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Question Number 37357 by math khazana by abdo last updated on 12/Jun/18
let a>0 b from C and Re(b)>0 1) calculate ∫_(−∞) ^(+∞) ((b cos(ax))/(x^2 +b^2 ))dx 2) find the value of ∫_(−∞) ^(+∞) ((x sin(ax))/(x^2 +b^2 )) dx.
$${let}\:{a}>\mathrm{0}\:{b}\:{from}\:{C}\:{and}\:\:{Re}\left({b}\right)>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{b}\:{cos}\left({ax}\right)}{{x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}\:{sin}\left({ax}\right)}{{x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:{dx}. \\ $$
Commented by abdo.msup.com last updated on 13/Jun/18
1) let I =∫_(−∞) ^(+∞) ((b cos(ax))/(x^(2 ) +b^2 ))dx I = Re( ∫_(−∞) ^(+∞) ((b e^(iax) )/(x^2 +b^2 ))dx) let consider ϕ(z) = ((b e^(iaz) )/(z^2 +b^2 )) ϕ(z) =((be^(iax) )/((z−ib)(z+ib))) so the poles of ϕ are ib and −ib we have Re(b)>0 ⇒ Im(ib)>0 so ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,ib) Res(ϕ,ib) =lim_(z→ib) (z−ib)ϕ(z) = ((b e^(ia(ib)) )/(2ib)) = (e^(−ab) /(2i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(−ab) /(2i)) =π e^(−ab) ⇒ I = π e^(−ab) .
$$\left.\mathrm{1}\right)\:{let}\:{I}\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{b}\:{cos}\left({ax}\right)}{{x}^{\mathrm{2}\:} \:+{b}^{\mathrm{2}} }{dx} \\ $$$${I}\:=\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{b}\:{e}^{{iax}} }{{x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }{dx}\right)\:{let}\:{consider} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{b}\:{e}^{{iaz}} }{{z}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\: \\ $$$$\varphi\left({z}\right)\:=\frac{{be}^{{iax}} }{\left({z}−{ib}\right)\left({z}+{ib}\right)}\:{so}\:{the}\:{poles}\:{of}\:\varphi \\ $$$${are}\:{ib}\:{and}\:−{ib}\:\:{we}\:{have}\:{Re}\left({b}\right)>\mathrm{0}\:\Rightarrow \\ $$$${Im}\left({ib}\right)>\mathrm{0}\:{so} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{ib}\right) \\ $$$${Res}\left(\varphi,{ib}\right)\:={lim}_{{z}\rightarrow{ib}} \:\left({z}−{ib}\right)\varphi\left({z}\right) \\ $$$$=\:\frac{{b}\:{e}^{{ia}\left({ib}\right)} }{\mathrm{2}{ib}}\:=\:\frac{{e}^{−{ab}} }{\mathrm{2}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−{ab}} }{\mathrm{2}{i}}\:=\pi\:{e}^{−{ab}} \:\Rightarrow \\ $$$${I}\:\:=\:\pi\:{e}^{−{ab}} \:\:. \\ $$
Commented by abdo.msup.com last updated on 13/Jun/18
2) let J = ∫_(−∞) ^(+∞) ((x sin(ax))/(x^2 +b^2 ))dx J = Im( ∫_(−∞) ^(+∞) ((x e^(iax) )/(x^2 +b^2 )) dx) let ψ(z) = ((z e^(iaz) )/(z^2 +b^2 )) the poles of ψ are ib and −ib ∫_(−∞) ^(+∞) ψ(z)dz =2iπ Res(ψ,ib) =2iπ ((ib e^(ia(ib)) )/(2ib)) = iπ e^(−ab) ⇒J =πe^(−ab)
$$\left.\mathrm{2}\right)\:{let}\:{J}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}\:{sin}\left({ax}\right)}{{x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }{dx}\: \\ $$$${J}\:=\:{Im}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}\:{e}^{{iax}} }{{x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:{dx}\right)\:\:{let} \\ $$$$\psi\left({z}\right)\:=\:\frac{{z}\:{e}^{{iaz}} }{{z}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:{the}\:{poles}\:{of}\:\psi\:{are} \\ $$$${ib}\:{and}\:−{ib} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\psi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\psi,{ib}\right) \\ $$$$=\mathrm{2}{i}\pi\:\:\frac{{ib}\:{e}^{{ia}\left({ib}\right)} }{\mathrm{2}{ib}}\:=\:{i}\pi\:{e}^{−{ab}} \:\:\Rightarrow{J}\:=\pi{e}^{−{ab}} \: \\ $$

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