Question Number 82816 by M±th+et£s last updated on 24/Feb/20
$${calculate}\:{the}\:{exact}\:{value}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$
Commented by msup trace by abdo last updated on 24/Feb/20
$${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{cosx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{cosx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\Rightarrow\mathrm{2}{I}={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{ix}} }{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\right)\:{let} \\ $$$$\varphi\left({z}\right)=\frac{{e}^{{iz}} }{{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\varphi\left({z}\right)=\frac{{e}^{{iz}} }{\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$$${redidus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Re}\left(\varphi,{i}\right) \\ $$$$=\mathrm{2}{i}\pi×\frac{{e}^{−\mathrm{1}} }{\mathrm{2}{i}}\:=\pi{e}^{−\mathrm{1}} \:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{2}}{e}^{−\mathrm{1}} \:\Rightarrow{I}=\frac{\pi}{\mathrm{2}{e}} \\ $$
Commented by M±th+et£s last updated on 24/Feb/20
$${thank}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 24/Feb/20
$${you}\:{are}\left[{welcome}\right. \\ $$