrm{2}} \right)}\:{the}\:{poles}\:\:{of}\:{f}\:{are}\:{i},−{i} \\ $$$$\sqrt{\mathrm{3}}\:{i}\:{and}−\sqrt{\mathrm{3}}\:{i}\:\:{and} \\ $$$$\int_{−\infty} ^{+\infty} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi\left(\:{Res}\left({f},{i}\right)\:+{Res}\left({f},{i}\sqrt{\mathrm{3}}\right)\right) \\ $$$${Res}\left({f},{i}\right)={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right){f}\left({z}\right)=\:\:\frac{−\mathrm{1}}{\left(\mathrm{2}{i}\right)\left(\mathrm{3}−\mathrm{1}\right)}=\frac{−\mathrm{1}}{\mathrm{4}{i}} \\ $$$${Res}\left({f},{i}\sqrt{\mathrm{3}}\right)={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \left({z}−{i}\sqrt{\mathrm{3}}\:\right){f}\left({z}\right)=\:\frac{−\mathrm{3}}{\mathrm{2}{i}\sqrt{\mathrm{3}}\left(−\mathrm{2}\right)}=\frac{\mathrm{3}}{\mathrm{4}{i}\sqrt{\mathrm{3}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{f}\left({z}\right){dz}=\mathrm{2}{i}\pi\left(\:\frac{−\mathrm{1}}{\mathrm{4}{i}}\:+\frac{\mathrm{3}}{\mathrm{4}{i}\sqrt{\mathrm{3}}}\right)\:=−\frac{\pi}{\mathrm{2}}\:+\frac{\mathrm{3}\pi}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\:\sqrt{\mathrm{3}}\:−\mathrm{1}\right)\:\:\Rightarrow\:\:{I}\:\:=\:\frac{\pi}{\mathrm{4}}\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right)\:. \\ $$

Answered by ajfour last updated on 13/Feb/18

= −(1/2)∫(dx/(1+x^2 )) +(3/2)∫(dx/(3+x^2 )) = −((tan^(−1) x)/2)+((√3)/2)tan^(−1) ((x/(√3)))+c . with limits 0 to ∞ I = −(π/4)+(((√3)π)/4) = ((((√3)−1)π)/4) .

$$\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{3}+{x}^{\mathrm{2}} } \\ $$$$\:\:\:=\:−\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{\sqrt{\mathrm{3}}}\right)+{c}\:. \\ $$$${with}\:{limits}\:\mathrm{0}\:{to}\:\infty \\ $$$$\:\:{I}\:=\:−\frac{\pi}{\mathrm{4}}+\frac{\sqrt{\mathrm{3}}\pi}{\mathrm{4}}\:\:=\:\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\pi}{\mathrm{4}}\:. \\ $$

Commented by abdo imad last updated on 14/Feb/18

yes correct thanks...

$${yes}\:{correct}\:{thanks}... \\ $$

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