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cos-x-1-x-2-dx-pi-e-




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

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