Question Number 37287 by math khazana by abdo last updated on 11/Jun/18
$${calculate}\:\:{f}\left({t}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left({tx}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$
Commented by prof Abdo imad last updated on 16/Jun/18
$${f}\left({t}\right)={Re}\left(\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{itx}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\right)\:{let} \\ $$$$\varphi\left({z}\right)=\:\frac{{e}^{{itz}} }{\mathrm{1}+{z}^{\mathrm{2}} }\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:=\mathrm{2}{i}\pi\:\frac{{e}^{{it}\left({i}\right)} }{\mathrm{2}{i}} \\ $$$$=\pi\:{e}^{−{t}} \:\:\:\:\:{with}\:{t}\geqslant\mathrm{0}\:. \\ $$