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Question Number 103593 by mathmax by abdo last updated on 16/Jul/20
calculate ∫_(−∞) ^∞ (dx/((x^2 +x +1)^2 (2x^2 +5)^2 ))
$$\mathrm{calculate}\:\:\int_{−\infty} ^{\infty} \:\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}\:+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2x}^{\mathrm{2}} \:+\mathrm{5}\right)^{\mathrm{2}} } \\ $$
Answered by mathmax by abdo last updated on 18/Jul/20
A =∫_(−∞) ^(+∞) (dx/((x^2 +x+1)^2 (2x^2 +5)^2 )) let ϕ(z) =(1/((z^2 +z+1)^2 (2z^2 +5)^2 )) poles of ϕ? z^2 +z +1 =0→Δ =−3 ⇒z_1 =((−1+i(√3))/2) =e^((i2π)/3) and z_2 =e^(−((i2π)/3)) 2z^2 +5 =0 ⇒z^2 +(5/2) =0 ⇒z^2 =−(5/2) ⇒z =+^− i(√(5/2)) ⇒ϕ(z) =(1/(4(z−e^((i2π)/3) )^2 (z+e^((i2π)/3) )^2 (z−i(√(5/2)))^2 (z+i(√(5/2)))^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,e^((i2π)/3) )+Res(ϕ,i(√(5/2)))} Res(ϕ,e^((i2π)/3) ) =lim_(z→e^((i2π)/3) ) (1/((2−1)!)){ (z−e^((i2π)/3) )^2 ϕ(z)}^((1)) =lim_(z→e((i2π)/3)) {(1/(4(z+e^((i2π)/3) )^2 (z^2 +(5/2))^2 ))}^((1)) =lim_(z→e^((i2π)/3) ) −(1/4)×((2(z+e^((i2π)/3) )(z^2 +(5/2))^2 +4z (z^2 +(5/2))(z+e^((i2π)/3) )^2 )/((z+e^((i2π)/3) )^4 (z^2 +(5/2))^4 )) =−(1/4)lim_(z→e^((i2π)/3) ) {(2/((z+e^((i2π)/3) )^3 (z^2 +(5/2))^2 )) +((4z)/((z+e^((i2π)/3) )^2 (z^2 +(5/2))^3 ))} ...be continued...
$$\mathrm{A}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2x}^{\mathrm{2}} \:+\mathrm{5}\right)^{\mathrm{2}} }\:\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{z}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2z}^{\mathrm{2}} \:+\mathrm{5}\right)^{\mathrm{2}} }\:\mathrm{poles}\:\mathrm{of}\:\varphi? \\ $$$$\mathrm{z}^{\mathrm{2}} \:+\mathrm{z}\:+\mathrm{1}\:=\mathrm{0}\rightarrow\Delta\:=−\mathrm{3}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \: \\ $$$$\mathrm{2z}^{\mathrm{2}} \:+\mathrm{5}\:=\mathrm{0}\:\Rightarrow\mathrm{z}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{2}}\:=\mathrm{0}\:\Rightarrow\mathrm{z}^{\mathrm{2}} \:=−\frac{\mathrm{5}}{\mathrm{2}}\:\Rightarrow\mathrm{z}\:=\overset{−} {+}\mathrm{i}\sqrt{\frac{\mathrm{5}}{\mathrm{2}}} \\ $$$$\Rightarrow\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left(\mathrm{z}−\mathrm{i}\sqrt{\frac{\mathrm{5}}{\mathrm{2}}}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{i}\sqrt{\frac{\mathrm{5}}{\mathrm{2}}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{residus}\:\mathrm{theorem}\:\mathrm{give} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\left\{\:\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)+\mathrm{Res}\left(\varphi,\mathrm{i}\sqrt{\frac{\mathrm{5}}{\mathrm{2}}}\right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} } \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \:\varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}\frac{\mathrm{i2}\pi}{\mathrm{3}}} \:\:\left\{\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} \:+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} } \:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{2}\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}^{\mathrm{2}} \:+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\mathrm{4z}\:\left(\mathrm{z}^{\mathrm{2}} \:+\frac{\mathrm{5}}{\mathrm{2}}\right)\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} }{\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{4}} \left(\mathrm{z}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{4}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} } \:\:\:\:\:\left\{\frac{\mathrm{2}}{\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{3}} \left(\mathrm{z}^{\mathrm{2}} \:+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} }\:+\frac{\mathrm{4z}}{\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} \:+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{3}} }\right\} \\ $$$$…\mathrm{be}\:\mathrm{continued}… \\ $$

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