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calculate-I-x-1-x-2-1-2-dx-




Question Number 36432 by prof Abdo imad last updated on 02/Jun/18
calculate  I = ∫_(−∞) ^(+∞)    ((x+1)/((x^2  +1)^2 ))dx .
$${calculate}\:\:{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 03/Jun/18
let introduce the complex function  ϕ(z) = ((z+1)/((z^2  +1)^2 ))  we have   ϕ(z) = ((z+1)/( (z−i)^2 (z+i)^2 )) so the poles of ϕ are i and−i  (doubles)   ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ Res(ϕ,i)  Res(ϕ,i) =lim_(z→i) (1/((2−1)!)){ (z−i)^2 ϕ(z)}^′   =lim_(z→i)  { ((z+1)/((z+i)^2 ))}^′   = lim_(z→i)   (((z+i)^2  −2(z+i)(z+1))/((z+i)^4 ))  =lim_(z→i)  ((z+i −2z−2)/((z+i)^3 ))  =  ((−2)/((2i)^3 ))  = ((−2)/(−8i)) = (1/(4i))  ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ (1/(4i)) = (π/2)  so  ★ I = (π/2) ★
$${let}\:{introduce}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}+\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{have}\: \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}+\mathrm{1}}{\:\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}−{i} \\ $$$$\left({doubles}\right)\: \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{'} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\left\{\:\frac{{z}+\mathrm{1}}{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{'} \\ $$$$=\:{lim}_{{z}\rightarrow{i}} \:\:\frac{\left({z}+{i}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{i}\right)\left({z}+\mathrm{1}\right)}{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\frac{{z}+{i}\:−\mathrm{2}{z}−\mathrm{2}}{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\:\:\frac{−\mathrm{2}}{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:\:=\:\frac{−\mathrm{2}}{−\mathrm{8}{i}}\:=\:\frac{\mathrm{1}}{\mathrm{4}{i}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{1}}{\mathrm{4}{i}}\:=\:\frac{\pi}{\mathrm{2}}\:\:{so} \\ $$$$\bigstar\:{I}\:=\:\frac{\pi}{\mathrm{2}}\:\bigstar \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18
x=tan(θ  dx=sec^2 θdθ  ∫_((−Π)/2) ^(Π/2) (((tanθ +1)×sec^2 θ)/(sec^2 θ×sec^2 θ)) dθ  ∫_(−(Π/2)) ^(Π/2) { ((sinθ)/(cosθ))×cos^2 θ+cos^2 θ}dθ  ∫_(−(Π/2)) ^(Π/2) sinθd(sinθ)+∫_(−(Π/2)) ^(Π/2)  ((1+cos2θ)/2)dθ  =∣((sin^2 θ)/2)∣_(−(Π/2)) ^(Π/2)  +(1/2)∣θ∣_(−(Π/2)) ^(Π/2) +(1/2)×(1/2)∣sin2θ∣_(−(Π/2)) ^(Π/2)   =0+(1/2)(Π)+(1/4)×0  =(Π/2)
$${x}={tan}\left(\theta\right. \\ $$$${dx}={sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$\int_{\frac{−\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} \frac{\left({tan}\theta\:+\mathrm{1}\right)×{sec}^{\mathrm{2}} \theta}{{sec}^{\mathrm{2}} \theta×{sec}^{\mathrm{2}} \theta}\:{d}\theta \\ $$$$\int_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} \left\{\:\frac{{sin}\theta}{{cos}\theta}×{cos}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta\right\}{d}\theta \\ $$$$\int_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} {sin}\theta{d}\left({sin}\theta\right)+\int_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{cos}\mathrm{2}\theta}{\mathrm{2}}{d}\theta \\ $$$$=\mid\frac{{sin}^{\mathrm{2}} \theta}{\mathrm{2}}\mid_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} \:+\frac{\mathrm{1}}{\mathrm{2}}\mid\theta\mid_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} +\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\mid{sin}\mathrm{2}\theta\mid_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} \\ $$$$=\mathrm{0}+\frac{\mathrm{1}}{\mathrm{2}}\left(\Pi\right)+\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{0} \\ $$$$=\frac{\Pi}{\mathrm{2}} \\ $$
Commented by abdo mathsup 649 cc last updated on 03/Jun/18
correct answer sir Tanmay  thanks..
$${correct}\:{answer}\:{sir}\:{Tanmay}\:\:{thanks}.. \\ $$

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