calculate ∫_0 ^∞ (dx/(x^4 +x^2 +1))


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Question Number 93634 by abdomathmax last updated on 14/May/20

$c a l c u l a t e \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1}$
Commented by mathmax by abdo last updated on 14/May/20
$I =∫_0 ^∞ (dx/(x^4 +x^2 +1)) ⇒2A =∫_(−∞) ^(+∞) (dx/(x^4 +x^2 +1)) let ϕ(z) =(1/(z^4 +z^2 +1)) poles of ϕ? z^4 +z^2 +1 =0 →t^2 +t +1 =0 (t=z^2 ) Δ=−3 ⇒t_1 =((−1+i(√3))/2) =e^((i2π)/3) , t_2 =((−1−i(√3))/2) =e^(−((i2π)/3)) ⇒ ϕ(z) =(1/((z^2 −e^((i2π)/3) )(z^2 −e^(−((i2π)/3)) ))) =(1/((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^((−iπ)/3) )(z+e^(−((iπ)/3)) ))) rdsidus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,e^((iπ)/3) )+Res(ϕ,−e^(−((iπ)/3)) )} Res(ϕ,e^((iπ)/3) ) =(1/(2e^((iπ)/3) (2isin(((2π)/3))))) =(e^(−((iπ)/3)) /(4i×((√3)/2))) =(e^(−((iπ)/3)) /(2i(√3))) Res(ϕ,−e^(−((iπ)/3)) ) =(1/(−2e^(−((iπ)/3)) (−2i sin(((2π)/3))))) =(e^((iπ)/3) /(4i ×((√3)/2))) =(e^((iπ)/3) /(2i(√3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/(2i(√3))){ e^(−((iπ)/3)) +e^((iπ)/3) } =(π/(√3))(2cos((π/3))) =(π/(√3)) 2∫_(−∞) ^(+∞) (dx/(x^4 +x^2 +1)) =(π/(√3)) ⇒★∫_(−∞) ^(+∞) (dx/(x^4 +x^2 +1)) =(π/(2(√3))) ★$
$I = \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + x^{2} + 1} l e t φ (z) = \frac{1}{z^{4} + z^{2} + 1}$ $p o l e s o f φ ?$ $z^{4} + z^{2} + 1 = 0 \to t^{2} + t + 1 = 0 (t = z^{2})$ $Δ = - 3 \Rightarrow t_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}}, t_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}} \Rightarrow$ $φ (z) = \frac{1}{(z^{2} - e^{\frac{i 2 π}{3}}) (z^{2} - e^{- \frac{i 2 π}{3}})} = \frac{1}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{\frac{- i π}{3}}) (z + e^{- \frac{i π}{3}})}$ $r d s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{3}}) + R e s (φ, - e^{- \frac{i π}{3}})}$ $R e s (φ, e^{\frac{i π}{3}}) = \frac{1}{2 e^{\frac{i π}{3}} (2 i s i n (\frac{2 π}{3}))} = \frac{e^{- \frac{i π}{3}}}{4 i \times \frac{\sqrt{3}}{2}} = \frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}}$ $R e s (φ, - e^{- \frac{i π}{3}}) = \frac{1}{- 2 e^{- \frac{i π}{3}} (- 2 i s i n (\frac{2 π}{3}))} = \frac{e^{\frac{i π}{3}}}{4 i \times \frac{\sqrt{3}}{2}} = \frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{2 i \sqrt{3}} {e^{- \frac{i π}{3}} + e^{\frac{i π}{3}}} = \frac{π}{\sqrt{3}} (2 c o s (\frac{π}{3})) = \frac{π}{\sqrt{3}}$ $2 \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π}{\sqrt{3}} \Rightarrow ★ \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π}{2 \sqrt{3}} ★$
Commented by mathmax by abdo last updated on 14/May/20

