Question Number 33202 by prof Abdo imad last updated on 12/Apr/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}\:+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:. \\ $$
Commented by prof Abdo imad last updated on 13/Apr/18
![let put I = ∫_(−∞) ^(+∞) (dt/((1+t+t^2 )^2 )) we have I = ∫_(−∞) ^(+∞) (dt/(((t+(1/2))^2 +(3/4))^2 )) .changement t+(1/2) =((√3)/2) tanθ give I = ∫_(−(π/2)) ^(π/2) (1/(( (3/4)tan^2 θ +(3/4))^2 )) ((√3)/2)(1+tan^2 θ)dθ =((√3)/2) .((16)/9) ∫_(−(π/2)) ^(π/2) (dθ/(1+tan^2 θ)) = ((16(√3))/9) ∫_0 ^(π/2) cos^2 θ dθ = ((16(√3))/9) ∫_0 ^(π/2) ((1+cos(2θ))/2)dθ =((8(√3))/9) ∫_0 ^(π/2) (1 +cos(2θ))dθ = ((4(√3) π)/9) + ((4(√3))/9) [ sin(2θ)]_0 ^(π/2) = ((4π(√3))/9) +0 ⇒ I = ((4π(√3))/9) .](https://www.tinkutara.com/question/Q33226.png)
$${let}\:{put}\:\:{I}\:\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{we}\:{have} \\ $$$${I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dt}}{\left(\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} }\:\:.{changement} \\ $$$${t}+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{tan}\theta\:{give} \\ $$$${I}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{\mathrm{1}}{\left(\:\frac{\mathrm{3}}{\mathrm{4}}{tan}^{\mathrm{2}} \theta\:+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:.\frac{\mathrm{16}}{\mathrm{9}}\:\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{{d}\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:=\:\frac{\mathrm{16}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}} \theta\:{d}\theta \\ $$$$=\:\frac{\mathrm{16}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}{d}\theta \\ $$$$=\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}\:+{cos}\left(\mathrm{2}\theta\right)\right){d}\theta \\ $$$$=\:\frac{\mathrm{4}\sqrt{\mathrm{3}}\:\pi}{\mathrm{9}}\:\:+\:\:\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\left[\:{sin}\left(\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\:\frac{\mathrm{4}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:+\mathrm{0}\:\Rightarrow \\ $$$${I}\:=\:\frac{\mathrm{4}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:\:. \\ $$