∫_0 ^( ∞) (dx/((x^4 −x^2 +1)^2 ))


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Question Number 132410 by bramlexs22 last updated on 14/Feb/21

$\int_{0}^{\infty} \frac{dx}{{(x^{4} - x^{2} + 1)}^{2}}$
	Answered by EDWIN88 last updated on 14/Feb/21
	$I=∫_0 ^∞ (dx/((x^4 −x^2 +1))) ; replace x by (1/x) I=∫_∞ ^0 (((−(1/x^2 ))/(((1/x^4 )−(1/x^2 )+1)^2 )))dx=∫_0 ^∞ ((x^6 dx)/((x^4 −x^2 +1)^2 )) 2I=∫_0 ^∞ ((x^6 +1)/((x^4 −x^2 +1)^2 ))dx ; Q(x)=(x^4 −x^2 +1)^2 Q′(x)=2(4x^3 −2x)(x^4 −x^2 +1) gcd Q′(x) and Q(x) is Q_1 (x)=x^4 −x^2 +1 and Q_2 (x)=((Q(x))/(Q_1 (x)))= x^4 −x^2 +1 I=(1/2)∫((x^6 +1)/((x^4 −x^2 +1)^2 )) dx = (1/2){ [((ax^3 +bx^2 +cx+d)/(x^4 −x^2 +1)) ]+∫ ((ex^3 +fx^2 +gx+h)/(x^4 −x^2 +1)) dx } differentiating both sides and solving for coefficients$
	$I = \int_{0}^{\infty} \frac{dx}{(x^{4} - x^{2} + 1)}; replace x by \frac{1}{x}$ $I = \int_{\infty}^{0} (\frac{- \frac{1}{x^{2}}}{{(\frac{1}{x^{4}} - \frac{1}{x^{2}} + 1)}^{2}}) dx = \int_{0}^{\infty} \frac{x^{6} dx}{{(x^{4} - x^{2} + 1)}^{2}}$ $2 I = \int_{0}^{\infty} \frac{x^{6} + 1}{{(x^{4} - x^{2} + 1)}^{2}} dx; Q (x) = {(x^{4} - x^{2} + 1)}^{2}$ $Q^{'} (x) = 2 ({4 x}^{3} - 2 x) (x^{4} - x^{2} + 1)$ $\gcd Q^{'} (x) and Q (x) is Q_{1} (x) = x^{4} - x^{2} + 1$ $and Q_{2} (x) = \frac{Q (x)}{Q_{1} (x)} = x^{4} - x^{2} + 1$ $I = \frac{1}{2} \int \frac{x^{6} + 1}{{(x^{4} - x^{2} + 1)}^{2}} dx = \frac{1}{2} {[\frac{{ax}^{3} + {bx}^{2} + cx + d}{x^{4} - x^{2} + 1}] + \int \frac{{ex}^{3} + {fx}^{2} + gx + h}{x^{4} - x^{2} + 1} dx}$ $differentiating both sides and solving$ $for coefficients$
	Answered by Lordose last updated on 14/Feb/21

	$Ω = \int_{0}^{\infty} \frac{1}{{(x^{4} - x^{2} + 1)}^{2}} dx = \int_{0}^{\infty} \frac{{(1 + x^{2})}^{2}}{{(1 + x^{6})}^{2}} dx$ $Ω = \int_{0}^{\infty} \frac{1}{{(1 + x^{6})}^{2}} dx + \int_{0}^{\infty} \frac{{2 x}^{2}}{{(1 + x^{6})}^{2}} dx + \int_{0}^{\infty} \frac{x^{4}}{{(1 + x^{6})}^{2}} dx$ $Ω \overset{x = u^{\frac{1}{6}}}{=} \frac{1}{6} (\int_{0}^{\infty} \frac{u^{\frac{1}{6} - 1}}{{(1 + u)}^{2}} du + 2 \int_{0}^{\infty} \frac{u^{\frac{1}{3} + \frac{1}{6} - 1}}{{(1 + u)}^{2}} + \int_{0}^{\infty} \frac{u^{\frac{2}{3} + \frac{1}{6} - 1}}{{(1 + u)}^{2}} du)$ $Ω = \frac{1}{6} (β (\frac{1}{6}, \frac{11}{6}) + 2 β (\frac{1}{2}, \frac{3}{2}) + β (\frac{5}{6}, \frac{7}{6}))$
