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