calculate ∫_(−∞) ^(+∞) ((x^2 −3)/(x^4 +x^2 +1))dx .


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Question Number 63023 by mathmax by abdo last updated on 27/Jun/19

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{x^{2} - 3}{x^{4} + x^{2} + 1} d x .$
Commented by mathmax by abdo last updated on 28/Jun/19
$let A =∫_(−∞) ^(+∞) ((x^2 −3)/(x^4 +x^2 +1))dx let W(z) =((z^2 −3)/(z^4 +z^2 +1)) poles of W? z^4 +z^2 +1 =0 →t^2 +t +1 =0( with z^2 =t) → Δ =1−4 =(i(√3))^2 ⇒t_1 =((−1+i(√3))/2) and t_2 =((−1−i(√3))/2) ⇒ t_1 =e^((i2π)/3) and t_2 =e^(−((i2π)/3)) ⇒z^4 +z^2 +1 =(t−t_1 )(t−t_2 ) =(z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) ) =(z−e^(i(π/3)) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z +e^((iπ)/3) )⇒W(z) =((z^2 −3)/((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^(−((iπ)/3)) ))) =((z^2 −3)/((z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) ))) the poles of W are +^− e^((iπ)/3) and +^− e^(−((iπ)/3)) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ { Res(W,e^((iπ)/3) ) +Res(W,−e^(−((iπ)/3)) )} Res(W,e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (z−e^((iπ)/3) )W(z) =((e^(i((2π)/3)) −3)/(2e^((iπ)/3) (e^((2iπ)/3) −e^(−((i2π)/3)) ))) =(1/2) e^(−((iπ)/3)) ((e^(i((2π)/3)) −3)/(2isin(((2π)/3)))) =(1/(4i((√3)/2))) (e^((iπ)/3) −3 e^(−((iπ)/3)) ) =(1/(2i(√3))){ e^((iπ)/3) −3 e^(−((iπ)/3)) } Res(W,−e^(−((iπ)/3)) ) =lim_(z→−e^(−((iπ)/3)) ) (z+e^(−((iπ)/3)) )W(z) =((e^(−((2iπ)/3)) −3)/((e^(−((2iπ)/3)) −e^((i2π)/3) )(−2e^(−((iπ)/3)) ))) =(e^((iπ)/3) /(2(2i sin(((2π)/3))))){ e^(−((2iπ)/3)) −3} =(1/(4i ((√3)/2))){ e^(−((iπ)/3)) −3 e^((iπ)/3) } =(1/(2i(√3))){ e^(−((iπ)/3)) −3 e^((iπ)/3) } ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ (1/(2i(√3))){ e^((iπ)/3) −3 e^(−((iπ)/3)) +e^(−((iπ)/3)) −3 e^((iπ)/3) } =(π/(√3)){ −2 e^((iπ)/3) −2 e^(−((iπ)/3)) } =((−2π)/(√3)){ e^((iπ)/3) +e^(−((iπ)/3)) } =((−2π)/(√3)){ 2cos((π/3))} =((−2π)/(√3))2(1/2) =((−2π)/(√3)) ⇒ ∫_(−∞) ^(+∞) ((x^2 −3)/(x^4 +x^2 +1))dx =((−2π)/(√3)) .$
$l e t A = \int_{- \infty}^{+ \infty} \frac{x^{2} - 3}{x^{4} + x^{2} + 1} d x l e t W (z) = \frac{z^{2} - 3}{z^{4} + z^{2} + 1} p o l e s o f W ?$ $z^{4} + z^{2} + 1 = 0 \to t^{2} + t + 1 = 0 (w i t h z^{2} = t) \to$ $Δ = 1 - 4 = {(i \sqrt{3})}^{2} \Rightarrow t_{1} = \frac{- 1 + i \sqrt{3}}{2} a n d t_{2} = \frac{- 1 - i \sqrt{3}}{2} \Rightarrow$ $t_{1} = e^{\frac{i 2 π}{3}} a n d t_{2} = e^{- \frac{i 2 π}{3}} \Rightarrow z^{4} + z^{2} + 1 = (t - t_{1}) (t - t_{2}) = (z^{2} - e^{\frac{i 2 π}{3}}) (z^{2} - e^{- \frac{i 2 π}{3}})$ $= (z - e^{i \frac{π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + e^{\frac{i π}{3}}) \Rightarrow W (z) = \frac{z^{2} - 3}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + e^{- \frac{i π}{3}})}$ $= \frac{z^{2} - 3}{(z^{2} - e^{\frac{i 2 π}{3}}) (z^{2} - e^{- \frac{i 2 π}{3}})}$ $t h e p o l e s o f W a r e \bar{+} e^{\frac{i π}{3}} a n d \bar{+} e^{- \frac{i π}{3}} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, e^{\frac{i π}{3}}) + R e s (W, - e^{- \frac{i π}{3}})}$ $R e s (W, e^{\frac{i π}{3}}) = {l i m}_{z \to e^{\frac{i π}{3}}} (z - e^{\frac{i π}{3}}) W (z) = \frac{e^{i \frac{2 π}{3}} - 3}{2 e^{\frac{i π}{3}} (e^{\frac{2 i π}{3}} - e^{- \frac{i 2 π}{3}})} = \frac{1}{2} e^{- \frac{i π}{3}} \frac{e^{i \frac{2 π}{3}} - 3}{2 i s i n (\frac{2 π}{3})}$ $= \frac{1}{4 i \frac{\sqrt{3}}{2}} (e^{\frac{i π}{3}} - 3 e^{- \frac{i π}{3}}) = \frac{1}{2 i \sqrt{3}} {e^{\frac{i π}{3}} - 3 e^{- \frac{i π}{3}}}$ $R e s (W, - e^{- \frac{i π}{3}}) = {l i m}_{z \to - e^{- \frac{i π}{3}}} (z + e^{- \frac{i π}{3}}) W (z) = \frac{e^{- \frac{2 i π}{3}} - 3}{(e^{- \frac{2 i π}{3}} - e^{\frac{i 2 π}{3}}) (- 2 e^{- \frac{i π}{3}})}$ $= \frac{e^{\frac{i π}{3}}}{2 (2 i s i n (\frac{2 π}{3}))} {e^{- \frac{2 i π}{3}} - 3} = \frac{1}{4 i \frac{\sqrt{3}}{2}} {e^{- \frac{i π}{3}} - 3 e^{\frac{i π}{3}}}$ $= \frac{1}{2 i \sqrt{3}} {e^{- \frac{i π}{3}} - 3 e^{\frac{i π}{3}}} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π \frac{1}{2 i \sqrt{3}} {e^{\frac{i π}{3}} - 3 e^{- \frac{i π}{3}} + e^{- \frac{i π}{3}} - 3 e^{\frac{i π}{3}}}$ $= \frac{π}{\sqrt{3}} {- 2 e^{\frac{i π}{3}} - 2 e^{- \frac{i π}{3}}} = \frac{- 2 π}{\sqrt{3}} {e^{\frac{i π}{3}} + e^{- \frac{i π}{3}}} = \frac{- 2 π}{\sqrt{3}} {2 c o s (\frac{π}{3})} = \frac{- 2 π}{\sqrt{3}} 2 \frac{1}{2}$ $= \frac{- 2 π}{\sqrt{3}} \Rightarrow \int_{- \infty}^{+ \infty} \frac{x^{2} - 3}{x^{4} + x^{2} + 1} d x = \frac{- 2 π}{\sqrt{3}} .$
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