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Question Number 84956 by M±th+et£s last updated on 17/Mar/20
show that ∫_0 ^(+∞) (1/(x^4 +2x^2 cos(((2π)/5))+1)) dx=(π/(2φ))
$${show}\:{that}\: \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} {cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)+\mathrm{1}}\:{dx}=\frac{\pi}{\mathrm{2}\phi} \\ $$
Commented by mathmax by abdo last updated on 18/Mar/20
2I =∫_(−∞) ^(+∞) (dx/(x^4 +2cos(((2π)/5))x^2 +1)) let W(z)=(1/(z^4 +2cos(((2π)/5))z^2 +1)) poles of W? z^4 +2cos(((2π)/5))z^2 +1=0 ⇒t^2 +2cos(((2π)/5))t+1=0 (t=z^2 ) Δ^′ =cos^2 (((2π)/5))−1 =(isin(((2π)/5)))^2 ⇒t_1 =−cos(((2π)/5))+isin(((2π)/5)) =e^(i(π−((2π)/5))) =e^(i(((3π)/5))) t_2 =e^(i(π+((2π)/5))) =e^(i(((7π)/5))) ⇒W(z)=(1/((z^2 −e^(i((3π)/5)) )(z^2 −e^(i(((7π)/5))) ))) =(1/((z−e^(i((3π)/(10))) )(z+e^(i((3π)/(10))) )(z−e^(i(((7π)/(10)))) )(z+e^((i7π)/(10)) ))) ∫_(−∞) ^(+∞) W(z)dz =2iπ{Res(W,e^((i3π)/(10)) )+Res(W,e^(i((7π)/(10))) )}...be continued...
$$\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right){x}^{\mathrm{2}} \:+\mathrm{1}}\:{let}\:{W}\left({z}\right)=\frac{\mathrm{1}}{{z}^{\mathrm{4}} \:+\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right){z}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${poles}\:{of}\:{W}? \\ $$$${z}^{\mathrm{4}} \:+\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right){z}^{\mathrm{2}} \:+\mathrm{1}=\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} +\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right){t}+\mathrm{1}=\mathrm{0}\:\:\left({t}={z}^{\mathrm{2}} \right) \\ $$$$\Delta^{'} ={cos}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)−\mathrm{1}\:=\left({isin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\right)^{\mathrm{2}} \:\Rightarrow{t}_{\mathrm{1}} =−{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)+{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right) \\ $$$$={e}^{{i}\left(\pi−\frac{\mathrm{2}\pi}{\mathrm{5}}\right)} \:={e}^{{i}\left(\frac{\mathrm{3}\pi}{\mathrm{5}}\right)} \:\:\:{t}_{\mathrm{2}} ={e}^{{i}\left(\pi+\frac{\mathrm{2}\pi}{\mathrm{5}}\right)} ={e}^{{i}\left(\frac{\mathrm{7}\pi}{\mathrm{5}}\right)} \:\Rightarrow{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{5}}} \right)\left({z}^{\mathrm{2}} −{e}^{{i}\left(\frac{\mathrm{7}\pi}{\mathrm{5}}\right)} \right)} \\ $$$$=\frac{\mathrm{1}}{\left({z}−{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{10}}} \right)\left({z}+{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{10}}} \right)\left({z}−{e}^{{i}\left(\frac{\mathrm{7}\pi}{\mathrm{10}}\right)} \right)\left({z}+{e}^{\frac{{i}\mathrm{7}\pi}{\mathrm{10}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{{Res}\left({W},{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{10}}} \right)+{Res}\left({W},{e}^{{i}\frac{\mathrm{7}\pi}{\mathrm{10}}} \right)\right\}…{be}\:{continued}… \\ $$
Commented by M±th+et£s last updated on 18/Mar/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 18/Mar/20
you are welcome
$${you}\:{are}\:{welcome} \\ $$
Answered by mind is power last updated on 18/Mar/20
∫_0 ^(+∞) (dx/((x^2 −e^(i((2π)/5)) )(x^2 −e^((8iπ)/5) ))) =(1/2)∫_(−∞) ^(+∞) (dz/((z^2 −e^((2iπ)/5) )(z^2 −e^((8iπ)/5) ))) =iπ.{(1/(2e^(i(π/5)) (e^((2iπ)/5) −e^(−((2iπ)/5)) )))+(1/((e^(i((8π)/5)) −e^((2iπ)/5) ).2e^(i((4π)/5)) ))} =iπ.((e^(−((iπ)/5)) /(4isin(((2π)/5))))−(e^(−((4iπ)/5)) /(4isin(((2π)/5))))) =(π/(4sin(((2π)/5))))(2cos((π/5)))=(π/(4sin((π/5))))=(π/(2∅))
$$\int_{\mathrm{0}} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} −{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{5}}} \right)\left({x}^{\mathrm{2}} −{e}^{\frac{\mathrm{8}{i}\pi}{\mathrm{5}}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{dz}}{\left({z}^{\mathrm{2}} −{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{5}}} \right)\left({z}^{\mathrm{2}} −{e}^{\frac{\mathrm{8}{i}\pi}{\mathrm{5}}} \right)} \\ $$$$={i}\pi.\left\{\frac{\mathrm{1}}{\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{5}}} \left({e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{5}}} −{e}^{−\frac{\mathrm{2}{i}\pi}{\mathrm{5}}} \right)}+\frac{\mathrm{1}}{\left({e}^{{i}\frac{\mathrm{8}\pi}{\mathrm{5}}} −{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{5}}} \right).\mathrm{2}{e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{5}}} }\right\} \\ $$$$={i}\pi.\left(\frac{{e}^{−\frac{{i}\pi}{\mathrm{5}}} }{\mathrm{4}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)}−\frac{{e}^{−\frac{\mathrm{4}{i}\pi}{\mathrm{5}}} }{\mathrm{4}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)}\right) \\ $$$$=\frac{\pi}{\mathrm{4}{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)}\left(\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{5}}\right)\right)=\frac{\pi}{\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{5}}\right)}=\frac{\pi}{\mathrm{2}\emptyset} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 18/Mar/20
god bless you sir
$${god}\:{bless}\:{you}\:{sir} \\ $$

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