1) decompose inside C(x) the fraction F(x)= (3/(4+x^4 )) 2) find ∫_(−∞) ^(+∞) (dx/(x−z)) with z from C 3) find the value of ∫


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Question Number 39967 by math khazana by abdo last updated on 14/Jul/18
$1) decompose inside C(x) the fraction F(x)= (3/(4+x^4 )) 2) find ∫_(−∞) ^(+∞) (dx/(x−z)) with z from C 3) find the value of ∫_(−∞) ^(+∞) ((3dx)/(4+x^4 )) .$
$1) d e c o m p o s e i n s i d e C (x) t h e f r a c t i o n$ $F (x) = \frac{3}{4 + x^{4}}$ $2) f i n d \int_{- \infty}^{+ \infty} \frac{d x}{x - z} w i t h z f r o m C$ $3) f i n d t h e v a l u e o f \int_{- \infty}^{+ \infty} \frac{3 d x}{4 + x^{4}} .$
Commented by math khazana by abdo last updated on 15/Jul/18

$3) f o r Q 3) i h a v e u s e d t h e f o r m u l a e$ $\int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)} i f 0 < a < 1 .$
Commented by abdo mathsup 649 cc last updated on 15/Jul/18

$1) w e h a v e F (x) = \frac{3}{{(\sqrt{2})}^{4} + x^{4}} = \frac{3}{({(2)}^{2} + {(x^{2})}^{2}}$ $= \frac{3}{{(x^{2} + 2)}^{2} - 4 x^{2}} = \frac{3}{(x^{2} - 2 x + 2) (x^{2} + 2 x + 2)}$ $r o o t s o f x^{2} - 2 x + 2$ $Δ^{^{'}} = 1 - 2 = - 1 = {(i)}^{2} \Rightarrow x_{0} = 1 + i = \frac{1}{\sqrt{2}} e^{\frac{i π}{4}}$ $x_{1} = 1 - i = \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}$ $r o o t s o f x^{2} + 2 x + 2$ $Δ^{^{'}} = 1 - 2 = - 1 \Rightarrow x_{2} = - 1 + i = \frac{1}{\sqrt{2}} e^{\frac{i 3 π}{4}}$ $x_{3} = - 1 - i = \frac{1}{\sqrt{2}} e^{- \frac{i 3 π}{4}} \Rightarrow$ $F (x) = \frac{3}{(x - x_{0}) (x - x_{1}) (x - x_{2}) (x - x_{3})}$ $= \frac{3}{(x - \frac{1}{\sqrt{2}} e^{\frac{i π}{4}}) (x - \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}) (x - \frac{1}{\sqrt{2}} e^{\frac{i 3 π}{4}}) (x - \frac{1}{\sqrt{2}} e^{- \frac{i 3 π}{4}})}$ $= \frac{a}{x - \frac{1}{\sqrt{2}} e^{\frac{i π}{4}}} + \frac{b}{x - \frac{1}{\sqrt{2}} e^{- \frac{i π}{4}}} + \frac{c}{x - \frac{1}{\sqrt{2}} e^{\frac{i 3 π}{4}}} + \frac{d}{x - \frac{1}{\sqrt{2}} e^{- \frac{i 3 π}{4}}}$ $w e h a v e λ_{i} = \frac{3}{4 x_{i}^{3}} \Rightarrow$ $a = \frac{3}{4 x_{0}^{3}} = \frac{3}{4 . \frac{1}{2 \sqrt{2}} e^{\frac{i 3 π}{4}}} = 3 \frac{\sqrt{2}}{2} e^{- \frac{i 3 π}{4}}$ $b = \frac{3}{4 x_{1}^{3}} = \frac{3}{4 . \frac{1}{2 \sqrt{2}} e^{- \frac{i 3 π}{4}}} = \frac{3 \sqrt{2}}{2} e^{\frac{i 3 π}{4}}$ $c = \frac{3}{4 x_{2}^{3}} = \frac{3}{4 . \frac{1}{2 \sqrt{2}} e^{i \frac{9 π}{4}}} = \frac{3 \sqrt{2}}{2} e^{- \frac{i 9 π}{4}} = \frac{3 \sqrt{2}}{2} e^{- \frac{i π}{4}}$ $d = \frac{3}{4 x_{3}^{3}} = \frac{3}{4 . \frac{1}{2 \sqrt{2}} e^{- \frac{i 9 π}{4}}} = \frac{3 \sqrt{2}}{2} e^{\frac{i 9 π}{4}} = \frac{3 \sqrt{2}}{2} e^{\frac{i π}{4}}$
Commented by abdo mathsup 649 cc last updated on 15/Jul/18
$2) let put A(ξ) = ∫_(−ξ) ^(+ξ) (dx/(x−z)) we have lim_(ξ→+∞) A(ξ)= ∫_(−∞) ^(+∞) (dx/(x−z)) let z = α +iβ A(ξ) = ∫_(−ξ) ^ξ (dx/(x−α −iβ)) = ∫_(−ξ) ^(+ξ) ((x−α +iβ)/((x−α)^(2 ) +β^2 )) dx = ∫_(−ξ) ^ξ ((x−α)/((x−α)^2 +β^2 ))dx +iβ ∫_(−ξ) ^ξ (dx/((x−α)^2 +β^2 )) but ∫_(−ξ) ^ξ ((x−α)/((x−α)^2 +β^2 )) dx = (1/2)[ln∣(x−α)^2 +β^2 ∣]_(−ξ) ^(+ξ) = (1/2)ln((((ξ−α)^2 +β^2 )/((ξ+α)^2 +β^2 )))→0 when ξ→+∞ changement x−α = βt give ∫_(−ξ) ^ξ (dx/((x−α)^2 +β^2 )) = ∫_((−ξ−α)/β) ^((ξ−α)/β) (1/(β^2 (1+t^2 ))) β dt = (1/β) [ arctant]_((−ξ−α)/β) ^((ξ−α)/β) =(1/β) { arctan(((ξ−α)/β))+arctan(((ξ+α)/β))} ⇒iβ ∫_(−ξ) ^ξ (dx/((x−α)^2 +β^2 )) =i{ arctan(((ξ−α)/β)) +arctan(((ξ+α)/β))} so if β>0 arctan(((ξ −α)/β)) +arctan(((ξ +α)/β))_(ξ→+∞) →π if β<0 arctan(((ξ−α)/β)) +arctan(((ξ +α)/β))→−π so A(ξ) →iπ if β>0 and A(ξ)→−iπ if β<0 finally ∫_(−∞) ^(+∞) (dx/(x−z)) =iπ if Im(z)>0 and ∫_(−∞) ^(+∞) (dx/(x−z)) =−iπ if Im(z)<0 .$
$2) l e t p u t A (ξ) = \int_{- ξ}^{+ ξ} \frac{d x}{x - z} w e h a v e$ ${l i m}_{ξ \to + \infty} A (ξ) = \int_{- \infty}^{+ \infty} \frac{d x}{x - z} l e t z = α + i β$ $A (ξ) = \int_{- ξ}^{ξ} \frac{d x}{x - α - i β}$ $= \int_{- ξ}^{+ ξ} \frac{x - α + i β}{{(x - α)}^{2} + β^{2}} d x$ $= \int_{- ξ}^{ξ} \frac{x - α}{{(x - α)}^{2} + β^{2}} d x + i β \int_{- ξ}^{ξ} \frac{d x}{{(x - α)}^{2} + β^{2}} b u t$ $\int_{- ξ}^{ξ} \frac{x - α}{{(x - α)}^{2} + β^{2}} d x = \frac{1}{2} {[l n ∣ {(x - α)}^{2} + β^{2} ∣]}_{- ξ}^{+ ξ}$ $= \frac{1}{2} l n (\frac{{(ξ - α)}^{2} + β^{2}}{{(ξ + α)}^{2} + β^{2}}) \to 0 w h e n ξ \to + \infty$ $c h a n g e m e n t x - α = β t g i v e$ $\int_{- ξ}^{ξ} \frac{d x}{{(x - α)}^{2} + β^{2}} = \int_{\frac{- ξ - α}{β}}^{\frac{ξ - α}{β}} \frac{1}{β^{2} (1 + t^{2})} β d t$ $= \frac{1}{β} {[a r c t a n t]}_{\frac{- ξ - α}{β}}^{\frac{ξ - α}{β}} = \frac{1}{β} {a r c t a n (\frac{ξ - α}{β}) + a r c t a n (\frac{ξ + α}{β})}$ $\Rightarrow i β \int_{- ξ}^{ξ} \frac{d x}{{(x - α)}^{2} + β^{2}} = i {a r c t a n (\frac{ξ - α}{β}) + a r c t a n (\frac{ξ + α}{β})}$ $s o i f β > 0 a r c t a n (\frac{ξ - α}{β}) + a r c t a n {(\frac{ξ + α}{β})}_{ξ \to + \infty} \to π$ $i f β < 0 a r c t a n (\frac{ξ - α}{β}) + a r c t a n (\frac{ξ + α}{β}) \to - π$ $s o A (ξ) \to i π i f β > 0 a n d A (ξ) \to - i π i f β < 0$ $f i n a l l y \int_{- \infty}^{+ \infty} \frac{d x}{x - z} = i π i f I m (z) > 0 a n d$ $\int_{- \infty}^{+ \infty} \frac{d x}{x - z} = - i π i f I m (z) < 0 .$
Commented by math khazana by abdo last updated on 15/Jul/18

