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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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