Menu Close

calculate-the-exact-value-0-cos-x-x-2-1-dx-




Question Number 82816 by M±th+et£s last updated on 24/Feb/20
calculate the exact value   ∫_0 ^∞ ((cos(x))/(x^2 +1)) dx
$${calculate}\:{the}\:{exact}\:{value}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$
Commented by msup trace by abdo last updated on 24/Feb/20
I=∫_0 ^∞  ((cosx)/(1+x^2 ))dx ⇒I=(1/2)∫_(−∞) ^(+∞) ((cosx)/(1+x^2 ))dx  ⇒2I=Re(∫_(−∞) ^(+∞)  (e^(ix) /(x^2  +1))dx) let  ϕ(z)=(e^(iz) /(z^2  +1)) ⇒ϕ(z)=(e^(iz) /((z−i)(z+i)))  redidus theorem give  ∫_(−∞) ^(+∞)  ϕ(z)dz=2iπ Re(ϕ,i)  =2iπ×(e^(−1) /(2i)) =πe^(−1)  ⇒  I =(π/2)e^(−1)  ⇒I=(π/(2e))
$${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{cosx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{cosx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\Rightarrow\mathrm{2}{I}={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{ix}} }{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\right)\:{let} \\ $$$$\varphi\left({z}\right)=\frac{{e}^{{iz}} }{{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\varphi\left({z}\right)=\frac{{e}^{{iz}} }{\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$$${redidus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Re}\left(\varphi,{i}\right) \\ $$$$=\mathrm{2}{i}\pi×\frac{{e}^{−\mathrm{1}} }{\mathrm{2}{i}}\:=\pi{e}^{−\mathrm{1}} \:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{2}}{e}^{−\mathrm{1}} \:\Rightarrow{I}=\frac{\pi}{\mathrm{2}{e}} \\ $$
Commented by M±th+et£s last updated on 24/Feb/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 24/Feb/20
you are[welcome
$${you}\:{are}\left[{welcome}\right. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *