Question Number 56827 by maxmathsup by imad last updated on 24/Mar/19
$${find}\:{the}\:{value}\:{of}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{3}} \\ $$
Commented by maxmathsup by imad last updated on 25/Mar/19
$${residus}\:{method}\:{let}\:\varphi\left({z}\right)=\frac{\mathrm{1}}{{z}^{\mathrm{4}} −{z}^{\mathrm{2}} \:+\mathrm{3}}\:\:{poles}\:{of}\:\varphi \\ $$$${z}^{\mathrm{4}} −{z}^{\mathrm{2}} \:+\mathrm{3}\:=\mathrm{0}\:\:\Rightarrow{t}^{\mathrm{2}} −{t}+\mathrm{3}=\mathrm{0}\:{with}\:{t}={z}^{\mathrm{2}} \\ $$$$\Delta\:=\mathrm{1}−\mathrm{12}\:=−\mathrm{11}\:\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{11}}}{\mathrm{2}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{11}}}{\mathrm{2}}\:\Rightarrow\varphi\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({z}^{\mathrm{2}} −{t}_{\mathrm{2}} \right)} \\ $$$$=\:\frac{\mathrm{1}}{\left({z}−\sqrt{{t}_{\mathrm{1}} }\right)\left({z}+\sqrt{{t}_{\mathrm{1}} }\right)\left({z}−\sqrt{{t}_{\mathrm{2}} }\right)\left({z}+\sqrt{{t}_{\mathrm{2}} }\right)}\:\:{we}\:{have} \\ $$$$\mid{t}_{\mathrm{1}} \mid\:=\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{11}}{\mathrm{4}}}=\sqrt{\mathrm{3}}\:\Rightarrow{t}_{\mathrm{1}} =\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:+\frac{{i}\sqrt{\mathrm{11}}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)={r}\:{e}^{{i}\theta} \:\Rightarrow{cos}\theta\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:{and}\:{sin}\theta=\frac{\sqrt{\mathrm{11}}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{tan}\theta\:=\sqrt{\mathrm{11}}\:\Rightarrow\theta={arctan}\left(\sqrt{\mathrm{11}}\right)\:\Rightarrow{t}_{\mathrm{1}} =\sqrt{\mathrm{3}}{e}^{{iarctan}\left(\sqrt{\mathrm{11}}\right)} \\ $$$$\Rightarrow\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−\sqrt{\sqrt{\mathrm{3}}}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)\left({z}+\sqrt{\sqrt{\mathrm{3}}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)\left({z}−\sqrt{\sqrt{\mathrm{3}}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)\left({z}+\sqrt{\sqrt{\mathrm{3}}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)} \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,\sqrt{\sqrt{\mathrm{3}}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} +{Res}\left(\varphi,−\sqrt{\sqrt{\mathrm{3}}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)\right\}\right. \\ $$$${Res}\left(\varphi,\sqrt{\sqrt{\mathrm{3}}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\sqrt{\mathrm{3}}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \mathrm{2}{i}\sqrt{\mathrm{3}}\:{sin}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)\mathrm{2}{cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)} \\ $$$$=\:\frac{{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} }{\mathrm{4}{i}\sqrt{\mathrm{3}}\sqrt{\sqrt{\mathrm{3}}}{sin}\left({arctan}\left(\sqrt{\mathrm{11}}\right)\right)} \\ $$$${Res}\left(\varphi,−\sqrt{\sqrt{\mathrm{3}}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \right)\:=\frac{\mathrm{1}}{−\mathrm{2}\sqrt{\sqrt{\mathrm{3}}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \left(−\sqrt{\sqrt{\mathrm{3}}}\right)\mathrm{2}{cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)\mathrm{2}{i}\sqrt{\sqrt{\mathrm{3}}}{sin}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)} \\ $$$$=\:\frac{{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} }{\mathrm{4}{i}\sqrt{\mathrm{3}}\sqrt{\sqrt{\mathrm{3}}}{sin}\left({arctan}\left(\sqrt{\mathrm{11}}\right)\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\mathrm{2}{i}\pi}{\mathrm{4}{i}\sqrt{\mathrm{3}}\sqrt{\sqrt{\mathrm{3}}}{sin}\left({arctan}\left(\sqrt{\mathrm{11}}\right)\right)}\:\left\{\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{11}}\right)} \:+{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\left.