calculate ∫_0 ^∞ (dx/((x^4 −x^2 +1)^2 ))


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Question Number 83253 by mathmax by abdo last updated on 29/Feb/20

$c a l c u l a t e \int_{0}^{\infty} \frac{d x}{{(x^{4} - x^{2} + 1)}^{2}}$
Commented by abdomathmax last updated on 01/Mar/20

$s o r r y t h e Q i s c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{4} - x^{2} + 1)}^{2}}$
Commented by mathmax by abdo last updated on 01/Mar/20
$let I =∫_(−∞) ^(+∞) (dx/((x^4 −x^2 +1)^2 )) let ϕ(z)=(1/((z^4 −z^2 +1)^2 )) poles of ϕ? z^4 −z^2 +1=0⇒t^2 −t+1=0 (t=z^2 ) Δ=1−4=−3 ⇒t_1 =((1+i(√3))/2) =e^((iπ)/3) and t_2 =((1−i(√3))/2)=e^(−((iπ)/3)) t^2 −t+1 =(t−e^((iπ)/3) )(t−e^(−((iπ)/3)) ) ⇒z^4 −z^2 +1 =(z^2 −e^((iπ)/3) )(z^2 −e^((−iπ)/3) ) =(z−e^((iπ)/6) )(z+e^((iπ)/6) )(z−e^(−((iπ)/6)) )(z+e^((−iπ)/6) )⇒ ϕ(z)=(1/((z−e^((iπ)/6) )^2 (z+e^((iπ)/6) )^2 (z−e^(−((iπ)/6)) )^2 (z+e^(−((iπ)/6)) )^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ{Res(ϕ,e^((iπ)/6) )+Res(ϕ,−e^(−((iπ)/6)) )} Res(ϕ,e^((iπ)/6) )=lim_(z→e^((iπ)/6) ) (1/((2−1)!)){(z−e^((iπ)/6) )^2 ϕ(z)}^((1)) =lim_(z→e^((iπ)/6) ) {(1/((z+e^((iπ)/6) )(z^2 −e^(−((iπ)/3)) )^2 ))}^((1)) =lim_(z→e^((iπ)/6) ) −(((z^2 −e^(−((iπ)/3)) )^2 +(z+e^((iπ)/6) )×(4z)(z^2 −e^(−((iπ)/3)) ))/((z+e^((iπ)/6) )^2 (z^2 −e^(−((iπ)/3)) )^4 )) =−lim_(z→e^((iπ)/6) ) (((z^2 −e^(−((iπ)/3)) )+4z(z+e^((iπ)/6) ))/((z+e^((iπ)/6) )^2 (z^2 −e^(−((iπ)/3)) )^3 )) =−((e^((iπ)/3) −e^(−((iπ)/3)) +4e^((iπ)/6) ×2e^((iπ)/6) )/((2e^((iπ)/6) )^2 (e^((iπ)/3) −e^(−((iπ)/3)) )^3 )) =−((2isin((π/3))+8 e^((iπ)/3) )/(2e^((iπ)/3) (2i)^3 sin^3 ((π/3)))) =−((isin((π/3))+4e^((iπ)/3) )/(−8i e^((iπ)/3) sin^3 ((π/3)))) =e^(−((iπ)/3)) ×((sin((π/3))−4ie^((iπ)/3) )/(sin^3 ((π/3)))) =((sin((π/3))e^(−((iπ)/3)) −4i)/(8(((√3)/2))^3 )) =((((√3)/2)e^(−((iπ)/3)) −4i)/(3(√3))) ....be continued...$
$l e t I = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{4} - x^{2} + 1)}^{2}} l e t φ (z) = \frac{1}{{(z^{4} - z^{2} + 1)}^{2}} p o l e s o f φ ?$ $z^{4} - z^{2} + 1 = 0 \Rightarrow t^{2} - t + 1 = 0 (t = z^{2})$ $Δ = 1 - 4 = - 3 \Rightarrow t_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}} a n d t_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}}$ $t^{2} - t + 1 = (t - e^{\frac{i π}{3}}) (t - e^{- \frac{i π}{3}}) \Rightarrow z^{4} - z^{2} + 1 = (z^{2} - e^{\frac{i π}{3}}) (z^{2} - e^{\frac{- i π}{3}})$ $= (z - e^{\frac{i π}{6}}) (z + e^{\frac{i π}{6}}) (z - e^{- \frac{i π}{6}}) (z + e^{\frac{- i π}{6}}) \Rightarrow$ $φ (z) = \frac{1}{{(z - e^{\frac{i π}{6}})}^{2} {(z + e^{\frac{i π}{6}})}^{2} {(z - e^{- \frac{i π}{6}})}^{2} {(z + e^{- \frac{i π}{6}})}^{2}}$ $r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{6}}) + R e s (φ, - e^{- \frac{i π}{6}})}$ $R e s (φ, e^{\frac{i π}{6}}) = {l i m}_{z \to e^{\frac{i π}{6}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{6}})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to e^{\frac{i π}{6}}} {\frac{1}{(z + e^{\frac{i π}{6}}) {(z^{2} - e^{- \frac{i π}{3}})}^{2}}}^{(1)}$ $= {l i m}_{z \to e^{\frac{i π}{6}}} - \frac{{(z^{2} - e^{- \frac{i π}{3}})}^{2} + (z + e^{\frac{i π}{6}}) \times (4 z) (z^{2} - e^{- \frac{i π}{3}})}{{(z + e^{\frac{i π}{6}})}^{2} {(z^{2} - e^{- \frac{i π}{3}})}^{4}}$ $= - {l i m}_{z \to e^{\frac{i π}{6}}} \frac{(z^{2} - e^{- \frac{i π}{3}}) + 4 z (z + e^{\frac{i π}{6}})}{{(z + e^{\frac{i π}{6}})}^{2} {(z^{2} - e^{- \frac{i π}{3}})}^{3}}$ $= - \frac{e^{\frac{i π}{3}} - e^{- \frac{i π}{3}} + 4 e^{\frac{i π}{6}} \times 2 e^{\frac{i π}{6}}}{{(2 e^{\frac{i π}{6}})}^{2} {(e^{\frac{i π}{3}} - e^{- \frac{i π}{3}})}^{3}} = - \frac{2 i s i n (\frac{π}{3}) + 8 e^{\frac{i π}{3}}}{2 e^{\frac{i π}{3}} {(2 i)}^{3} {s i n}^{3} (\frac{π}{3})}$ $= - \frac{i s i n (\frac{π}{3}) + 4 e^{\frac{i π}{3}}}{- 8 i e^{\frac{i π}{3}} {s i n}^{3} (\frac{π}{3})} = e^{- \frac{i π}{3}} \times \frac{s i n (\frac{π}{3}) - 4 {i e}^{\frac{i π}{3}}}{{s i n}^{3} (\frac{π}{3})}$ $= \frac{s i n (\frac{π}{3}) e^{- \frac{i π}{3}} - 4 i}{8 {(\frac{\sqrt{3}}{2})}^{3}} = \frac{\frac{\sqrt{3}}{2} e^{- \frac{i π}{3}} - 4 i}{3 \sqrt{3}} . . . . b e c o n t i n u e d . . .$
Commented by mathmax by abdo last updated on 01/Mar/20
$pareametric method let ϕ(a) =∫_0 ^∞ (dx/(x^4 −x^2 +a)) witha>(1/4) we have ϕ^′ (a) =−∫_0 ^∞ (dx/((x^4 −x^2 +a)^2 )) ⇒∫_0 ^∞ (dx/((x^4 −x^2 +a)^2 ))=−ϕ^′ (a) we have 2ϕ(a) =∫_(−∞) ^(+∞) (dx/(x^4 −x^2 +a)) let W(z)=(1/(z^4 −z^2 +a)) poles of W? z^4 −z^2 +a =0 ⇒t^2 −t +a =0 (t=z^2 ) Δ=1−4a<0 ⇒Δ=(i(√(4a−1)))^2 ⇒z_1 =((1+i(√(4a−1)))/2) z_2 =((1−i(√(4a−1)))/2) ⇒W(z) =(1/((z^2 −z_1 )(z^2 −z_2 ))) =(1/((z−(√z_1 ))(z+(√z_1 ))(z−(√z_2 ))(z+(√z_2 )))) ∫_(−∞) ^(+∞) W(z)dz =2iπ {Res(W,(√z_1 )) +Res(W,−(√z_2 ))} Res(W,(√z_1 )) =(1/(2(√z_1 )(z_1 ^2 −z_2 ))) Res(W,−(√z_2 )) =(1/(−2(√z_2 )(z_2 ^2 −z_1 ))) =−(1/(2(√z_2 )(z_2 ^2 −z_1 ))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =iπ{ (1/((√z_1 )(z_1 ^2 −z_2 )))−(1/((√z_2 )(z_2 ^2 −z_1 )))} ∣z_1 ∣=(1/2)(√(1+4a−1))=(√a) ⇒z_1 =(√a)e^(iarctan(√(4a−1))) z_2 =(√a)e^(−iarctan((√(4a−1)))) ⇒(√z_1 )=^4 (√a)e^((i/2)arctan((√(4a−1)))) z_1 ^2 −z_2 =a e^(2iarctan((√(4a−1)))) −(√a)e^(−iarctan((√(4a−1)))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =iπ{ (1/((^4 (√a))e^(iarctan((√(4a−1)))) (a e^(2iarctan((√(4a−1)))) −(√a)e^(−iarctan((√(4a−1)))) ))−(1/((^4 (√a))e^(−iarctan((√(4a−1)))) (ae^(−2iarctan((√(4a−1)))) −(√a)e^(iarctan((√(4a−1)))) )))} ...be continued...$
$p a r e a m e t r i c m e t h o d l e t φ (a) = \int_{0}^{\infty} \frac{d x}{x^{4} - x^{2} + a} w i t h a > \frac{1}{4}$ $w e h a v e φ^{^{'}} (a) = - \int_{0}^{\infty} \frac{d x}{{(x^{4} - x^{2} + a)}^{2}} \Rightarrow \int_{0}^{\infty} \frac{d x}{{(x^{4} - x^{2} + a)}^{2}} = - φ^{^{'}} (a)$ $w e h a v e 2 φ (a) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} - x^{2} + a} l e t W (z) = \frac{1}{z^{4} - z^{2} + a}$ $p o l e s o f W ?$ $z^{4} - z^{2} + a = 0 \Rightarrow t^{2} - t + a = 0 (t = z^{2})$ $Δ = 1 - 4 a < 0 \Rightarrow Δ = {(i \sqrt{4 a - 1})}^{2} \Rightarrow z_{1} = \frac{1 + i \sqrt{4 a - 1}}{2}$ $z_{2} = \frac{1 - i \sqrt{4 a - 1}}{2} \Rightarrow W (z) = \frac{1}{(z^{2} - z_{1}) (z^{2} - z_{2})} = \frac{1}{(z - \sqrt{z_{1}}) (z + \sqrt{z_{1}}) (z - \sqrt{z_{2}}) (z + \sqrt{z_{2}})}$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, \sqrt{z_{1}}) + R e s (W, - \sqrt{z_{2}})}$ $R e s (W, \sqrt{z_{1}}) = \frac{1}{2 \sqrt{z_{1}} (z_{1}^{2} - z_{2})}$ $R e s (W, - \sqrt{z_{2}}) = \frac{1}{- 2 \sqrt{z_{2}} (z_{2}^{2} - z_{1})} = - \frac{1}{2 \sqrt{z_{2}} (z_{2}^{2} - z_{1})} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = i π {\frac{1}{\sqrt{z_{1}} (z_{1}^{2} - z_{2})} - \frac{1}{\sqrt{z_{2}} (z_{2}^{2} - z_{1})}}$ $∣ z_{1} ∣= \frac{1}{2} \sqrt{1 + 4 a - 1} = \sqrt{a} \Rightarrow z_{1} = \sqrt{a} e^{i a r c t a n \sqrt{4 a - 1}}$ $z_{2} = \sqrt{a} e^{- i a r c t a n (\sqrt{4 a - 1})} \Rightarrow \sqrt{z_{1}} =^{4} \sqrt{a} e^{\frac{i}{2} a r c t a n (\sqrt{4 a - 1})}$ $z_{1}^{2} - z_{2} = a e^{2 i a r c t a n (\sqrt{4 a - 1})} - \sqrt{a} e^{- i a r c t a n (\sqrt{4 a - 1})} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = i π {\frac{1}{(^{4} \sqrt{a}) e^{i a r c t a n (\sqrt{4 a - 1})} (a e^{2 i a r c t a n (\sqrt{4 a - 1})} - \sqrt{a} e^{- i a r c t a n (\sqrt{4 a - 1})}} - \frac{1}{(^{4} \sqrt{a}) e^{- i a r c t a n (\sqrt{4 a - 1})} ({a e}^{- 2 i a r c t a n (\sqrt{4 a - 1})} - \sqrt{a} e^{i a r c t a n (\sqrt{4 a - 1})})}}$ $. . . b e c o n t i n u e d . . .$
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