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Question Number 98721 by mathmax by abdo last updated on 15/Jun/20

calculate ∫_0 ^∞ (dx/(x^4 +x^2 +1)) 1) by using residue theorem 2) by using complex decomposition

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{by}\:\mathrm{using}\:\mathrm{residue}\:\mathrm{theorem} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{by}\:\mathrm{using}\:\mathrm{complex}\:\mathrm{decomposition} \\ $$

Answered by mathmax by abdo last updated on 16/Jun/20

let I =∫_0 ^∞ (dx/(x^4 +x^2 +1)) ⇒2I =∫_(−∞) ^(+∞) (dx/(x^4 +x^2 +1)) let ϕ(z) =(1/(z^4 +z^2 +1)) poles of ϕ? z^4 +z^2 +1 =0 ⇒u^2 +u+1 =0 (u=z^2 ) Δ =−3 ⇒u_1 =((−1+i(√3))/2) =e^(i((2π)/3)) , u_2 =((−1−i(√3))/2) =e^(−((i2π)/3)) ⇒ ϕ(z) =(1/((z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) ))) =(1/((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^(−((iπ)/3)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,e^((iπ)/3) ) +Res(ϕ,−e^(−((iπ)/3)) )} Res(ϕ,e^((iπ)/3) ) =(1/(2e^((iπ)/3) (2isin(((2π)/3))))) =(e^(−((iπ)/3)) /(4i (((√3)/2)))) =(e^(−((iπ)/3)) /(2i(√3))) Res(ϕ,−e^(−((iπ)/3)) ) =(1/(−2e^(−((iπ)/3)) (−2isin(((2π)/3))))) =(e^((iπ)/3) /(4i(((√3)/2)))) =(e^((iπ)/3) /(2i(√3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(e^(−((iπ)/3)) /(2i(√3))) +(e^((iπ)/3) /(2i(√3)))} =(π/(√3))(2cos((π/3))) =(π/(√3))(2×(1/2)) =(π/(√3)) ⇒I =(π/(2(√3)))

$$\mathrm{let}\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\mathrm{2I}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{4}} \:+\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\:\mathrm{poles}\:\mathrm{of}\:\varphi? \\ $$$$\mathrm{z}^{\mathrm{4}} \:+\mathrm{z}^{\mathrm{2}} +\mathrm{1}\:=\mathrm{0}\:\Rightarrow\mathrm{u}^{\mathrm{2}} \:+\mathrm{u}+\mathrm{1}\:=\mathrm{0}\:\left(\mathrm{u}=\mathrm{z}^{\mathrm{2}} \right) \\ $$$$\Delta\:=−\mathrm{3}\:\Rightarrow\mathrm{u}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:\:,\:\:\mathrm{u}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \:\Rightarrow \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}^{\mathrm{2}} −\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)}\:=\frac{\mathrm{1}}{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:+\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:=\frac{\mathrm{1}}{\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \left(\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)}\:=\frac{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{4i}\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}\:=\frac{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2i}\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:=\frac{\mathrm{1}}{−\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \left(−\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)}\:=\frac{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{4i}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}\:=\frac{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\frac{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2i}\sqrt{\mathrm{3}}}\:+\frac{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2i}\sqrt{\mathrm{3}}}\right\}\:=\frac{\pi}{\sqrt{\mathrm{3}}}\left(\mathrm{2cos}\left(\frac{\pi}{\mathrm{3}}\right)\right)\:=\frac{\pi}{\sqrt{\mathrm{3}}}\left(\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\frac{\pi}{\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\mathrm{I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 16/Jun/20

