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Question Number 44309 by abdo.msup.com last updated on 26/Sep/18
find the value of I =∫_(−∞) ^(+∞) ((cos(αt))/((x^2 +x +1)^2 ))dx α from R. 2)calculate ∫_(−∞) ^(+∞) (dx/((x^2 +x +1)^2 ))
$${find}\:{the}\:{value}\:{of}\: \\ $$$${I}\:=\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{cos}\left(\alpha{t}\right)}{\left({x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\alpha\:{from}\:{R}. \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 28/Sep/18
1) we have I =∫_(−∞) ^(+∞) ((cos(αx))/({(x+(1/2))^2 +(3/4)}^2 ))dx =_(x+(1/2)=((√3)/2) t) ∫_(−∞) ^(+∞) ((cos(α(((√3)/2)t−(1/2))))/(((3/4))^2 {t^2 +1}^2 ))((√3)/2) dt =((16)/9) ((√3)/2) ∫_(−∞) ^(+∞) ((cos{α(((t(√3)−1)/2))})/((t^2 +1)^2 ))dt =((8(√3))/9) Re( ∫_(−∞) ^(+∞) (e^(iα(((t(√3)−1)/2))) /((t^2 +1)^2 ))dt) let ϕ(z) = (e^(iα(((z(√3)−1)/2))) /((z^2 +1)^2 )) the poles f ϕ are i and −i(doubles) so ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ ,i) =lim_(z→i) (1/((2−1)!)){ (z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { (e^(iα(((z(√3)−1)/2))) /((z+i)^2 ))}^((1)) =lim_(z→i) e^(−((iα)/2)) { (e^(i((√3)/2)α z) /((z+i)^2 ))}^((1)) =lim_(z→i) e^(−((iα)/2)) ((i((√3)/2)α e^(i((√3)/2)αz) (z+i)^2 −2(z+i) e^(i((√3)/2)αz) )/((z+i)^4 )) =e^(−((iα)/2)) lim_(z→i) ((i((√3)/2)(z+i)e^(i((√3)/2)αz) −2 e^(i((√3)/2)αz) )/((z+i)^3 )) =e^(−((iα)/2)) ((−(√3)e^(−((α(√3))/2)) − 2 e^(−((α(√3))/2)) )/(−8i)) =((e^(−((iα)/2)) (2+(√3))e^(−((α(√3))/2)) )/(8i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2i π (2+(√3)) e^(−((iα)/2)) (e^(−((α(√3))/2)) /(8i)) =(π/4)(2+(√3)) e^(−((iα)/2)) e^(−((α(√3))/2)) = (π/4) (2+(√3)) e^(−((α(√3))/2)) (cos((α/2))−i sin((α/2))) ⇒ I =((8(√3))/9) (π/4)(2+(√3)) e^(−((α(√3))/2)) cos((α/2)) ⇒I =((2π (√3))/9)(2+(√3)) e^(−((α(√3))/2)) cos((α/2)). π
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{I}\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\alpha{x}\right)}{\left\{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\right\}^{\mathrm{2}} }{dx}\:=_{{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{t}} \:\:\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\alpha\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right)}{\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \left\{{t}^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{dt} \\ $$$$=\frac{\mathrm{16}}{\mathrm{9}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left\{\alpha\left(\frac{{t}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\right)\right\}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:=\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{9}}\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{i}\alpha\left(\frac{{t}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\right)} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\right) \\ $$$${let}\:\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\alpha\left(\frac{{z}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\right)} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:{the}\:{poles}\:{f}\:\varphi\:{are}\:{i}\:{and}\:−{i}\left({doubles}\right)\:{so} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi\:,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\left\{\:\frac{{e}^{{i}\alpha\left(\frac{{z}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\right)} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{i}} \:\:{e}^{−\frac{{i}\alpha}{\mathrm{2}}} \:\left\{\:\:\:\frac{{e}^{{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\alpha\:{z}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:{e}^{−\frac{{i}\alpha}{\mathrm{2}}} \:\:\:\frac{{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\alpha\:{e}^{{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\alpha{z}} \left({z}+{i}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{i}\right)\:{e}^{{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\alpha{z}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={e}^{−\frac{{i}\alpha}{\mathrm{2}}} \:{lim}_{{z}\rightarrow{i}} \:\:\:\:\frac{{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({z}+{i}\right){e}^{{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\alpha{z}} \:−\mathrm{2}\:{e}^{{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\alpha{z}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$={e}^{−\frac{{i}\alpha}{\mathrm{2}}} \:\frac{−\sqrt{\mathrm{3}}{e}^{−\frac{\alpha\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:−\:\mathrm{2}\:{e}^{−\frac{\alpha\sqrt{\mathrm{3}}}{\mathrm{2}}} }{−\mathrm{8}{i}}\:=\frac{{e}^{−\frac{{i}\alpha}{\mathrm{2}}} \left(\mathrm{2}+\sqrt{\mathrm{3}}\right){e}^{−\frac{\alpha\sqrt{\mathrm{3}}}{\mathrm{2}}} }{\mathrm{8}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\:\pi\:\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:{e}^{−\frac{{i}\alpha}{\mathrm{2}}} \:\:\frac{{e}^{−\frac{\alpha\sqrt{\mathrm{3}}}{\mathrm{2}}} }{\mathrm{8}{i}}\:=\frac{\pi}{\mathrm{4}}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:{e}^{−\frac{{i}\alpha}{\mathrm{2}}} \:{e}^{−\frac{\alpha\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$$$=\:\frac{\pi}{\mathrm{4}}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:{e}^{−\frac{\alpha\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\left({cos}\left(\frac{\alpha}{\mathrm{2}}\right)−{i}\:{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\right)\:\Rightarrow \\ $$$${I}\:\:=\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\frac{\pi}{\mathrm{4}}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:{e}^{−\frac{\alpha\sqrt{\mathrm{3}}}{\mathrm{2}}} \:{cos}\left(\frac{\alpha}{\mathrm{2}}\right)\:\Rightarrow{I}\:=\frac{\mathrm{2}\pi\:\sqrt{\mathrm{3}}}{\mathrm{9}}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:{e}^{−\frac{\alpha\sqrt{\mathrm{3}}}{\mathrm{2}}} \:{cos}\left(\frac{\alpha}{\mathrm{2}}\right). \\ $$$$ \\ $$$$\pi \\ $$
Commented by maxmathsup by imad last updated on 28/Sep/18
2) let take α =0 ⇒ ∫_(−∞) ^(+∞) (dx/((x^2 +x+1)^2 )) =((2π(√3))/9)(2+(√3)) .
$$\left.\mathrm{2}\right)\:{let}\:{take}\:\alpha\:=\mathrm{0}\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:. \\ $$

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