∫_0 ^∞ (x^2 /(x^4 +x^2 +1))dx


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Question Number 65203 by arcana last updated on 26/Jul/19

$\int_{0}^{\infty} \frac{x^{2}}{x^{4} + x^{2} + 1} d x$
Commented by mathmax by abdo last updated on 26/Jul/19
$let I =∫_0 ^∞ (x^2 /(x^4 +x^2 +1))dx ⇒2I =∫_(−∞) ^(+∞) (x^2 /(x^4 +x^2 +1))dx let ϕ(z) =(z^2 /(z^4 +z^2 +1)) poles of ϕ? z^4 +z^2 +1 =0 ⇒t^2 +t +1 =0 (t=z^2 ) Δ =1−4 =−3 =(i(√3))^2 ⇒t_1 =((−1+i(√3))/2) and t_2 =((−1−i(√3))/2) t_1 =e^((i2π)/3) and t_2 =e^(−((2iπ)/3)) ⇒t^2 +t+1 =(t−e^((i2π)/3) )(t−e^(−((i2π)/3)) ) =(z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) ) ⇒ϕ(z) =(z^2 /((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z+e^(−((iπ)/3)) ))) the poles of ϕ are +^− e^((iπ)/3) and +^− e^(−((iπ)/3)) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,e^((iπ)/3) ) +Res(ϕ,−e^(−((iπ)/3)) )} Res(ϕ,e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (z−e^((iπ)/3) )ϕ(z)=(e^((2iπ)/3) /(2e^((iπ)/3) (e^((2iπ)/3) −e^(−((2iπ)/3)) ))) =(e^((iπ)/3) /(2(2isin(((2π)/3))))) =(e^((iπ)/3) /(4i((√3)/2))) =(e^((iπ)/3) /(2i(√3))) Res(ϕ,−e^(−((iπ)/3)) ) =lim_(z→e^(−((iπ)/3)) ) (z+e^(−((iπ)/3)) ) = (e^(−((2iπ)/3)) /((e^(−((2iπ)/3)) −e^((i2π)/3) )(−2e^(−((iπ)/3)) ))) =(e^(−((iπ)/3)) /(−2i((√3)/2)(−2))) =(e^(−((iπ)/3)) /(2i(√3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(e^((iπ)/3) /(2i(√3))) +(e^(−((iπ)/3)) /(2i(√3)))} =(π/(√3)){2cos((π/3)) =(π/(√3)) (2.(1/2)) =(π/(√3)) ⇒ I =(1/2) ∫_(−∞) ^(+∞) ϕ(z)dz =(π/(2(√3))) .$
$l e t I = \int_{0}^{\infty} \frac{x^{2}}{x^{4} + x^{2} + 1} d x \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{x^{2}}{x^{4} + x^{2} + 1} d x l e t$ $φ (z) = \frac{z^{2}}{z^{4} + z^{2} + 1} p o l e s o f φ ?$ $z^{4} + z^{2} + 1 = 0 \Rightarrow t^{2} + t + 1 = 0 (t = z^{2})$ $Δ = 1 - 4 = - 3 = {(i \sqrt{3})}^{2} \Rightarrow t_{1} = \frac{- 1 + i \sqrt{3}}{2} a n d t_{2} = \frac{- 1 - i \sqrt{3}}{2}$ $t_{1} = e^{\frac{i 2 π}{3}} a n d t_{2} = e^{- \frac{2 i π}{3}} \Rightarrow t^{2} + t + 1 = (t - e^{\frac{i 2 π}{3}}) (t - e^{- \frac{i 2 π}{3}})$ $= (z^{2} - e^{\frac{i 2 π}{3}}) (z^{2} - e^{- \frac{i 2 π}{3}}) \Rightarrow φ (z) = \frac{z^{2}}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + e^{- \frac{i π}{3}})}$ $t h e p o l e s o f φ a r e \bar{+} e^{\frac{i π}{3}} a n d \bar{+} e^{- \frac{i π}{3}} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{3}}) + R e s (φ, - e^{- \frac{i π}{3}})}$ $R e s (φ, e^{\frac{i π}{3}}) = {l i m}_{z \to e^{\frac{i π}{3}}} (z - e^{\frac{i π}{3}}) φ (z) = \frac{e^{\frac{2 i π}{3}}}{2 e^{\frac{i π}{3}} (e^{\frac{2 i π}{3}} - e^{- \frac{2 i π}{3}})}$ $= \frac{e^{\frac{i π}{3}}}{2 (2 i s i n (\frac{2 π}{3}))} = \frac{e^{\frac{i π}{3}}}{4 i \frac{\sqrt{3}}{2}} = \frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}}$ $R e s (φ, - e^{- \frac{i π}{3}}) = {l i m}_{z \to e^{- \frac{i π}{3}}} (z + e^{- \frac{i π}{3}}) = \frac{e^{- \frac{2 i π}{3}}}{(e^{- \frac{2 i π}{3}} - e^{\frac{i 2 π}{3}}) (- 2 e^{- \frac{i π}{3}})}$ $= \frac{e^{- \frac{i π}{3}}}{- 2 i \frac{\sqrt{3}}{2} (- 2)} = \frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}} + \frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}}}$ $= \frac{π}{\sqrt{3}} {2 c o s (\frac{π}{3}) = \frac{π}{\sqrt{3}} (2 . \frac{1}{2}) = \frac{π}{\sqrt{3}} \Rightarrow I = \frac{1}{2} \int_{- \infty}^{+ \infty} φ (z) d z = \frac{π}{2 \sqrt{3}} .$
	Answered by Tanmay chaudhury last updated on 26/Jul/19

	$\int_{0}^{\infty} \frac{d x}{x^{2} + 1 + \frac{1}{x^{2}}}$ $\frac{1}{2} \int_{0}^{\infty} \frac{1 + \frac{1}{x^{2}} + 1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} + 1} d x$ $\frac{1}{2} \int_{0}^{\infty} \frac{d (x - \frac{1}{x})}{{(x - \frac{1}{x})}^{2} + 3} + \frac{1}{2} \int_{0}^{\infty} \frac{d (x + \frac{1}{x})}{{(x + \frac{1}{x})}^{2} - 1}$ $\frac{1}{2} \times \frac{1}{\sqrt{3}} ∣ {t a n}^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + \frac{1}{2} \times \frac{1}{2 \times 1} l n (\frac{x + \frac{1}{x} - 1}{x + \frac{1}{x} + 1}) ∣_{0}^{\infty}$ $= \frac{1}{2 \sqrt{3}} {t a n}^{- 1} (\infty) - \frac{1}{2 \sqrt{3}} {t a n}^{- 1} (- \infty) + \frac{1}{4} l n {(\frac{1 + \frac{1}{x^{2}} - \frac{1}{x}}{1 + \frac{1}{x^{2}} + \frac{1}{x}})}_{x = \infty} - \frac{1}{4} l n {(\frac{x^{2} + 1 - x}{x^{2} + 1 + x})}_{x = 0}$ $= \frac{1}{2 \sqrt{3}} \times \frac{π}{2} + \frac{1}{2 \sqrt{3}} \times \frac{π}{2}) + \frac{1}{4} \times l n 1 - \frac{1}{4} l n 1$ $= \frac{π}{2 \sqrt{3}}$
	Commented by arcana last updated on 26/Jul/19

	$thankss . try solve it with residuals theorem$ $(its less difficult)$
	Commented by Tanmay chaudhury last updated on 26/Jul/19

	$o k s i r . . .$
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