$s o r r y 2 \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π}{\sqrt{3}} \Rightarrow ★ \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π}{2 \sqrt{3}} ★$
	Answered by Kunal12588 last updated on 14/May/20
	$(x^4 +x^2 +1)=(ax^2 +bx+c)(px^2 +qx+r) x^4 +x^2 +1=apx^4 +(aq+bp)x^3 +(ar+bq+cp)x^2 +(br+cq)x+cr a=1, b=±1, c=1 p=1, q=∓1, r=1 x^4 +x^2 +1=(x^2 +x+1)(x^2 −x+1) (1/(x^4 +x^2 +1))=((ax+b)/(x^2 +x+1))+((px+q)/(x^2 −x+1)) 1=(a+p)x^3 +(−a+b+p+q)x^2 +(a−b+p+q)x +(b+q) a=(1/2), b=(1/2) p=((−1)/2), q=(1/2) (1/(x^4 +x^2 +1))=((x+1)/(x^2 +x+1))−((x−1)/(x^4 +x+1)) I=∫(dx/(x^4 +x^2 +1))=(1/2)∫((x+1)/(x^2 +x+1))dx−(1/2)∫((x−1)/(x^2 +x+1))dx I_1 =(1/2)∫((x+1)/(x^2 +x+1))dx =(1/4)∫((2x+1)/(x^2 +x+1))dx+(1/4)∫(dx/(x^2 +x+1)) =(1/4)ln∣x^2 +x+1∣+(1/4)∫(dx/((x+(1/2))^2 +(3/4))) =(1/4)ln∣x^2 +x+1∣+(1/4)×(2/(√3)) tan^(−1) (((x+(1/2)))/((√3)/2))+c I_1 =(1/4)ln∣x^2 +x+1∣+(1/(2(√3))) tan^(−1) ((2x+1)/(√3))+m I_2 =(1/2)∫((x−1)/(x^2 −x+1))dx =(1/4)∫((2x−1)/(x^2 −x+1))dx−(1/4)∫(dx/(x^2 −x+1)) =(1/4)ln∣x^2 −x+1∣−(1/4)∫(dx/((x−(1/2))^2 +(((√3)/2))^2 )) I_2 =(1/4)ln∣x^2 −x+1∣−(1/(2(√3)))tan^(−1) ((2x−1)/(√3))+n I=(1/4)ln∣x^2 +x+1∣+(1/(2(√3))) tan^(−1) ((2x+1)/(√3)) −(1/4)ln∣x^2 −x+1∣+(1/(2(√3)))tan^(−1) ((2x−1)/(√3))+C =(1/4)ln∣((x^2 +x+1)/(x^2 −x+1))∣+(1/(2(√3)))[tan^(−1) ((2x+1)/(√3))+tan^(−1) ((2x−1)/(√3))]+C ∫_0 ^∞ (dx/(x^4 +x^2 +1))=(1/4)[ln∣((x^2 +x+1)/(x^2 −x+1))∣]_0 ^∞ + +(1/(2(√3)))[tan^(−1) ((2x+1)/(√3))+tan^(−1) ((2x−1)/(√3))]_0 ^∞ =(1/4){ln(((1+(1/∞)+(1/∞^2 ))/(1−(1/∞)+(1/∞^2 ))))−ln(((0^2 +0+1)/(0^2 −0+1)))} +(1/(2(√3))){tan^(−1) (∞)+tan^(−1) (∞)−tan^(−1) ((1/(√3)))−tan^(−1) (((−1)/(√3)))} =(1/4)[ln(1)−ln(1)]+(1/(2(√3)))[(π/2)+(π/2)−(π/6)+(π/6)] =(π/(2(√3))) ∫_0 ^∞ (dx/(x^4 +x^2 +1))=(π/(2(√3)))$
	$(x^{4} + x^{2} + 1) = ({a x}^{2} + b x + c) ({p x}^{2} + q x + r)$ $x^{4} + x^{2} + 1 = {a p x}^{4} + (a q + b p) x^{3} + (a r + b q + c p) x^{2}$ $+ (b r + c q) x + c r$ $a = 1, b = \pm 1, c = 1$ $p = 1, q = \mp 1, r = 1$ $x^{4} + x^{2} + 1 = (x^{2} + x + 1) (x^{2} - x + 1)$ $\frac{1}{x^{4} + x^{2} + 1} = \frac{a x + b}{x^{2} + x + 1} + \frac{p x + q}{x^{2} - x + 1}$ $1 = (a + p) x^{3} + (- a + b + p + q) x^{2} + (a - b + p + q) x$ $+ (b + q)$ $a = \frac{1}{2}, b = \frac{1}{2}$ $p = \frac{- 1}{2}, q = \frac{1}{2}$ $\frac{1}{x^{4} + x^{2} + 1} = \frac{x + 1}{x^{2} + x + 1} - \frac{x - 1}{x^{4} + x + 1}$ $I = \int \frac{d x}{x^{4} + x^{2} + 1} = \frac{1}{2} \int \frac{x + 1}{x^{2} + x + 1} d x - \frac{1}{2} \int \frac{x - 1}{x^{2} + x + 1} d x$ $I_{1} = \frac{1}{2} \int \frac{x + 1}{x^{2} + x + 1} d x$ $= \frac{1}{4} \int \frac{2 x + 1}{x^{2} + x + 1} d x + \frac{1}{4} \int \frac{d x}{x^{2} + x + 1}$ $= \frac{1}{4} l n ∣ x^{2} + x + 1 ∣ + \frac{1}{4} \int \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}$ $= \frac{1}{4} l n ∣ x^{2} + x + 1 ∣ + \frac{1}{4} \times \frac{2}{\sqrt{3}} {t a n}^{- 1} \frac{(x + \frac{1}{2})}{\frac{\sqrt{3}}{2}} + c$ $I_{1} = \frac{1}{4} l n ∣ x^{2} + x + 1 ∣ + \frac{1}{2 \sqrt{3}} {t a n}^{- 1} \frac{2 x + 1}{\sqrt{3}} + m$ $I_{2} = \frac{1}{2} \int \frac{x - 1}{x^{2} - x + 1} d x$ $= \frac{1}{4} \int \frac{2 x - 1}{x^{2} - x + 1} d x - \frac{1}{4} \int \frac{d x}{x^{2} - x + 1}$ $= \frac{1}{4} l n ∣ x^{2} - x + 1 ∣ - \frac{1}{4} \int \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}$ $I_{2} = \frac{1}{4} l n ∣ x^{2} - x + 1 ∣ - \frac{1}{2 \sqrt{3}} {t a n}^{- 1} \frac{2 x - 1}{\sqrt{3}} + n$ $I = \frac{1}{4} l n ∣ x^{2} + x + 1 ∣ + \frac{1}{2 \sqrt{3}} {t a n}^{- 1} \frac{2 x + 1}{\sqrt{3}}$ $- \frac{1}{4} l n ∣ x^{2} - x + 1 ∣ + \frac{1}{2 \sqrt{3}} {t a n}^{- 1} \frac{2 x - 1}{\sqrt{3}} + C$ $= \frac{1}{4} l n ∣ \frac{x^{2} + x + 1}{x^{2} - x + 1} ∣ + \frac{1}{2 \sqrt{3}} [{t a n}^{- 1} \frac{2 x + 1}{\sqrt{3}} + {t a n}^{- 1} \frac{2 x - 1}{\sqrt{3}}] + C$ $\int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{1}{4} {[l n ∣ \frac{x^{2} + x + 1}{x^{2} - x + 1} ∣]}_{0}^{\infty} +$ $+ \frac{1}{2 \sqrt{3}} {[{t a n}^{- 1} \frac{2 x + 1}{\sqrt{3}} + {t a n}^{- 1} \frac{2 x - 1}{\sqrt{3}}]}_{0}^{\infty}$ $= \frac{1}{4} {l n (\frac{1 + \frac{1}{\infty} + \frac{1}{\infty^{2}}}{1 - \frac{1}{\infty} + \frac{1}{\infty^{2}}}) - l n (\frac{0^{2} + 0 + 1}{0^{2} - 0 + 1})}$ $+ \frac{1}{2 \sqrt{3}} {{t a n}^{- 1} (\infty) + {t a n}^{- 1} (\infty) - {t a n}^{- 1} (\frac{1}{\sqrt{3}}) - {t a n}^{- 1} (\frac{- 1}{\sqrt{3}})}$ $= \frac{1}{4} [l n (1) - l n (1)] + \frac{1}{2 \sqrt{3}} [\frac{π}{2} + \frac{π}{2} - \frac{π}{6} + \frac{π}{6}]$ $= \frac{π}{2 \sqrt{3}}$ $\int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π}{2 \sqrt{3}}$
	Commented by niroj last updated on 14/May/20
	Great hardwork��
	Commented by prakash jain last updated on 14/May/20
	��
	Commented by Kunal12588 last updated on 14/May/20
	thanks sir!
	Commented by mathmax by abdo last updated on 14/May/20

