find the value of ∫_0 ^(+∞) (x^4 /((1+x^2 +x^4 )^2 ))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 57233 by maxmathsup by imad last updated on 31/Mar/19

$f i n d t h e v a l u e o f \int_{0}^{+ \infty} \frac{x^{4}}{{(1 + x^{2} + x^{4})}^{2}} d x$
Commented by maxmathsup by imad last updated on 04/Apr/19
$let I =∫_0 ^∞ (x^4 /((1+x^2 +x^4 )^2 )) dx ⇒2I =∫_(−∞) ^(+∞) (x^4 /((x^4 +x^2 +1)^2 )) dx let ϕ(z) =(z^4 /((z^4 +z^2 +1)^2 )) .poles of ϕ? z^4 +z^2 +1 =0 ⇒t^2 +t +1 =0 (witht=z^2 ) →Δ=1−4=−3 ⇒ t_1 =((−1+i(√3))/2) =e^((i2π)/3) and t_2 =((−1−i(√3))/2) =e^(−((i2π)/3)) z^2 =t_1 ⇒z =+^− e^((iπ)/3) z^2 =t_2 ⇒z =+^− e^(−((iπ)/3)) ⇒the poles of ϕ are +^− e^((iπ)/3) and +^− e^(−((iπ)/3)) (doubles) and ϕ(z) =(z^4 /((z−e^((iπ)/3) )^2 (z+e^((iπ)/3) )^2 (z−e^(−((iπ)/3)) )^2 (z+e^(−((iπ)/3)) )^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,e^((iπ)/3) ) +Res(ϕ,−e^(−((iπ)/3)) )} Res(ϕ,e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (1/((2−1)!)) {(z−e^((iπ)/3) )^2 ϕ(z)}^((1)) =lim_(z→e^((iπ)/3) ) { (z^4 /((z+e^((iπ)/3) )^2 (z−e^(−((iπ)/3)) )^2 (z+e^(−((iπ)/3)) )^2 ))}^((1)) =lim_(z→e^((iπ)/3) ) ((4z^3 (z+e^((iπ)/3) )^2 (z−e^(−((iπ)/3)) )^2 (z+e^(−((iπ)/3)) )^2 −z^4 {(z+e^((iπ)/3) )^2 (z−e^(−((iπ)/3)) )^2 (z+e^(−((iπ)/3)) )^2 }^((1)) )/((z+e^((iπ)/3) )^4 (z−e^(−((iπ)/3)) )^4 (z+e^(−((iπ)/3)) )^4 )) ....be continued...$
$l e t I = \int_{0}^{\infty} \frac{x^{4}}{{(1 + x^{2} + x^{4})}^{2}} d x \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{x^{4}}{{(x^{4} + x^{2} + 1)}^{2}} d x l e t$ $φ (z) = \frac{z^{4}}{{(z^{4} + z^{2} + 1)}^{2}} . p o l e s o f φ ?$ $z^{4} + z^{2} + 1 = 0 \Rightarrow t^{2} + t + 1 = 0 (w i t h t = z^{2}) \to Δ = 1 - 4 = - 3 \Rightarrow$ $t_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}} a n d t_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}}$ $z^{2} = t_{1} \Rightarrow z = \bar{+} e^{\frac{i π}{3}}$ $z^{2} = t_{2} \Rightarrow z = \bar{+} e^{- \frac{i π}{3}} \Rightarrow t h e p o l e s o f φ a r e \bar{+} e^{\frac{i π}{3}} a n d \bar{+} e^{- \frac{i π}{3}} (d o u b l e s) a n d$ $φ (z) = \frac{z^{4}}{{(z - e^{\frac{i π}{3}})}^{2} {(z + e^{\frac{i π}{3}})}^{2} {(z - e^{- \frac{i π}{3}})}^{2} {(z + e^{- \frac{i π}{3}})}^{2}} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{3}}) + R e s (φ, - e^{- \frac{i π}{3}})}$ $R e s (φ, e^{\frac{i π}{3}}) = {l i m}_{z \to e^{\frac{i π}{3}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{3}})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to e^{\frac{i π}{3}}} {\frac{z^{4}}{{(z + e^{\frac{i π}{3}})}^{2} {(z - e^{- \frac{i π}{3}})}^{2} {(z + e^{- \frac{i π}{3}})}^{2}}}^{(1)}$ $= {l i m}_{z \to e^{\frac{i π}{3}}} \frac{4 z^{3} {(z + e^{\frac{i π}{3}})}^{2} {(z - e^{- \frac{i π}{3}})}^{2} {(z + e^{- \frac{i π}{3}})}^{2} - z^{4} {{(z + e^{\frac{i π}{3}})}^{2} {(z - e^{- \frac{i π}{3}})}^{2} {(z + e^{- \frac{i π}{3}})}^{2}}^{(1)}}{{(z + e^{\frac{i π}{3}})}^{4} {(z - e^{- \frac{i π}{3}})}^{4} {(z + e^{- \frac{i π}{3}})}^{4}}$ $. . . . b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum