show that ∫_0 ^(+∞) (1/(x^4 +2x^2 cos(((2π)/5))+1)) dx=(π/(2φ))


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Question Number 84956 by M±th+et£s last updated on 17/Mar/20

$s h o w t h a t$ $\int_{0}^{+ \infty} \frac{1}{x^{4} + 2 x^{2} c o s (\frac{2 π}{5}) + 1} d x = \frac{π}{2 ϕ}$
Commented by mathmax by abdo last updated on 18/Mar/20
$2I =∫_(−∞) ^(+∞) (dx/(x^4 +2cos(((2π)/5))x^2 +1)) let W(z)=(1/(z^4 +2cos(((2π)/5))z^2 +1)) poles of W? z^4 +2cos(((2π)/5))z^2 +1=0 ⇒t^2 +2cos(((2π)/5))t+1=0 (t=z^2 ) Δ^′ =cos^2 (((2π)/5))−1 =(isin(((2π)/5)))^2 ⇒t_1 =−cos(((2π)/5))+isin(((2π)/5)) =e^(i(π−((2π)/5))) =e^(i(((3π)/5))) t_2 =e^(i(π+((2π)/5))) =e^(i(((7π)/5))) ⇒W(z)=(1/((z^2 −e^(i((3π)/5)) )(z^2 −e^(i(((7π)/5))) ))) =(1/((z−e^(i((3π)/(10))) )(z+e^(i((3π)/(10))) )(z−e^(i(((7π)/(10)))) )(z+e^((i7π)/(10)) ))) ∫_(−∞) ^(+∞) W(z)dz =2iπ{Res(W,e^((i3π)/(10)) )+Res(W,e^(i((7π)/(10))) )}...be continued...$
$2 I = \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + 2 c o s (\frac{2 π}{5}) x^{2} + 1} l e t W (z) = \frac{1}{z^{4} + 2 c o s (\frac{2 π}{5}) z^{2} + 1}$ $p o l e s o f W ?$ $z^{4} + 2 c o s (\frac{2 π}{5}) z^{2} + 1 = 0 \Rightarrow t^{2} + 2 c o s (\frac{2 π}{5}) t + 1 = 0 (t = z^{2})$ $Δ^{^{'}} = {c o s}^{2} (\frac{2 π}{5}) - 1 = {(i s i n (\frac{2 π}{5}))}^{2} \Rightarrow t_{1} = - c o s (\frac{2 π}{5}) + i s i n (\frac{2 π}{5})$ $= e^{i (π - \frac{2 π}{5})} = e^{i (\frac{3 π}{5})} t_{2} = e^{i (π + \frac{2 π}{5})} = e^{i (\frac{7 π}{5})} \Rightarrow W (z) = \frac{1}{(z^{2} - e^{i \frac{3 π}{5}}) (z^{2} - e^{i (\frac{7 π}{5})})}$ $= \frac{1}{(z - e^{i \frac{3 π}{10}}) (z + e^{i \frac{3 π}{10}}) (z - e^{i (\frac{7 π}{10})}) (z + e^{\frac{i 7 π}{10}})}$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, e^{\frac{i 3 π}{10}}) + R e s (W, e^{i \frac{7 π}{10}})} . . . b e c o n t i n u e d . . .$
Commented by M±th+et£s last updated on 18/Mar/20

$t h a n k y o u s i r$
Commented by mathmax by abdo last updated on 18/Mar/20

$y o u a r e w e l c o m e$
	Answered by mind is power last updated on 18/Mar/20
	$∫_0 ^(+∞) (dx/((x^2 −e^(i((2π)/5)) )(x^2 −e^((8iπ)/5) ))) =(1/2)∫_(−∞) ^(+∞) (dz/((z^2 −e^((2iπ)/5) )(z^2 −e^((8iπ)/5) ))) =iπ.{(1/(2e^(i(π/5)) (e^((2iπ)/5) −e^(−((2iπ)/5)) )))+(1/((e^(i((8π)/5)) −e^((2iπ)/5) ).2e^(i((4π)/5)) ))} =iπ.((e^(−((iπ)/5)) /(4isin(((2π)/5))))−(e^(−((4iπ)/5)) /(4isin(((2π)/5))))) =(π/(4sin(((2π)/5))))(2cos((π/5)))=(π/(4sin((π/5))))=(π/(2∅))$
	$\int_{0}^{+ \infty} \frac{d x}{(x^{2} - e^{i \frac{2 π}{5}}) (x^{2} - e^{\frac{8 i π}{5}})}$ $= \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d z}{(z^{2} - e^{\frac{2 i π}{5}}) (z^{2} - e^{\frac{8 i π}{5}})}$ $= i π . {\frac{1}{2 e^{i \frac{π}{5}} (e^{\frac{2 i π}{5}} - e^{- \frac{2 i π}{5}})} + \frac{1}{(e^{i \frac{8 π}{5}} - e^{\frac{2 i π}{5}}) . 2 e^{i \frac{4 π}{5}}}}$ $= i π . (\frac{e^{- \frac{i π}{5}}}{4 i s i n (\frac{2 π}{5})} - \frac{e^{- \frac{4 i π}{5}}}{4 i s i n (\frac{2 π}{5})})$ $= \frac{π}{4 s i n (\frac{2 π}{5})} (2 c o s (\frac{π}{5})) = \frac{π}{4 s i n (\frac{π}{5})} = \frac{π}{2 \emptyset}$
	Commented by M±th+et£s last updated on 18/Mar/20

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