Question Number 63023 by mathmax by abdo last updated on 27/Jun/19
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}^{\mathrm{2}} −\mathrm{3}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:. \\ $$
Commented by mathmax by abdo last updated on 28/Jun/19
$${let}\:{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{x}^{\mathrm{2}} −\mathrm{3}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:{let}\:{W}\left({z}\right)\:=\frac{{z}^{\mathrm{2}} −\mathrm{3}}{{z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{1}}\:\:{poles}\:{of}\:{W}? \\ $$$${z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{0}\:\rightarrow{t}^{\mathrm{2}} \:+{t}\:+\mathrm{1}\:=\mathrm{0}\left(\:{with}\:{z}^{\mathrm{2}} \:={t}\right)\:\rightarrow \\ $$$$\Delta\:=\mathrm{1}−\mathrm{4}\:=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\Rightarrow{t}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow \\ $$$${t}_{\mathrm{1}} ={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:{and}\:{t}_{\mathrm{2}} ={e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:\Rightarrow{z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{1}\:=\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)\:=\left({z}^{\mathrm{2}} −{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)\left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right) \\ $$$$=\left({z}−{e}^{{i}\frac{\pi}{\mathrm{3}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}\:+{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\Rightarrow{W}\left({z}\right)\:=\frac{{z}^{\mathrm{2}} −\mathrm{3}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{{z}^{\mathrm{2}} −\mathrm{3}}{\left({z}^{\mathrm{2}} −{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)\left({z}^{\mathrm{2}} −{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)} \\ $$$${the}\:{poles}\:{of}\:\:{W}\:{are}\:\overset{−} {+}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \:{and}\:\:\overset{−} {+}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:\:\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left({W},{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\:+{Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\right\} \\ $$$${Res}\left({W},{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\:={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{3}}} } \:\:\left({z}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right){W}\left({z}\right)\:=\frac{{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} −\mathrm{3}}{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{3}}} \left({e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \:−{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:\:\frac{{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} −\mathrm{3}}{\mathrm{2}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:\left({e}^{\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{3}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\left\{\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{3}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right\} \\ $$$${Res}\left({W},−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\:={lim}_{{z}\rightarrow−{e}^{−\frac{{i}\pi}{\mathrm{3}}} } \:\:\:\:\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right){W}\left({z}\right)\:=\frac{{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} −\mathrm{3}}{\left({e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} \:−{e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \right)\left(−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{{e}^{\frac{{i}\pi}{\mathrm{3}}} }{\mathrm{2}\left(\mathrm{2}{i}\:{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)}\left\{\:{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} −\mathrm{3}\right\}\:=\frac{\mathrm{1}}{\mathrm{4}{i}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\left\{\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{3}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\left\{\:\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{3}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \right\}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\left\{\:\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{3}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:−\mathrm{3}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \right\} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{3}}}\left\{\:−\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} −\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right\}\:=\frac{−\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\left\{\:\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \:+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right\}\:=\frac{−\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\left\{\:\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}\right)\right\}\:=\frac{−\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{−\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} −\mathrm{3}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:=\frac{−\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\:. \\ $$$$ \\ $$