calculate ∫_0 ^∞ (dt/(1+t^(4 ) +t^6 ))


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Question Number 39165 by math khazana by abdo last updated on 03/Jul/18

$c a l c u l a t e \int_{0}^{\infty} \frac{d t}{1 + t^{4} + t^{6}}$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jul/18

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jul/18

$1.5 \times 1 > \int_{0}^{\infty} \frac{d t}{1 + t^{4} + t^{6}} > 3 \times {(0.5)}^{2}$ $1.5 > \int_{0}^{\infty} \frac{d t}{1 + t^{4} + t^{6}} > 0.75$
Commented by math khazana by abdo last updated on 04/Jul/18

$l e t I = \int_{0}^{\infty} \frac{d t}{1 + t^{4} + t^{6}}$ $2 I = \int_{- \infty}^{+ \infty} \frac{d t}{1 + t^{4} + t^{6}} l e t φ (z) = \frac{1}{1 + z^{4} + z^{8}}$ $p o l e s o f φ ?$ $1 + z^{4} + z^{8} = 0 \Leftrightarrow 1 + z^{4} + {(z^{4})}^{2} = 0$ $\Leftrightarrow \frac{1 - {(z^{4})}^{3}}{1 - z^{4}} = 0 a n d z^{4} \neq 1 \Leftrightarrow z^{12} = 1 a n d z^{4} \neq 1$ $z^{4} = 1 \Rightarrow 4 θ = 2 k π \Rightarrow θ_{k} = \frac{k π}{2} a n d k \in [[0, 3]]$ $z_{k} = e^{i \frac{k π}{2}} l e t z = {r e}^{i α} s o z^{12} = 1 \Rightarrow r = 1 a n d$ $12 α = 2 k π \Rightarrow α_{k} = \frac{k π}{6} w i t h k \in [[0, 11]] \Rightarrow$ $z_{k} = e^{i \frac{k π}{6}} a n d k \in [[0, 5]]$ $z_{0} = 1 (n o n p o l e o f φ)$ $z_{1} = e^{i \frac{π}{6}} z_{1}^{4} = e^{i \frac{2 π}{3}} \neq 1 \Rightarrow z_{1} p o l e o f φ$ $z_{2} = e^{i \frac{π}{3}} (z_{2}^{4} \neq 1 \Rightarrow z_{2} p o l e o f φ)$ $z_{3} = e^{i \frac{π}{2}} (z_{3}^{4} = 1 s o z_{3} n o n p o l e o f φ)$ $z_{4} = e^{i \frac{2 π}{3}} (z_{4}^{4} \neq 1 s o z_{4} p o l e o f φ)$ $z_{5} = e^{i \frac{5 π}{6}} (z_{5}^{4} \neq 1 s o z_{5} i s p o l e o f φ)$ $z_{6} = e^{i π} (z_{6}^{4} = 1 n o n p o l e o f φ)$ $z_{7} = e^{i \frac{7 π}{6}} = - e^{i \frac{π}{6}} (p o l e o f φ)$ $z_{8} = e^{i \frac{8 π}{6}} = e^{i \frac{4 π}{3}} (p o l e o f φ)$ $z_{9} = e^{i \frac{9 π}{6}} = e^{i \frac{3 π}{2}} (n o n p o l e o f φ)$ $z_{10} = e^{i \frac{10 π}{6}} = e^{i \frac{5 π}{3}} (p o l e o f φ)$ $z_{11} = e^{i \frac{11 π}{6}} = - e^{- i \frac{π}{6}} (p o l e o f φ) b e c o n t i n u e d . . .$
Commented by math khazana by abdo last updated on 04/Jul/18
$∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,z_1 ) +Res(ϕ,z_2 ) +Res(ϕ,z_4 ) +Res(ϕ,z_5 ) +Res(ϕ, z_7 ) +Res(ϕ^� ,z_8 ) +Res(ϕ,z_(10) ) +Res(ϕ,z_(11) ) } Res(ϕ,z_k ) = (1/(4z_k ^3 + 6z_k ^5 )) ⇒ Res(ϕ,z_1 ) = (1/(4(e^(i(π/6)) )^3 +6(e^((iπ)/6) )^5 )) = (1/(4i +6 e^((i5π)/6) )) Res(ϕ,z_2 ) = (1/(4(e^((iπ)/3) )^3 +6 (e^((iπ)/3) )^5 )) = (1/(−4 +6 e^((i5π)/3) )) Res(ϕ,z_4 )= (1/(4(e^((i2π)/3) )^3 +6(e^((i2π)/3) )^5 )) =(1/(4 +6 e^((i10π)/3) )) =....$
$\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, z_{1}) + R e s (φ, z_{2}) + R e s (φ, z_{4})$ $+ R e s (φ, z_{5}) + R e s (φ, z_{7}) + R e s (\bar{φ}, z_{8}) + R e s (φ, z_{10})$ $+ R e s (φ, z_{11})}$ $R e s (φ, z_{k}) = \frac{1}{4 z_{k}^{3} + 6 z_{k}^{5}} \Rightarrow$ $R e s (φ, z_{1}) = \frac{1}{4 {(e^{i \frac{π}{6}})}^{3} + 6 {(e^{\frac{i π}{6}})}^{5}}$ $= \frac{1}{4 i + 6 e^{\frac{i 5 π}{6}}}$ $R e s (φ, z_{2}) = \frac{1}{4 {(e^{\frac{i π}{3}})}^{3} + 6 {(e^{\frac{i π}{3}})}^{5}} = \frac{1}{- 4 + 6 e^{\frac{i 5 π}{3}}}$ $R e s (φ, z_{4}) = \frac{1}{4 {(e^{\frac{i 2 π}{3}})}^{3} + 6 {(e^{\frac{i 2 π}{3}})}^{5}} = \frac{1}{4 + 6 e^{\frac{i 10 π}{3}}}$ $= . . . .$
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