Question Number 80925 by M±th+et£s last updated on 08/Feb/20
$$\int_{−\infty} ^{\infty} \frac{{cos}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:=\frac{\pi}{{e}} \\ $$
Commented by mathmax by abdo last updated on 10/Feb/20
$${I}=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\Rightarrow{I}={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{ix}} }{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\right)\:{let}\: \\ $$$$\varphi\left({z}\right)=\frac{{e}^{{iz}} }{{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\varphi\left({z}\right)=\frac{{e}^{{iz}} }{\left({z}−{i}\right)\left({z}+{i}\right)}{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({zdz}?=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)=\mathrm{2}{i}\pi\:\:\frac{{e}^{−\mathrm{1}} }{\mathrm{2}{i}}\:=\frac{\pi}{{e}}\:\Rightarrow\:{I}=\frac{\pi}{{e}}\right. \\ $$