Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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 byabdo.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 byabdo.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)

$$\left.\mathrm{2}\right)\:{let}\:{J}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}\:{sin}\left({ax}\right)}{{x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }{dx}\: \\ $$ $${J}\:=\:{Im}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}\:{e}^{{iax}} }{{x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:{dx}\right)\:\:{let} \\ $$ $$\psi\left({z}\right)\:=\:\frac{{z}\:{e}^{{iaz}} }{{z}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:{the}\:{poles}\:{of}\:\psi\:{are} \\ $$ $${ib}\:{and}\:−{ib} \\ $$ $$\int_{−\infty} ^{+\infty} \:\:\psi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\psi,{ib}\right) \\ $$ $$=\mathrm{2}{i}\pi\:\:\frac{{ib}\:{e}^{{ia}\left({ib}\right)} }{\mathrm{2}{ib}}\:=\:{i}\pi\:{e}^{−{ab}} \:\:\Rightarrow{J}\:=\pi{e}^{−{ab}} \: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com