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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

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}})} \dots b e c o n t i n u e d \dots

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

\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

g o d b l e s s y o u s i r