$3) l e t I = \int_{- \infty}^{+ \infty} \frac{3 d x}{4 + x^{4}}$ $I = \frac{6}{4} \int_{0}^{\infty} \frac{d x}{1 + {(\frac{x}{\sqrt{2}})}^{4}} = \frac{3}{2} \int_{0}^{\infty} \frac{d x}{1 + {(\frac{x}{\sqrt{2}})}^{4}}$ $c h a n g e m e n t {(\frac{x}{\sqrt{2}})}^{4} = t g i v e x = \sqrt{2} t^{\frac{1}{4}}$ $\int_{0}^{\infty} \frac{d x}{1 + {(\frac{x}{\sqrt{2}})}^{4}} = \frac{1}{4} \int_{0}^{\infty} \frac{\sqrt{2}}{1 + t} t^{\frac{1}{4} - 1} d t$ $= \frac{\sqrt{2}}{4} \int_{0}^{\infty} \frac{t^{\frac{1}{4} - 1}}{1 + t} d t = \frac{\sqrt{2}}{4} \frac{π}{s i n (\frac{π}{4})} = \frac{π \sqrt{2}}{4 . \frac{\sqrt{2}}{2}}$ $= \frac{π}{2} \Rightarrow I = \frac{3}{2} \frac{π}{2} = \frac{3 π}{4}$
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