Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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..

$${correct}\:{answer}\:{sir}\:{Tanmay}\:\:{thanks}.. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com