calculate ∫_0 ^∞ (dx/((x^4 +x^2 +3)^2 ))


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Question Number 88928 by mathmax by abdo last updated on 13/Apr/20

$c a l c u l a t e \int_{0}^{\infty} \frac{d x}{{(x^{4} + x^{2} + 3)}^{2}}$
Commented by mathmax by abdo last updated on 15/Apr/20
$A =∫_0 ^∞ (dx/((x^4 +x^2 +3)^2 )) ⇒2A =∫_(−∞) ^(+∞) (dx/((x^4 +x^2 +3)^2 )) let ϕ(z)=(1/((z^4 +z^2 +3)^2 )) poles of ϕ? z^4 +z^2 +3 =0⇒t^2 +t+3=0 (t=z^2 ) Δ=1−12 =−11 ⇒t_1 =((−1+i(√(11)))/2) and t_2 =((−1−i(√(11)))/2) ∣t_1 ∣=(1/2)(√(1+11))=(1/2)(2(√3))=(√(3 )) ⇒t_1 =(√3)e^(−iarctan((√(11)))) t_2 =(√3)e^(iarctan((√(11)))) ⇒ϕ(z)=(1/((t−t_1 )^2 (t−t_2 )^2 )) =(1/((z^2 −(√3)e^(−iarctan((√(11)))) )^2 (z^2 −(√3)e^(iarctan((√(11)))) )^2 )) =(1/((z−αe^(−(i/2)arctan((√(11)))) )^2 (z+α e^(−(i/2)arctan((√(11)))) )^2 (z−α e^((i/2)arctan((√(11)))) )^2 (z+α e^((i/2)arctan((√(11)))) )^2 )) (with α =^4 (√3)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,α e^((i/2)arctan((√(11)))) )+Res(ϕ,−α e^(−(i/2)arctan((√(11)))) } Res(ϕ,α e^((i/2)arctan((√(11)))) ) =lim_(z→αe^((i/2)arctan((√(11)))) ) (1/((2−1)!)){(z−αe^((i/2)arcan((√(11)))) )^2 ϕ(z)}^((1)) =lim_(z→α e^((i/2)arctan((√(11)))) ) {(1/((z+α e^(−(i/2)arctan((√(11)))) )^2 (z^2 −(√3)e^(iarctan((√(11)))) )^2 )) }^((1)) ....be continued...$
$A = \int_{0}^{\infty} \frac{d x}{{(x^{4} + x^{2} + 3)}^{2}} \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{4} + x^{2} + 3)}^{2}}$ $l e t φ (z) = \frac{1}{{(z^{4} + z^{2} + 3)}^{2}} p o l e s o f φ ?$ $z^{4} + z^{2} + 3 = 0 \Rightarrow t^{2} + t + 3 = 0 (t = z^{2})$ $Δ = 1 - 12 = - 11 \Rightarrow t_{1} = \frac{- 1 + i \sqrt{11}}{2} a n d t_{2} = \frac{- 1 - i \sqrt{11}}{2}$ $∣ t_{1} ∣= \frac{1}{2} \sqrt{1 + 11} = \frac{1}{2} (2 \sqrt{3}) = \sqrt{3} \Rightarrow t_{1} = \sqrt{3} e^{- i a r c t a n (\sqrt{11})}$ $t_{2} = \sqrt{3} e^{i a r c t a n (\sqrt{11})} \Rightarrow φ (z) = \frac{1}{{(t - t_{1})}^{2} {(t - t_{2})}^{2}}$ $= \frac{1}{{(z^{2} - \sqrt{3} e^{- i a r c t a n (\sqrt{11})})}^{2} {(z^{2} - \sqrt{3} e^{i a r c t a n (\sqrt{11})})}^{2}}$ $= \frac{1}{{(z - α e^{- \frac{i}{2} a r c t a n (\sqrt{11})})}^{2} {(z + α e^{- \frac{i}{2} a r c t a n (\sqrt{11})})}^{2} {(z - α e^{\frac{i}{2} a r c t a n (\sqrt{11})})}^{2} {(z + α e^{\frac{i}{2} a r c t a n (\sqrt{11})})}^{2}}$ $(w i t h α =^{4} \sqrt{3})$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, α e^{\frac{i}{2} a r c t a n (\sqrt{11})}) + R e s (φ, - α e^{- \frac{i}{2} a r c t a n (\sqrt{11})}}$ $R e s (φ, α e^{\frac{i}{2} a r c t a n (\sqrt{11})}) = {l i m}_{z \to α e^{\frac{i}{2} a r c t a n (\sqrt{11})}} \frac{1}{(2 - 1)!} {{(z - α e^{\frac{i}{2} a r c a n (\sqrt{11})})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to α e^{\frac{i}{2} a r c t a n (\sqrt{11})}} {\frac{1}{{(z + α e^{- \frac{i}{2} a r c t a n (\sqrt{11})})}^{2} {(z^{2} - \sqrt{3} e^{i a r c t a n (\sqrt{11})})}^{2}}}^{(1)}$ $. . . . b e c o n t i n u e d . . .$
