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Question Number 88930 by mathmax by abdo last updated on 13/Apr/20
find A_λ =∫_0 ^∞ ((cos(λx))/((x^2 −x+1)^2 ))dx with λ>0 2)find the value of ∫_0 ^∞ ((cos(3x))/((x^2 −x+1)^2 ))dx
$${find}\:{A}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\lambda{x}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:{with}\:\lambda>\mathrm{0} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 14/Apr/20
A_λ =∫_0 ^∞ ((cos(λx))/((x^2 −x+1)^2 ))dx ⇒2A_λ =∫_(−∞) ^(+∞) ((cos(λx))/((x^2 −x+1)^2 ))dx =Re(∫_(−∞) ^(+∞) (e^(iλx) /((x^2 −x+1)^2 ))dx) let ϕ(z) =(e^(iλz) /((z^2 −z +1)^2 )) poles of ϕ? z^2 −z +1 =→Δ =1−4 =−3 ⇒z_1 =((1+i(√3))/2) =e^((iπ)/3) z_2 =((1−i(√3))/2) =e^(−((iπ)/3)) ⇒ϕ(z) =(e^(iλz) /((z−e^((iπ)/3) )^2 (z−e^(−((iπ)/3)) )^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^((iπ)/3) ) Res(ϕ,e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (1/((2−1)!)){(z−e^((iπ)/3) )^2 ϕ(z)}^((1)) =lim_(z→e^((iπ)/3) ) {(z−e^(−((iπ)/3)) )^(−2) e^(iλz) }^((1)) =lim_(z→e^((iπ)/3) ) {−2(z−e^(−((iπ)/3)) )^(−3) e^(iλz) +iλ(z−e^(−((iπ)/3)) )^(−2) e^(iλz) } ={−2 (2isin((π/3))^(−3) +iλ (2i sin((π/3)))^(−2) } e^(iλe^((iπ)/3) ) ={((−2)/((i(√3))^3 )) +((iλ)/((i(√3))^2 ))} e^(iλ((1/2)+((i(√3))/2))) ={ (2/(3i(√3))) −((iλ)/3)} e^(−((√3)/2)λ) {cos((λ/2))+isin((λ/2))} ={((−2i)/(3(√3)))−((iλ)/3)}{ cos((λ/2))+isin((λ/2))}e^(−((√3)/2)λ) =((2/(3(√3)))+(λ/3))(−icos((λ/2))+sin((λ/2))e^(−((√3)/2)λ) ⇒ 2A_λ =((2/(3(√3)))+(λ/3))sin((λ/2))e^(−((√3)/2)λ) ⇒ A_λ =((1/(3(√3))) +(λ/6))sin((λ/2))e^(−((√3)/2)λ)
$${A}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left(\lambda{x}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\Rightarrow\mathrm{2}{A}_{\lambda} =\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\lambda{x}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\lambda{x}} }{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\right)\:{let}\:\varphi\left({z}\right)\:=\frac{{e}^{{i}\lambda{z}} }{\left({z}^{\mathrm{2}} −{z}\:+\mathrm{1}\right)^{\mathrm{2}} }\:\:{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{2}} −{z}\:+\mathrm{1}\:=\rightarrow\Delta\:=\mathrm{1}−\mathrm{4}\:=−\mathrm{3}\:\Rightarrow{z}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={e}^{\frac{{i}\pi}{\mathrm{3}}} \\ $$$${z}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={e}^{−\frac{{i}\pi}{\mathrm{3}}} \:\Rightarrow\varphi\left({z}\right)\:=\frac{{e}^{{i}\lambda{z}} }{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left({z}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} } \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{3}}} \right) \\ $$$${Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\:={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{3}}} } \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{3}}} } \:\left\{\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)^{−\mathrm{2}} \:{e}^{{i}\lambda{z}} \right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{3}}} } \:\:\:\:\left\{−\mathrm{2}\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)^{−\mathrm{3}} \:{e}^{{i}\lambda{z}} \:\:+{i}\lambda\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)^{−\mathrm{2}} \:{e}^{{i}\lambda{z}} \right\} \\ $$$$=\left\{−\mathrm{2}\:\:\left(\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{3}}\right)^{−\mathrm{3}} \:\:+{i}\lambda\:\left(\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{3}}\right)\right)^{−\mathrm{2}} \right\}\:{e}^{{i}\lambda{e}^{\frac{{i}\pi}{\mathrm{3}}} } \right. \\ $$$$=\left\{\frac{−\mathrm{2}}{\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }\:+\frac{{i}\lambda}{\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\right\}\:{e}^{{i}\lambda\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \\ $$$$=\left\{\:\:\frac{\mathrm{2}}{\mathrm{3}{i}\sqrt{\mathrm{3}}}\:−\frac{{i}\lambda}{\mathrm{3}}\right\}\:{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\lambda} \left\{{cos}\left(\frac{\lambda}{\mathrm{2}}\right)+{isin}\left(\frac{\lambda}{\mathrm{2}}\right)\right\} \\ $$$$=\left\{\frac{−\mathrm{2}{i}}{\mathrm{3}\sqrt{\mathrm{3}}}−\frac{{i}\lambda}{\mathrm{3}}\right\}\left\{\:{cos}\left(\frac{\lambda}{\mathrm{2}}\right)+{isin}\left(\frac{\lambda}{\mathrm{2}}\right)\right\}{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\lambda} \\ $$$$=\left(\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}+\frac{\lambda}{\mathrm{3}}\right)\left(−{icos}\left(\frac{\lambda}{\mathrm{2}}\right)+{sin}\left(\frac{\lambda}{\mathrm{2}}\right){e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\lambda} \:\Rightarrow\right. \\ $$$$\mathrm{2}{A}_{\lambda} =\left(\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}+\frac{\lambda}{\mathrm{3}}\right){sin}\left(\frac{\lambda}{\mathrm{2}}\right){e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\lambda} \:\Rightarrow \\ $$$${A}_{\lambda} =\left(\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}\:+\frac{\lambda}{\mathrm{6}}\right){sin}\left(\frac{\lambda}{\mathrm{2}}\right){e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\lambda} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 14/Apr/20
∫_0 ^∞ ((cos(3x))/((x^2 −x+1)^2 ))dx =A_3 =((1/(3(√3)))+(1/2))sin((3/2))e^(−((3(√3))/2))
$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:={A}_{\mathrm{3}} =\left(\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{3}}{\mathrm{2}}\right){e}^{−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$

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