\mathrm{11}\right)}\right.} \right\} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}\sqrt{\sqrt{\mathrm{3}}}{sin}\left({arctan}\left(\sqrt{\mathrm{11}}\right)\right.}\:.\mathrm{2}{cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\:\sqrt{\mathrm{3}}\sqrt{\sqrt{\mathrm{3}}}}\:\:\frac{{cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)}{\mathrm{2}{sin}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right){cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)} \\ $$$${I}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}\sqrt{\sqrt{\mathrm{3}}}{sin}\left(\frac{{arctan}\left(\sqrt{\mathrm{11}}\right)}{\mathrm{2}}\right)}\:. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Mar/19
$$\frac{\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{3}} \\ $$$$=\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{\frac{\mathrm{2}\sqrt{\mathrm{3}}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{{x}^{\mathrm{2}} }−\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{3}}{{x}^{\mathrm{2}} }−\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{{d}\left({x}−\frac{\sqrt{\mathrm{3}}}{{x}}\right)}{\left({x}−\frac{\sqrt{\mathrm{3}}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{{d}\left({x}+\frac{\sqrt{\mathrm{3}}}{{x}}\right)}{\left({x}+\frac{\sqrt{\mathrm{3}}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\int_{−\infty} ^{+\infty} \frac{{d}\left({x}−\frac{\sqrt{\mathrm{3}}}{{x}}\right)}{\left({x}−\frac{\sqrt{\mathrm{3}}}{{x}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}\:\right)^{\mathrm{2}} }\leftarrow\boldsymbol{{I}}_{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}}×\mid{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\sqrt{\mathrm{3}}}{{x}}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}}\right)\mid_{−\infty} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}}×\left\{\left(\frac{\pi}{\mathrm{2}}\right)−\left(\frac{−\pi}{\mathrm{2}}\right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}}\leftarrow{value}\:{of}\:{I}_{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\int_{−\infty} ^{+\infty} \frac{{d}\left({x}+\frac{\sqrt{\mathrm{3}}}{{x}}\right)}{\left({x}+\frac{\sqrt{\mathrm{3}}}{{x}}\right)^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{1}\right)}\:\leftarrow\boldsymbol{{I}}_{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\mathrm{2}\left(\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{1}}\:\right)}\mid{ln}\left(\frac{{x}+\frac{\sqrt{\mathrm{3}}}{{x}}−\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{1}}}{{x}+\frac{\sqrt{\mathrm{3}}}{{x}}+\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{1}}}\right)\mid_{−\infty} ^{+\infty} \\ $$$${now}\:\mid{ln}\left(\frac{{x}+\frac{{a}}{{x}}−{b}}{{x}+\frac{{a}}{{x}}+{b}}\right)\mid_{−\infty} ^{+\infty} \\ $$$$=\mid{ln}\left(\frac{{x}^{\mathrm{2}} +{a}−{bx}}{{x}^{\mathrm{2}} +{a}+{bx}}\right)\mid_{−\infty} ^{+\infty} \\ $$$$=\mid{ln}\left(\frac{\mathrm{1}+\frac{{a}}{{x}^{\mathrm{2}} }−\frac{{b}}{{x}}}{\mathrm{1}+\frac{{a}}{{x}^{\mathrm{2}} }+\frac{{b}}{{x}}}\right)\mid_{−\infty} ^{+\infty} =\mathrm{0} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{I}}_{\mathrm{2}} =\mathrm{0} \\ $$$$\boldsymbol{{hence}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}} \\ $$$$\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{1}}}\:\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{answer}} \\ $$$$ \\ $$