Commented by mathmax by abdo last updated on 15/Apr/20
$let use parametric method let f(t) =∫_0 ^∞ (dx/(x^4 +x^2 +t)) with t>(1/4) f^′ (t) =−∫_0 ^(+∞) (dx/((x^4 +x^2 +t)^2 )) ⇒ ∫_0 ^(+∞) (dx/((x^4 +x^2 +t)^2 )) =−f^′ (t) 2f(t)=∫_(−∞) ^(+∞) (dx/(x^4 +x^2 +t)) let h(z) =(1/(z^4 +z^2 +t)) polesf h? z^4 +z^2 +t =0 ⇒u^2 +u +t =o (u=z^2 ) Δ =1−4t =−(4t−1)<0 ⇒u_1 =((−1+i(√(4t−1)))/2) ∣u_1 ∣ =(1/2)(√(1+4t−1))=(√t) ⇒u_1 =(√t_ ) e^(−iarctan((√(4t−1)))) u_2 =((−1−i(√(4t−1)))/2)=(√t)e^(iarctan((√(4t−1)))) ⇒ h(z) =(1/((z^2 −(√t)e^(iarctan((√(4t−1))) )(z^2 −(√t)e^(−iarctan((√(4t))−1)) ))) =(1/((z−αe^((i/2)arctan((√(4t−1)))) )(z+α e^((i/2)arctan((√(4t−1)))) )(z−α e^(−(i/2)arctan((√(4t−1)))) )(z+αe^(−(i/2)arctan((√(4t−1)))) ))) ∫_(−∞) ^(+∞) h(z)dz =2iπ { Res(h,αe^((i/2)arctan((√(4t−1)))) ) +Res(h,−αe^(−(i/2)arctan((√(4t−1))) ) (α=^4 (√t)) Res(h,α e^((i/2)arctan((√(4t−1)))) )=(1/(2αe^((i/2)arctan((√(4t−1)))) (√t)(2i sin(arctan((√(4t−1)))))) =(e^(−(i/2)arctan((√(4t−1)))) /(4iα(√t)sin(arctan((√(4t−1))))) Res(h,−α e^(−(i/2)arctan((√(4t−1)))) ) =(1/(−2α e^(−(i/2)arctan((√(4t−1)))) (√t)(−2i sin(arctan((√(4t−1))))) =(e^((i/2)arctan((√(4t−1)))) /(4iα (√t)sin(arctan((√(4t−1)))))) ⇒ ∫_(−∞) ^(+∞) h(z)dz =(1/(2(√t)(^4 (√t))sin(arctan((√(4t−1))))){e^((i/2)arctan((√(4t−1)))) +e^(−(i/2)arctan((√(4t−1)))) ) =(1/(2(√t)(^4 (√t)) sin(arctan((√(4t−1)))))×2cos(arctan((√(4t−1))) =(1/((√t)(^4 (√t))tan(arctan((√(4t−1))))) =(1/((√t)(^4 (√t))(√(4t−1)))) =2f(t) ⇒ f(t)=(1/((^4 (√t))(√(4t^2 −t)))) ⇒f^′ (t)=−(({(^4 (√t))(√(4t^2 −1))}^′ )/((√t)(4t^2 −t))) {t^(1/4) (√(4t^2 −t))}^′ =(1/4)t^((1/4)−1) (√(4t^2 −1)) +t^(1/4) ×((8t−1)/(2(√(4t^2 −t))))$
$l e t u s e p a r a m e t r i c m e t h o d l e t f (t) = \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + t} w i t h t > \frac{1}{4}$ $f^{^{'}} (t) = - \int_{0}^{+ \infty} \frac{d x}{{(x^{4} + x^{2} + t)}^{2}} \Rightarrow \int_{0}^{+ \infty} \frac{d x}{{(x^{4} + x^{2} + t)}^{2}} = - f^{^{'}} (t)$ $2 f (t) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + x^{2} + t} l e t h (z) = \frac{1}{z^{4} + z^{2} + t} p o l e s f h ?