Question Number 32715 by caravan msup abdo. last updated on 31/Mar/18
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{it}\right)\left(\mathrm{1}+{it}^{\mathrm{2}} \right)}\:\:. \\ $$
Commented by abdo imad last updated on 03/Apr/18
$${let}\:{put}\:\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}+{iz}\right)\left(\mathrm{1}+{iz}^{\mathrm{2}} \right)} \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{\mathrm{1}}{\left({iz}\:−{i}^{\mathrm{2}} \right)\left({iz}^{\mathrm{2}} \:−{i}^{\mathrm{2}} \right)}\:\:\frac{−\mathrm{1}}{\left({z}−{i}\right)\left({z}^{\mathrm{2}} −{i}\right)} \\ $$$$=\:\frac{−\mathrm{1}}{\left({z}−{i}\right)\left({z}−\sqrt{{i}}\right)\left({z}+\sqrt{{i}}\right)}\:=\:\:\frac{−\mathrm{1}}{\left({z}−{i}\right)\left({z}\:−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\left({z}\:+{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right)\varphi\left({z}\right)=\:\frac{−\mathrm{1}}{−\mathrm{1}−{i}}\:=\:\frac{\mathrm{1}}{\mathrm{1}+{i}}=\frac{\mathrm{1}−{i}}{\mathrm{2}} \\ $$$${Res}\left(\varphi,{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow{e}^{{i}\frac{\pi}{\mathrm{4}}} } \:\:\left({z}−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\varphi\left({z}\right) \\ $$$$\:=\:\:\frac{−\mathrm{1}}{\left({e}^{{i}\frac{\pi}{\mathrm{4}}} \:−{i}\right)\left(\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)}\:\:=\:\frac{−\mathrm{1}}{\mathrm{2}{i}\:−\mathrm{2}{i}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:=\:\frac{{i}}{\mathrm{2}\:\left(\mathrm{1}−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:\frac{\mathrm{1}−{i}}{\mathrm{2}}\:+\:\frac{{i}}{\mathrm{2}\left(\mathrm{1}−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)}\right)\:\:. \\ $$