	Answered by Dwaipayan Shikari last updated on 14/Feb/21

	$\int_{0}^{\infty} \frac{{(x^{2} + 1)}^{2}}{{(x^{6} + 1)}^{2}} d x = \int_{0}^{\infty} \frac{x^{4}}{{(x^{6} + 1)}^{2}} d x + \frac{2 x^{2}}{{(x^{6} + 1)}^{2}} + \frac{1}{{(x^{6} + 1)}^{2}} d x x^{6} = u$ $= \frac{1}{6} \int_{0}^{\infty} \frac{u^{- \frac{1}{6}}}{{(u + 1)}^{2}} d u + \frac{1}{3} \int_{0}^{\infty} \frac{u^{- \frac{1}{2}}}{{(u + 1)}^{2}} + \frac{1}{6} \int_{0}^{\infty} \frac{u^{- \frac{5}{6}}}{{(u + 1)}^{2}} d u$ $= \frac{1}{6} \int_{0}^{\infty} \frac{u^{\frac{5}{6} - 1}}{{(u + 1)}^{\frac{7}{6} + \frac{5}{6}}} + \frac{1}{3} \int_{0}^{\infty} \frac{u^{\frac{1}{2} - 1}}{{(u + 1)}^{\frac{3}{2} + \frac{1}{2}}} + \frac{1}{6} \int_{0}^{\infty} \frac{u^{\frac{1}{6} - 1}}{{(u + 1)}^{\frac{1}{6} + \frac{11}{6}}} d u$ $= \frac{1}{6} . \frac{Γ (\frac{5}{6}) Γ (\frac{7}{6})}{Γ (2)} + \frac{1}{3} . \frac{Γ (\frac{1}{2}) Γ (\frac{3}{2})}{Γ (2)} + \frac{1}{6} . \frac{Γ (\frac{1}{6}) Γ (\frac{11}{6})}{Γ (2)}$ $= \frac{π}{18} + \frac{π}{6} + \frac{5 π}{18} = \frac{π}{2}$
	Answered by mathmax by abdo last updated on 14/Feb/21
	$I=∫_0 ^∞ (dx/((x^4 −x^2 +1)^2 )) ⇒2I=∫_(−∞) ^(+∞) (dx/((x^4 −x^2 +1)^2 )) ϕ(z)=(1/((z^4 −z^2 +1)^2 )) poles of ϕ! z^4 −z^2 +1=0 ⇒u^2 −u+1=0(u=z^2 ) Δ=−3 ⇒u_1 =((1+i(√3))/2) =e^((iπ)/3) and u_2 =e^(−((iπ)/3)) ⇒ z^4 −z^2 +1 =(z^2 −e^((iπ)/3) )(z^2 −e^(−((iπ)/3)) ) ⇒ϕ(z)=(1/((z^2 −e^((iπ)/3) )^2 (z^2 −e^(−((iπ)/3)) )^2 )) =(1/((z−e^((iπ)/6) )^2 (z+e^((iπ)/6) )^2 (z−e^(−((iπ)/6)) )^2 (z+e^(−((iπ)/6)) )^2 )) the poles of ϕ are +^− e^((iπ)/6) and +^− e^(−((iπ)/6)) (doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,e^((iπ)/6) )+Res(ϕ,−e^(−((iπ)/6)) )} Res(ϕ,e^((iπ)/6) )=lim_(z→e^((iπ)/6) ) (1/((2−1)!)){(z−e^((iπ)/6) )^2 ϕ(z)}^((1)) =lim_(z→e^((iπ)/6) ) {(1/((z+e^((iπ)/6) )^2 (z^2 −e^(−((iπ)/3)) )))}^((1)) =lim_(z→e^((iπ)/6) ) −((2(z+e^((iπ)/6) )(z^2 −e^(−((iπ)/3)) )+2z(z+e^((iπ)/6) )^2 )/((z+e^((iπ)/6) )^4 (z^2 −e^(−((iπ)/3)) )^4 )) =−lim_(z→e^((iπ)/6) ) (2/((z+e^((iπ)/6) )^3 (z^2 −e^(−((iπ)/3)) )^3 )) +((2z)/((z+e^((iπ)/6) )^2 (z^2 −e^(−((iπ)/3)) )^4 )) =−{(2/((2e^((iπ)/3) )^3 (2isin((π/3)))^3 )) +((2e^((iπ)/6) )/((2e^((iπ)/6) )^2 (2isin((π/3)))^4 ))} =−{(2/(8(−1)(−8i)(((√3)/2))^3 )) +((2e^((iπ)/6) )/(4e^((iπ)/3) .16(((√3)/2))^4 ))}....be continued...$