	$t h a n k x f o r t h i s h a r d w o r k$
	Answered by Ar Brandon last updated on 14/May/20

	$L = \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{1}{2} \int_{0}^{\infty} \frac{(x^{2} + 1) - (x^{2} - 1)}{x^{4} + x^{2} + 1} d x$ $\Rightarrow 2 L = \int_{0}^{\infty} \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x - \int_{0}^{\infty} \frac{x^{2} - 1}{x^{4} + x^{2} + 1} d x$ $\Rightarrow 2 L = \int_{0}^{\infty} \frac{1 + \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}} d x - \int_{0}^{\infty} \frac{1 - \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}} d x$ $\Rightarrow 2 L = \int_{0}^{\infty} \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 3} d x - \int_{0}^{\infty} \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2} - 1} d x$ $\Rightarrow 2 L \underset{u = x - \frac{1}{x}}{\overset{v = x + \frac{1}{x}}{=}} \int_{- \infty}^{+ \infty} \frac{du}{u^{2} + 3} - \int_{+ \infty}^{+ \infty} \frac{dv}{v^{2} - 1}$ $2 L = \frac{\sqrt{3}}{3} {[\arctan (\frac{u}{\sqrt{3}})]}_{- \infty}^{+ \infty}$ $2 L = \frac{π \sqrt{3}}{3} \Rightarrow L = \frac{π \sqrt{3}}{6}$ $\int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π \sqrt{3}}{6}$
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