2) let use complex decompositon 2I =∫_(−∞) ^(+∞) (dx/(x^4 +x^2 +1)) x^4 +x^2 +1 =0 ⇒t^2 +t +1 =0 (t=x^2 ) Δ =−3 ⇒t_1 =((−1+i(√3))/2) =e^((i2π)/3) and t_2 =((−1−i(√3))/2) =e^(−((i2π)/3)) ⇒F(x) =(1/(x^4 +x^2 +1)) =(1/((x^2 −e^((i2π)/3) )(x^2 −e^(−((i2π)/3)) ))) =(1/((x−e^((iπ)/3) )(x+e^((iπ)/3) )(x−e^(−((iπ)/3)) )(x+e^(−((iπ)/3)) ))) =(a/(x−e^((iπ)/3) )) +(b/(x+e^((iπ)/3) )) +(c/(x−e^(−((iπ)/3)) )) +(d/(x+e^(−((iπ)/3)) )) a =(1/(2e^((iπ)/3) (2isin(((2π)/3))))) =(e^(−((iπ)/3)) /(2i(√3))) , b =(1/(−2e^((iπ)/3) (2isin(((2π)/3))))) =−(e^(−((iπ)/3)) /(2i(√3))) c =(1/(2e^(−((iπ)/3)) (−2isin(((2π)/3))))) =−(e^((iπ)/3) /(2i(√3))) , d =(1/(−2e^(−((iπ)/3)) (−2isin(((2π)/3))))) =(e^((iπ)/3) /(2i(√3))) ⇒∫_(−∞) ^(+∞) F(x)dx =(1/(2i(√3))){ e^(−((iπ)/3)) ∫_(−∞) ^(+∞) (dx/(x−e^((iπ)/3) )) −e^(−((iπ)/3)) ∫_(−∞) ^(+∞) (dx/(x+e^((iπ)/3) ))−e^((iπ)/3) ∫_(−∞) ^(+∞) (dx/(x−e^(−((iπ)/3)) )) +e^((iπ)/3) ∫_(−∞) ^(+∞) (dx/(x+e^(−((iπ)/3)) ))} we have proved thst ∫_(−∞) ^(+∞) (dx/(x−a)) =iπ if im(a)>0 and −iπ if Im(a)<0 ⇒ ∫_(−∞) ^(+∞) F(x)dx =(1/(2i(√3))){(iπ)e^(−((iπ)/3)) −(−iπ)e^(−((iπ)/3)) −(−iπ)e^((iπ)/3) +(iπ)e^((iπ)/3) } =(π/(2(√3))){ e^(−((iπ)/3)) +e^(−((iπ)/3)) +e^((iπ)/3) +e^((iπ)/3) } =(π/(√3)){e^((iπ)/3) +e^(−((iπ)/3)) } =(π/(√3)){2cos((π/3))) =(π/(√3)) =2I ⇒ I =(π/(2(√3)))

$$\left.\mathrm{2}\right)\:\mathrm{let}\:\mathrm{use}\:\mathrm{complex}\:\mathrm{decompositon}\:\:\mathrm{2I}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\mathrm{x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow\mathrm{t}^{\mathrm{2}} \:+\mathrm{t}\:+\mathrm{1}\:=\mathrm{0}\:\:\left(\mathrm{t}=\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\Delta\:=−\mathrm{3}\:\Rightarrow\mathrm{t}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \:\:\mathrm{and}\:\:\mathrm{t}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \\ $$$$\Rightarrow\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)}\:=\frac{\mathrm{1}}{\left(\mathrm{x}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{x}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{x}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{x}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{\mathrm{a}}{\mathrm{x}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }\:+\frac{\mathrm{b}}{\mathrm{x}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }\:+\frac{\mathrm{c}}{\mathrm{x}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} }\:+\frac{\mathrm{d}}{\mathrm{x}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} } \\ $$$$\mathrm{a}\:=\frac{\mathrm{1}}{\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \left(\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)}\:=\frac{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2i}\sqrt{\mathrm{3}}}\:\:,\:\:\mathrm{b}\:=\frac{\mathrm{1}}{−\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \left(\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)}\:=−\frac{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2i}\sqrt{\mathrm{3}}} \\ $$$$\mathrm{c}\:=\frac{\mathrm{1}}{\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \left(−\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)}\:=−\frac{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2i}\sqrt{\mathrm{3}}}\:\:,\:\:\mathrm{d}\:=\frac{\mathrm{1}}{−\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \left(−\mathrm{2isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)}\:=\frac{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{2i}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2i}\sqrt{\mathrm{3}}}\left\{\:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }\:−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} }\right. \\ $$$$\left.+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} }\right\}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{proved}\:\mathrm{thst}\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}−\mathrm{a}}\:=\mathrm{i}\pi\:\mathrm{if}\:\mathrm{im}\left(\mathrm{a}\right)>\mathrm{0}\:\mathrm{and} \\ $$$$−\mathrm{i}\pi\:\mathrm{if}\:\mathrm{Im}\left(\mathrm{a}\right)<\mathrm{0}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2i}\sqrt{\mathrm{3}}}\left\{\left(\mathrm{i}\pi\right)\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\:−\left(−\mathrm{i}\pi\right)\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:−\left(−\mathrm{i}\pi\right)\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:+\left(\mathrm{i}\pi\right)\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right\} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\left\{\:\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right\}\:=\frac{\pi}{\sqrt{\mathrm{3}}}\left\{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right\}\:=\frac{\pi}{\sqrt{\mathrm{3}}}\left\{\mathrm{2cos}\left(\frac{\pi}{\mathrm{3}}\right)\right) \\ $$$$=\frac{\pi}{\sqrt{\mathrm{3}}}\:=\mathrm{2I}\:\Rightarrow\:\mathrm{I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$

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