	$I = \int_{0}^{\infty} \frac{dx}{{(x^{4} - x^{2} + 1)}^{2}} \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{4} - x^{2} + 1)}^{2}}$ $φ (z) = \frac{1}{{(z^{4} - z^{2} + 1)}^{2}} poles of φ!$ $z^{4} - z^{2} + 1 = 0 \Rightarrow u^{2} - u + 1 = 0 (u = z^{2})$ $Δ = - 3 \Rightarrow u_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}} and u_{2} = e^{- \frac{i π}{3}} \Rightarrow$ $z^{4} - z^{2} + 1 = (z^{2} - e^{\frac{i π}{3}}) (z^{2} - e^{- \frac{i π}{3}}) \Rightarrow φ (z) = \frac{1}{{(z^{2} - e^{\frac{i π}{3}})}^{2} {(z^{2} - e^{- \frac{i π}{3}})}^{2}}$ $= \frac{1}{{(z - e^{\frac{i π}{6}})}^{2} {(z + e^{\frac{i π}{6}})}^{2} {(z - e^{- \frac{i π}{6}})}^{2} {(z + e^{- \frac{i π}{6}})}^{2}}$ $the poles of φ are \bar{+} e^{\frac{i π}{6}} and \bar{+} e^{- \frac{i π}{6}} (doubles)$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, e^{\frac{i π}{6}}) + Res (φ, - e^{- \frac{i π}{6}})}$ $Res (φ, e^{\frac{i π}{6}}) = \lim_{z \to e^{\frac{i π}{6}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{6}})}^{2} φ (z)}^{(1)}$ $= \lim_{z \to e^{\frac{i π}{6}}} {\frac{1}{{(z + e^{\frac{i π}{6}})}^{2} (z^{2} - e^{- \frac{i π}{3}})}}^{(1)}$ $= \lim_{z \to e^{\frac{i π}{6}}} - \frac{2 (z + e^{\frac{i π}{6}}) (z^{2} - e^{- \frac{i π}{3}}) + 2 z {(z + e^{\frac{i π}{6}})}^{2}}{{(z + e^{\frac{i π}{6}})}^{4} {(z^{2} - e^{- \frac{i π}{3}})}^{4}}$ $= - \lim_{z \to e^{\frac{i π}{6}}} \frac{2}{{(z + e^{\frac{i π}{6}})}^{3} {(z^{2} - e^{- \frac{i π}{3}})}^{3}} + \frac{2 z}{{(z + e^{\frac{i π}{6}})}^{2} {(z^{2} - e^{- \frac{i π}{3}})}^{4}}$ $= - {\frac{2}{{({2 e}^{\frac{i π}{3}})}^{3} {(2 isin (\frac{π}{3}))}^{3}} + \frac{{2 e}^{\frac{i π}{6}}}{{({2 e}^{\frac{i π}{6}})}^{2} {(2 isin (\frac{π}{3}))}^{4}}}$ $= - {\frac{2}{8 (- 1) (- 8 i) {(\frac{\sqrt{3}}{2})}^{3}} + \frac{{2 e}^{\frac{i π}{6}}}{{4 e}^{\frac{i π}{3}} . 16 {(\frac{\sqrt{3}}{2})}^{4}}} . . . . be continued . . .$
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