Question Number 98883 by mathmax by abdo last updated on 16/Jun/20
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{8}} \:+\mathrm{x}^{\mathrm{4}} +\mathrm{1}} \\ $$
Answered by maths mind last updated on 18/Jun/20
$${X}^{\mathrm{8}} +{X}^{\mathrm{4}} +\mathrm{1} \\ $$$${X}^{\mathrm{4}} ={y} \\ $$$${y}^{\mathrm{2}} +{y}+\mathrm{1}=\mathrm{0}\Rightarrow{y}={e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{3}}} ,{k}\in\left\{\mathrm{1},\mathrm{2}\right\} \\ $$$${x}_{{k}} ={e}^{{i}\frac{\pi}{\mathrm{6}}+\frac{{i}\mathrm{2}{k}\pi}{\mathrm{4}}} ,{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\} \\ $$$${x}_{{k}} ={e}^{{i}\frac{\pi}{\mathrm{3}}+\frac{\mathrm{2}{ik}\pi}{\mathrm{4}}} \\ $$$${a}_{\mathrm{1}} ={e}^{{i}\frac{\pi}{\mathrm{6}}} ,{a}_{\mathrm{2}} ={e}^{{i}\frac{\pi}{\mathrm{3}}} ,{a}_{\mathrm{3}} ={e}^{{i}\left(\frac{\pi}{\mathrm{6}}+\frac{\pi}{\mathrm{2}}\right)} ,{a}_{\mathrm{4}} ={e}^{{i}\left(\frac{\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{2}}\right)} , \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{{dx}}{{x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{dx}}{{x}^{\mathrm{8}} +{x}^{\mathrm{4}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}{i}\pi\:\underset{{ak}} {\sum}{Res}\left({f}\left({z}\right),{a}_{{k}} \right) \\ $$$$={i}\pi.\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\frac{\mathrm{1}}{\mathrm{8}{a}_{{k}} ^{\mathrm{7}} +\mathrm{4}{a}_{{k}} ^{\mathrm{3}} } \\ $$$$ \\ $$$$ \\ $$