Menu Close

let-I-cos-x-1-ix-2-dx-1-extract-Re-I-and-Im-I-2-calculate-I-3-conclude-the-values-of-Re-I-and-Im-I-

Tinku Tara 0 Comments
Pin




Question Number 39370 by maxmathsup by imad last updated on 05/Jul/18
let I (λ) = ∫_(−∞) ^(+∞) ((cos(λx))/((1+ix)^2 ))dx 1) extract Re(I(λ)) and Im(I(λ)) 2) calculate I(λ) 3) conclude the values of Re(I(λ)) and Im(I(λ)).
$${let}\:{I}\:\left(\lambda\right)\:=\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\lambda{x}\right)}{\left(\mathrm{1}+{ix}\right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{1}\right)\:\:{extract}\:{Re}\left({I}\left(\lambda\right)\right)\:{and}\:{Im}\left({I}\left(\lambda\right)\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{I}\left(\lambda\right) \\ $$$$\left.\mathrm{3}\right)\:{conclude}\:\:{the}\:{values}\:{of}\:{Re}\left({I}\left(\lambda\right)\right)\:{and}\:{Im}\left({I}\left(\lambda\right)\right). \\ $$
Commented by abdo mathsup 649 cc last updated on 08/Jul/18
1) we have I(λ) = ∫_(−∞) ^(+∞) ((cos(λx)(1−ix)^2 )/((1+ix)^2 (1−ix)^2 )) = ∫_(−∞) ^(+∞) ((cos(λx)(1−ix)^2 )/((1+x^2 )^2 ))dx = ∫_(−∞) ^(+∞) ((cos(λx)(1−2ix −x^2 ))/((1+x^2 )^2 ))dx = ∫_(−∞) ^(+∞) (((1−x^2 )cos(λx) −2i x cos(λx))/((1+x^2 )^2 ))dx = ∫_(−∞) ^(+∞) (((1−x^2 )cos(λx))/((1+x^2 )^2 ))dx −i ∫_(−∞) ^(+∞) ((x cos(λx))/((1+x^2 )^2 ))dx⇒ Re( I(λ)) = ∫_(−∞) ^(+∞) (((1−x^2 )cos(λx))/((1+x^2 )^2 )) dx and Im( I(λ)) =−∫_(−∞) ^(+∞) ((x cos(λx))/((1+x^2 )^2 ))dx =0 2) I(λ) = Re( ∫_(−∞) ^(+∞) (e^(iλx) /((1+ix)^2 ))dx) let ϕ(z) = (e^(iλz) /((1+iz)^2 )) poles of ϕ? ϕ(z) = (e^(iλz) /((iz−i^2 )^2 )) = (e^(iλz) /((−1)(z−i)^2 )) =−(e^(iλz) /((z−i)^2 )) so is a double pole for ϕ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) = lim_(z→i) (1/((2−1)!)) { (z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {−e^(iλz) }^((1)) =lim_(z→i) −iλ e^(iλz) =−iλ e^(−λ) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ(−i λ e^(−λ) ) = 2π λ e^(−λ) ⇒ I(I(λ)) =2π λ e^(−λ) 3) Re(I(λ)) = ∫_(−∞) ^(+∞) (((1−x^2 )cos(λx))/((1+x^2 )^2 ))dx= 2πλ e^(−λ) and Im(I(λ))=0
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\:{I}\left(\lambda\right)\:=\:\int_{−\infty} ^{+\infty} \frac{{cos}\left(\lambda{x}\right)\left(\mathrm{1}−{ix}\right)^{\mathrm{2}} }{\left(\mathrm{1}+{ix}\right)^{\mathrm{2}} \left(\mathrm{1}−{ix}\right)^{\mathrm{2}} }\: \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\lambda{x}\right)\left(\mathrm{1}−{ix}\right)^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\lambda{x}\right)\left(\mathrm{1}−\mathrm{2}{ix}\:−{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right){cos}\left(\lambda{x}\right)\:−\mathrm{2}{i}\:{x}\:{cos}\left(\lambda{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right){cos}\left(\lambda{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:−{i}\:\int_{−\infty} ^{+\infty} \:\:\frac{{x}\:{cos}\left(\lambda{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\Rightarrow \\ $$$${Re}\left(\:{I}\left(\lambda\right)\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right){cos}\left(\lambda{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:{and} \\ $$$${Im}\left(\:{I}\left(\lambda\right)\right)\:=−\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}\:{cos}\left(\lambda{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{I}\left(\lambda\right)\:=\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\lambda{x}} }{\left(\mathrm{1}+{ix}\right)^{\mathrm{2}} }{dx}\right)\:{let} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\lambda{z}} }{\left(\mathrm{1}+{iz}\right)^{\mathrm{2}} }\:\:{poles}\:{of}\:\varphi? \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{{e}^{{i}\lambda{z}} }{\left({iz}−{i}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:\frac{{e}^{{i}\lambda{z}} }{\left(−\mathrm{1}\right)\left({z}−{i}\right)^{\mathrm{2}} }\:=−\frac{{e}^{{i}\lambda{z}} }{\left({z}−{i}\right)^{\mathrm{2}} } \\ $$$${so}\:{is}\:{a}\:{double}\:{pole}\:{for}\:\varphi \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\: \\ $$$${Res}\left(\varphi,{i}\right)\:=\:{lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\left\{\:\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\left\{−{e}^{{i}\lambda{z}} \right\}^{\left(\mathrm{1}\right)} ={lim}_{{z}\rightarrow{i}} \:−{i}\lambda\:{e}^{{i}\lambda{z}} \\ $$$$=−{i}\lambda\:{e}^{−\lambda} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(−{i}\:\lambda\:{e}^{−\lambda} \right)\:=\:\mathrm{2}\pi\:\lambda\:{e}^{−\lambda} \:\:\Rightarrow \\ $$$${I}\left({I}\left(\lambda\right)\right)\:=\mathrm{2}\pi\:\lambda\:{e}^{−\lambda} \\ $$$$\left.\mathrm{3}\right)\:{Re}\left({I}\left(\lambda\right)\right)\:=\:\int_{−\infty} ^{+\infty} \:\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right){cos}\left(\lambda{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\:\mathrm{2}\pi\lambda\:{e}^{−\lambda} \\ $$$${and}\:\:{Im}\left({I}\left(\lambda\right)\right)=\mathrm{0} \\ $$

Pin

Leave a Reply

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