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Question Number 82433 by mathmax by abdo last updated on 21/Feb/20
1)decompose inside C(x)and R(x) F=(1/((x^2 +x+1)^2 )) 2)calculate ∫_0 ^∞ (dx/((x^2 +x+1)^2 ))
$$\left.\mathrm{1}\right){decompose}\:{inside}\:{C}\left({x}\right){and}\:{R}\left({x}\right)\:{F}=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 24/Feb/20
x^2 +x+1 =0→Δ =1−4=−3 ⇒z_1 =((−1+i(√3))/2) =e^(i((2π)/3)) z_2 =e^(−((i2π)/3)) ⇒F =(1/((x−e^((i2π)/3) )^2 (x−e^(−((i2π)/3)) )^2 )) =(a/(x−e^((i2π)/3) ))+(b/((x−e^((i2π)/3) )^2 )) +(c/((x−e^(−i((2π)/3)) ))) +(d/((x−e^(−i((2π)/3)) )^2 )) b =(1/((2isin(((2π)/3)))^2 )) =−(1/(4i(((√3)/2))^2 )) =−(1/(3i)) d=(1/((2isin(((2π)/3)))^2 ))=−(1/(3i)) ⇒F(x)=(a/((x−e^((i2π)/3) )))−(1/(3i(x−e^((i2π)/3) )^2 )) +(c/(x−e^(−((i2π)/3)) )) −(1/(3i(x−e^(−((i2π)/3)) )^2 )) lim_(x→+∞) xF(x)=0 =a+c ⇒c=−a F(0)=1 =−ae^(−((i2π)/3)) −(1/(3i))e^(−((i4π)/3)) +ae^((i2π)/3) −(1/(3i))e^((i4π)/3) =2ia sin(((2π)/3))−(1/(3i))(2cos(((4π)/3)))=2ia ×((√3)/2)−(2/(3i))(−(1/2)) =ai(√3)+(1/(3i)) =ai(√3)−(i/3)=1 ⇒ai(√3)=1+(i/3) ⇒ a=(1/(i(√3)))(1+(i/3))=(1/(i(√3)))+(1/(3(√3))) =(1/(3(√3)))−(i/( (√3))) and c=−(1/(3(√3)))+(i/( (√3))) ⇒ F(x)=((1/(3(√3)))−(i/( (√3))))×(1/(x−e^((i2π)/3) )) +(i/(3(x−e^((i2π)/3) )^2 ))−((1/(3(√3)))−(i/( (√3))))×(1/(x−e^(−((i2π)/3)) )) +(i/(3(x−e^(−i((2π)/3)) )^2 ))
$${x}^{\mathrm{2}} \:+{x}+\mathrm{1}\:=\mathrm{0}\rightarrow\Delta\:=\mathrm{1}−\mathrm{4}=−\mathrm{3}\:\Rightarrow{z}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${z}_{\mathrm{2}} ={e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:\Rightarrow{F}\:=\frac{\mathrm{1}}{\left({x}−{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left({x}−{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} }\:=\frac{{a}}{{x}−{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} }+\frac{{b}}{\left({x}−{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} } \\ $$$$+\frac{{c}}{\left({x}−{e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)}\:+\frac{{d}}{\left({x}−{e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} } \\ $$$${b}\:=\frac{\mathrm{1}}{\left(\mathrm{2}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{4}{i}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{3}{i}} \\ $$$${d}=\frac{\mathrm{1}}{\left(\mathrm{2}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{3}{i}}\:\Rightarrow{F}\left({x}\right)=\frac{{a}}{\left({x}−{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)}−\frac{\mathrm{1}}{\mathrm{3}{i}\left({x}−{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} } \\ $$$$+\frac{{c}}{{x}−{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} }\:−\frac{\mathrm{1}}{\mathrm{3}{i}\left({x}−{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:={a}+{c}\:\Rightarrow{c}=−{a} \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{1}\:=−{ae}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} −\frac{\mathrm{1}}{\mathrm{3}{i}}{e}^{−\frac{{i}\mathrm{4}\pi}{\mathrm{3}}} +{ae}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:−\frac{\mathrm{1}}{\mathrm{3}{i}}{e}^{\frac{{i}\mathrm{4}\pi}{\mathrm{3}}} \\ $$$$=\mathrm{2}{ia}\:{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{3}{i}}\left(\mathrm{2}{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{3}}\right)\right)=\mathrm{2}{ia}\:×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}{i}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$={ai}\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}{i}}\:={ai}\sqrt{\mathrm{3}}−\frac{{i}}{\mathrm{3}}=\mathrm{1}\:\Rightarrow{ai}\sqrt{\mathrm{3}}=\mathrm{1}+\frac{{i}}{\mathrm{3}}\:\Rightarrow \\ $$$${a}=\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}}\left(\mathrm{1}+\frac{{i}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}\:=\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}−\frac{{i}}{\:\sqrt{\mathrm{3}}}\:{and}\:{c}=−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}+\frac{{i}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\left(\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}−\frac{{i}}{\:\sqrt{\mathrm{3}}}\right)×\frac{\mathrm{1}}{{x}−{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} }\:+\frac{{i}}{\mathrm{3}\left({x}−{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} }−\left(\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}−\frac{{i}}{\:\sqrt{\mathrm{3}}}\right)×\frac{\mathrm{1}}{{x}−{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} } \\ $$$$+\frac{{i}}{\mathrm{3}\left({x}−{e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} } \\ $$$$ \\ $$

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