$ $z^{4} + z^{2} + t = 0 \Rightarrow u^{2} + u + t = o (u = z^{2})$ $Δ = 1 - 4 t = - (4 t - 1) < 0 \Rightarrow u_{1} = \frac{- 1 + i \sqrt{4 t - 1}}{2}$ $∣ u_{1} ∣ = \frac{1}{2} \sqrt{1 + 4 t - 1} = \sqrt{t} \Rightarrow u_{1} = \sqrt{t_{}} e^{- i a r c t a n (\sqrt{4 t - 1})}$ $u_{2} = \frac{- 1 - i \sqrt{4 t - 1}}{2} = \sqrt{t} e^{i a r c t a n (\sqrt{4 t - 1})} \Rightarrow$ $h (z) = \frac{1}{(z^{2} - \sqrt{t} e^{i a r c t a n (\sqrt{4 t - 1}}) (z^{2} - \sqrt{t} e^{- i a r c t a n (\sqrt{4 t} - 1)})}$ $= \frac{1}{(z - α e^{\frac{i}{2} a r c t a n (\sqrt{4 t - 1})}) (z + α e^{\frac{i}{2} a r c t a n (\sqrt{4 t - 1})}) (z - α e^{- \frac{i}{2} a r c t a n (\sqrt{4 t - 1})}) (z + α e^{- \frac{i}{2} a r c t a n (\sqrt{4 t - 1})})}$ $\int_{- \infty}^{+ \infty} h (z) d z = 2 i π {R e s (h, α e^{\frac{i}{2} a r c t a n (\sqrt{4 t - 1})}) + R e s (h, - α e^{- \frac{i}{2} a r c t a n (\sqrt{4 t - 1}})$ $(α =^{4} \sqrt{t})$ $R e s (h, α e^{\frac{i}{2} a r c t a n (\sqrt{4 t - 1})}) = \frac{1}{2 α e^{\frac{i}{2} a r c t a n (\sqrt{4 t - 1})} \sqrt{t} (2 i s i n (a r c t a n (\sqrt{4 t - 1}))}$ $= \frac{e^{- \frac{i}{2} a r c t a n (\sqrt{4 t - 1})}}{4 i α \sqrt{t} s i n (a r c t a n (\sqrt{4 t - 1})}$ $R e s (h, - α e^{- \frac{i}{2} a r c t a n (\sqrt{4 t - 1})})$ $= \frac{1}{- 2 α e^{- \frac{i}{2} a r c t a n (\sqrt{4 t - 1})} \sqrt{t} (- 2 i s i n (a r c t a n (\sqrt{4 t - 1})}$ $= \frac{e^{\frac{i}{2} a r c t a n (\sqrt{4 t - 1})}}{4 i α \sqrt{t} s i n (a r c t a n (\sqrt{4 t - 1}))} \Rightarrow$ $\int_{- \infty}^{+ \infty} h (z) d z = \frac{1}{2 \sqrt{t} (^{4} \sqrt{t}) s i n (a r c t a n (\sqrt{4 t - 1})} {e^{\frac{i}{2} a r c t a n (\sqrt{4 t - 1})} + e^{- \frac{i}{2} a r c t a n (\sqrt{4 t - 1)}})$ $= \frac{1}{2 \sqrt{t} (^{4} \sqrt{t}) s i n (a r c t a n (\sqrt{4 t - 1)}} \times 2 c o s (a r c t a n (\sqrt{4 t - 1})$ $= \frac{1}{\sqrt{t} (^{4} \sqrt{t}) t a n (a r c t a n (\sqrt{4 t - 1})} = \frac{1}{\sqrt{t} (^{4} \sqrt{t}) \sqrt{4 t - 1}} = 2 f (t) \Rightarrow$ $f (t) = \frac{1}{(^{4} \sqrt{t}) \sqrt{4 t^{2} - t}} \Rightarrow f^{^{'}} (t) = - \frac{{(^{4} \sqrt{t}) \sqrt{4 t^{2} - 1}}^{^{'}}}{\sqrt{t} (4 t^{2} - t)}$ ${t^{\frac{1}{4}} \sqrt{4 t^{2} - t}}^{^{'}} = \frac{1}{4} t^{\frac{1}{4} - 1} \sqrt{4 t^{2} - 1} + t^{\frac{1}{4}} \times \frac{8 t - 1}{2 \sqrt{4 t^{2} - t}}$
Commented by mathmax by abdo last updated on 15/Apr/20

$f^{^{'}} (t) = \frac{\frac{1}{2} t^{- \frac{3}{4}} (4 t^{2} - t) + t^{\frac{1}{4}} (8 t - 1)}{2 \sqrt{t} \sqrt{4 t^{2} - t} (4 t^{2} - t)} . . . .$
Commented by mathmax by abdo last updated on 15/Apr/20

$\int_{0}^{\infty} \frac{d x}{{(x^{4} + x^{2} + 3)}^{2}} = - f^{^{'}} (3)$
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