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Question Number 61921 by maxmathsup by imad last updated on 11/Jun/19
let A =∫_(−∞) ^(+∞) ((x+1)/((x^2 +x+1)( x^2 −2i)))dx 1) calculate A 2) extract Re(A) and Im(A) and determine its values (i^2 =−1)
$${let}\:{A}\:=\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left(\:{x}^{\mathrm{2}} \:−\mathrm{2}{i}\right)}{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A} \\ $$$$\left.\mathrm{2}\right)\:{extract}\:{Re}\left({A}\right)\:{and}\:{Im}\left({A}\right)\:{and}\:{determine}\:{its}\:{values}\:\:\:\left({i}^{\mathrm{2}} =−\mathrm{1}\right) \\ $$
Commented by maxmathsup by imad last updated on 12/Jun/19
1) residus method let w(z) =((z+1)/((z^2 +z +1)(z^2 −2i))) poles of w? z^2 +z +1 =0 →Δ=1−4 =−3 =(i(√3))^2 ⇒z_1 =((−1+i(√3))/2) =e^(i((2π)/3)) z_2 =((−1−i(√3))/2) =e^(−i((2π)/3)) also z^2 −2i =z^2 −((√2)(√i))^2 =(z−(√2)e^((iπ)/4) )(z+(√2)e^((iπ)/4) ) ⇒ w(z) =((z+1)/((z−e^(i((2π)/3)) )(z+e^((i2π)/3) )(z−(√2)e^((iπ)/4) )(z+(√2)e^((iπ)/4) ))) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ{ Res(w,e^(i((2π)/3)) ) +Res(w,(√2)e^((iπ)/4) )} the poles are simples ⇒ Res(w,e^((i2π)/3) ) =lim_(z→e^((i2π)/3) ) (z−e^((i2π)/3) )w(z) =((1+e^((i2π)/3) )/(2e^((i2π)/3) ( e^(i((4π)/3)) −2i))) =((e^(−((i2π)/3)) +1)/(2( e^(−((i2π)/3)) −2i))) =−((1+e^(−((i2π)/3)) )/(2i(ie^(−((i2π)/3)) +2))) Res(w,(√2)e^((iπ)/4) ) =lim_(z→(√2)e^((iπ)/4) ) (z−(√2)e^((iπ)/4) ) w(z) =((1+(√2)e^((iπ)/4) )/(2(√2)e^((iπ)/4) ( 2 e^((iπ)/2) +(√2)e^((iπ)/4) +1))) =((e^(−((iπ)/4)) +(√2))/(2(√2)(2i +(√2)e^((iπ)/4) +1))) =((e^(−((iπ)/4)) +(√2))/(2(√2)i(2 −i(√2)e^((iπ)/4) −i))) ⇒ ∫_(−∞) ^(+∞) w(z)dz =2iπ{ −((1+e^(−((i2π)/3)) )/(2i(i e^(−((i2π)/3)) +2))) +((e^(−((iπ)/4)) +(√2))/(2(√2)i(2−i−i(√2)e^((iπ)/4) ))} =−π ((1+e^(−i((2π)/3)) )/(2+ie^(−((i2π)/3)) )) +(π/( (√2))) (((√2)+e^(−((iπ)/4)) )/(2−i −i(√2)e^((iπ)/4) )) =A rest to find A at form α+iλ
$$\left.\mathrm{1}\right)\:{residus}\:{method}\:\:\:{let}\:{w}\left({z}\right)\:=\frac{{z}+\mathrm{1}}{\left({z}^{\mathrm{2}} \:+{z}\:+\mathrm{1}\right)\left({z}^{\mathrm{2}} −\mathrm{2}{i}\right)}\:\:{poles}\:{of}\:{w}? \\ $$$${z}^{\mathrm{2}} \:+{z}\:+\mathrm{1}\:=\mathrm{0}\:\rightarrow\Delta=\mathrm{1}−\mathrm{4}\:=−\mathrm{3}\:=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \: \\ $$$${z}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:{also}\:\:{z}^{\mathrm{2}} −\mathrm{2}{i}\:={z}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}\sqrt{{i}}\right)^{\mathrm{2}} \:=\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:\Rightarrow \\ $$$${w}\left({z}\right)\:=\frac{{z}+\mathrm{1}}{\left({z}−{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)\left({z}+{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left({w},{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)\:+{Res}\left({w},\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\right\}\:\:{the}\:{poles}\:{are}\:{simples}\:\Rightarrow \\ $$$${Res}\left({w},{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)\:={lim}_{{z}\rightarrow{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} } \:\:\:\left({z}−{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right){w}\left({z}\right)\:=\frac{\mathrm{1}+{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} }{\mathrm{2}{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \left(\:{e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{3}}} −\mathrm{2}{i}\right)} \\ $$$$=\frac{{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:+\mathrm{1}}{\mathrm{2}\left(\:{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} −\mathrm{2}{i}\right)}\:=−\frac{\mathrm{1}+{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} }{\mathrm{2}{i}\left({ie}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:+\mathrm{2}\right)} \\ $$$${Res}\left({w},\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\:\left({z}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:{w}\left({z}\right)\:=\frac{\mathrm{1}+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{2}\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\:\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{2}}} \:+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:+\mathrm{1}\right)} \\ $$$$=\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:+\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{2}{i}\:+\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:+\mathrm{1}\right)}\:=\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:+\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}{i}\left(\mathrm{2}\:−{i}\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:−{i}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\:−\frac{\mathrm{1}+{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} }{\mathrm{2}{i}\left({i}\:{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:+\mathrm{2}\right)}\:+\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:+\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}{i}\left(\mathrm{2}−{i}−{i}\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right.}\right\} \\ $$$$=−\pi\:\frac{\mathrm{1}+{e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} }{\mathrm{2}+{ie}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} }\:+\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\frac{\sqrt{\mathrm{2}}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{2}−{i}\:−{i}\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:={A}\:{rest}\:{to}\:{find}\:{A}\:{at}\:{form}\:\alpha+{i}\lambda \\ $$$$ \\ $$
Answered by MJS last updated on 12/Jun/19
((x+1)/((x^2 +x+1)(x^2 −2i)))=((x^2 (x+1))/((x^2 +x+1)(x^4 +4)))+((2(x+1))/((x^2 +x+1)(x^4 +4)))i A=∫_(−∞) ^(+∞) ((x^2 (x+1))/((x^2 +x+1)(x^4 +4)))dx+i∫_(−∞) ^(+∞) ((2(x+1))/((x^2 +x+1)(x^4 +4)))dx ∫((x^2 (x+1))/((x^2 +x+1)(x^4 +4)))dx=∫((x^2 (x+1))/((x^2 +x+1)(x^2 −2x+2)(x^2 +2x+2)))dx= =−(1/(13))∫((4x+1)/(x^2 +x+1))dx+(1/(52))∫((3x+8)/(x^2 −2x+2))dx+(1/4)∫(x/(x^2 +2x+2))dx= [all are easy to solve I hope so I allow myself to not type the paths] =−(2/(13))ln (x^2 +x+1) +((2(√3))/(39))arctan (((√3)(2x+1))/3) + +(3/(104))ln (x^2 −2x+2) +((11)/(52))arctan (x−1) + +(1/8)ln (x^2 +2x+2) −(1/4)arctan (x+1) +C ⇒ real (A) =(π/(78))(−3+4(√3)) ∫((2(x+1))/((x^2 +x+1)(x^4 +4)))dx=∫((2(x+1))/((x^2 +x+1)(x^2 −2x+2)(x^2 +2x+2)))dx= =(2/(13))∫((3x+4)/(x^2 +x+1))dx−(1/(52))∫((11x−14)/(x^2 −2x+2))dx−(1/4)∫((x+2)/(x^2 +2x+2))dx= =(3/(12))ln (x^2 +x+1) +((10(√3))/(39))arctan (((√3)(2x+1))/3) − −((11)/(104))ln (x^2 −2x+2) +(3/(52))arctan (x−1) − −(1/8)ln (x^2 +2x+2) −(1/4)arctan (x+1) +C ⇒ imag (A) =((5π)/(78))(−3+4(√3)) A=(π/(78))(−3+4(√3))(1+5i)
$$\frac{{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2i}\right)}=\frac{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{4}\right)}+\frac{\mathrm{2}\left({x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{4}\right)}\mathrm{i} \\ $$$${A}=\underset{−\infty} {\overset{+\infty} {\int}}\frac{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{4}\right)}{dx}+\mathrm{i}\underset{−\infty} {\overset{+\infty} {\int}}\frac{\mathrm{2}\left({x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{4}\right)}{dx} \\ $$$$ \\ $$$$\int\frac{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{4}\right)}{dx}=\int\frac{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}{dx}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{13}}\int\frac{\mathrm{4}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}+\frac{\mathrm{1}}{\mathrm{52}}\int\frac{\mathrm{3}{x}+\mathrm{8}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}{dx}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{all}\:\mathrm{are}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{I}\:\mathrm{hope}\:\mathrm{so}\:\mathrm{I}\:\mathrm{allow}\:\mathrm{myself}\right. \\ $$$$\left.\:\:\:\:\:\:\mathrm{to}\:\mathrm{not}\:\mathrm{type}\:\mathrm{the}\:\mathrm{paths}\right] \\ $$$$=−\frac{\mathrm{2}}{\mathrm{13}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{39}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}}\:+ \\ $$$$\:\:\:\:\:+\frac{\mathrm{3}}{\mathrm{104}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\:+\frac{\mathrm{11}}{\mathrm{52}}\mathrm{arctan}\:\left({x}−\mathrm{1}\right)\:+ \\ $$$$\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)\:−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{arctan}\:\left({x}+\mathrm{1}\right)\:+{C} \\ $$$$\Rightarrow\:\mathrm{real}\:\left({A}\right)\:=\frac{\pi}{\mathrm{78}}\left(−\mathrm{3}+\mathrm{4}\sqrt{\mathrm{3}}\right) \\ $$$$ \\ $$$$\int\frac{\mathrm{2}\left({x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{4}\right)}{dx}=\int\frac{\mathrm{2}\left({x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}{dx}= \\ $$$$=\frac{\mathrm{2}}{\mathrm{13}}\int\frac{\mathrm{3}{x}+\mathrm{4}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{52}}\int\frac{\mathrm{11}{x}−\mathrm{14}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}{dx}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}{dx}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{12}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:+\frac{\mathrm{10}\sqrt{\mathrm{3}}}{\mathrm{39}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}}\:− \\ $$$$\:\:\:\:\:−\frac{\mathrm{11}}{\mathrm{104}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\:+\frac{\mathrm{3}}{\mathrm{52}}\mathrm{arctan}\:\left({x}−\mathrm{1}\right)\:− \\ $$$$\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)\:−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{arctan}\:\left({x}+\mathrm{1}\right)\:+{C} \\ $$$$\Rightarrow\:\mathrm{imag}\:\left({A}\right)\:=\frac{\mathrm{5}\pi}{\mathrm{78}}\left(−\mathrm{3}+\mathrm{4}\sqrt{\mathrm{3}}\right) \\ $$$${A}=\frac{\pi}{\mathrm{78}}\left(−\mathrm{3}+\mathrm{4}\sqrt{\mathrm{3}}\right)\left(\mathrm{1}+\mathrm{5i}\right) \\ $$
Commented by maxmathsup by imad last updated on 12/Jun/19
thank you sir mjs.
$${thank}\:{you}\:{sir}\:{mjs}. \\ $$
Commented by MJS last updated on 12/Jun/19
you′re welcome
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$

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