Question Number 48720 by Abdo msup. last updated on 27/Nov/18
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{2}} \:−\mathrm{2}{cosx}+\mathrm{1}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$
Commented by Abdo msup. last updated on 28/Nov/18
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{2}} −\mathrm{2}{cosx}\:+\mathrm{1}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:−\int_{−\infty} ^{+\infty} \:\:\frac{\mathrm{2}{cosx}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}={H}−{K} \\ $$$${let}\:{find}\:{H}\:{let}\:\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{{z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{1}}\:{poles}\:{of}\:\varphi? \\ $$$${z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{1}=\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} \:+{t}+\mathrm{1}=\mathrm{0}\:\:\left({t}={z}^{\mathrm{2}} \right) \\ $$$$\Delta=−\mathrm{3}=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:\:{and}\:{z}_{\mathrm{2}} ={e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${z}^{\mathrm{2}} ={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:\Rightarrow\:{z}\:=\overset{−} {+}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \\ $$$${z}^{\mathrm{2}} ={e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:\Rightarrow{z}\:=\overset{−} {+}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:\Rightarrow \\ $$$$\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\:+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\right. \\ $$$${Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)=\frac{{e}^{\mathrm{2}\frac{{i}\pi}{\mathrm{3}}} +\mathrm{1}}{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{3}}} \left(\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{3}}\right)\left(\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}\right)\right)\right.} \\ $$$$=\frac{{e}^{\frac{{i}\pi}{\mathrm{3}}} \:+{e}^{−\frac{{i}\pi}{\mathrm{3}}} }{\mathrm{8}{i}\:{sin}\left(\frac{\pi}{\mathrm{3}}\right){cos}\left(\frac{\pi}{\mathrm{3}}\right)}\:=\frac{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}\right)}{\mathrm{8}{isin}\left(\frac{\pi}{\mathrm{3}}\right){cos}\left(\frac{\pi}{\mathrm{3}}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}} \\ $$$${Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\:=\frac{{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} +\mathrm{1}}{−\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}\right)\left(\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{3}}\right)\right)\left(−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$=\frac{{e}^{\frac{{i}\pi}{\mathrm{3}}} \:+{e}^{\frac{−{i}\pi}{\mathrm{3}}} }{\mathrm{8}{i}\:{cos}\left(\frac{\pi}{\mathrm{3}}\right){sin}\left(\frac{\pi}{\mathrm{3}}\right)}\:=\frac{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}\right)}{\mathrm{8}{i}\:{cos}\left(\frac{\pi}{\mathrm{3}}\right){sin}\left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{isin}\left(\frac{\pi}{\mathrm{3}}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{4}{i}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\:+\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{3}}}\right\}\:=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\:={H} \\ $$
Commented by Abdo msup. last updated on 28/Nov/18
$${let}\:{find}\:{K} \\ $$$${K}\:=\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{2}{cosx}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{2}\:{e}^{{ix}} }{{x}^{\mathrm{4}\:} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\right) \\ $$$${let}\:{W}\left({z}\right)\:=\frac{\mathrm{2}\:{e}^{{iz}} }{{z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${W}\left({z}\right)=\frac{\mathrm{2}\:{e}^{{iz}} }{{z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow{W}\left({z}\right)\:=\frac{\mathrm{2}\:{e}^{{iz}} }{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}\:+{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\:+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\right\} \\ $$$${Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{3}}} \right)\:=\:\frac{\mathrm{2}\:{e}^{{i}\left({cos}\frac{\pi}{\mathrm{3}}+{isin}\left(\frac{\pi}{\mathrm{3}}\right)\right)} }{\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \left(\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{3}}\right)\right)\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$$=\:\frac{{e}^{{i}\left(\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} }{\mathrm{4}{isin}\left(\frac{\pi}{\mathrm{3}}\right){cos}\left(\frac{\pi}{\mathrm{3}}\right)}\:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \:\:=\frac{{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:{e}^{{i}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}\right)} }{\mathrm{4}{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} }{{i}\sqrt{\mathrm{3}}}\:{e}^{{i}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}\right)} \\ $$$${Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\:=\:\frac{\mathrm{2}\:{e}^{{i}\left({cos}\frac{\pi}{\mathrm{3}}−{isin}\left(\frac{\pi}{\mathrm{3}}\right)\right)} }{−\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}\right)\left(\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{3}}\right)\left(−\mathrm{2}\:{e}^{\frac{−{i}\pi}{\mathrm{3}}} \right)\right.} \\ $$$$=\frac{{e}^{{i}\left(\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} }{\mathrm{4}{i}\:{cos}\left(\frac{\pi}{\mathrm{3}}\right){sin}\left(\frac{\pi}{\mathrm{3}}\right)}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \:=\frac{{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:{e}^{{i}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)} }{\mathrm{4}{i}\:\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\:\frac{{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:}} \:\:{e}^{{i}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)} }{{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}}\left\{\:\:{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:{e}^{{i}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}\right)} \:+\:{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:{e}^{{i}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\left\{\right.\right.} \right\} \\ $$$$=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\left\{\:{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \left\{{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}\right)+{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}\right)+\right.\right. \\ $$$$\left.{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \left\{\:{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)+{i}\:{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)\right\}\right\} \\ $$$${K}\:={Re}\left(\int_{−\infty} ^{+\infty} {W}\left({z}\right){dz}\right) \\ $$$$=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\left\{\:{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} {cos}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}\right)\:+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} {cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\pi}{\mathrm{3}}\right)\right\} \\ $$$${so}\:{thevalue}\:{of}\:{I}\:{is}\:{determined}. \\ $$$$ \\ $$