1) calculate ∫_(−∞) ^(+∞) (dx/(x−a)) with a ∈C 2) find the values of ∫_0 ^∞ (dx/(x^4 +1)) and ∫_0 ^∞ (dx/(x^6 +1)) by using the decomposition inside C(x).


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Question Number 65293 by mathmax by abdo last updated on 27/Jul/19

$1) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{x - a} w i t h a \in C$ $2) f i n d t h e v a l u e s o f \int_{0}^{\infty} \frac{d x}{x^{4} + 1} a n d \int_{0}^{\infty} \frac{d x}{x^{6} + 1}$ $b y u s i n g t h e d e c o m p o s i t i o n i n s i d e C (x) .$
Commented by mathmax by abdo last updated on 01/Aug/19
$1) let I(ξ) =∫_(−ξ) ^ξ (dx/(x−a)) let a =α+iβ we have ∫_(−∞) ^(+∞) (dx/(x−a)) =lim_(ξ→+∞) I(ξ) I(ξ) =∫_(−ξ) ^ξ (dx/(x−α−iβ)) =∫_(−ξ) ^ξ ((x−α+iβ)/((x−α)^2 +β^2 ))dx =(1/2)[ln{(x−α)^2 +β^2 }]_(−ξ) ^ξ +iβ ∫_(−ξ) ^ξ (dx/((x−α)^2 +β^2 )) =(1/2)ln((((ξ−α)^2 +β^2 )/((ξ+α)^2 )+β^2 ))) +iβ ∫_(−ξ) ^ξ (dx/((x−α)^2 +β^2 )) cha7gement x−α =∣β∣u give ∫_(−ξ) ^ξ (dx/((x−α)^2 +β^2 )) = ∫_((−ξ−α)/(∣β∣)) ^((ξ−α)/(∣β∣)) ((∣β∣du)/(β^2 (1+u^2 ))) =(1/(∣β∣)){ arctan(((ξ−α)/(∣β∣)))+arctan(((ξ+α)/(∣β∣)))} case 1 β>0 ⇒lim_(ξ→+∞) I(ξ) =iβ(1/β)((π/2)+(π/2))=iπ case 2 β<0 ⇒lim_(ξ→+∞) I(ξ) =−iβ(1/β)((π/2)+(π/2))=−iπ .$
$1) l e t I (ξ) = \int_{- ξ}^{ξ} \frac{d x}{x - a} l e t a = α + i β w e h a v e \int_{- \infty}^{+ \infty} \frac{d x}{x - a} = {l i m}_{ξ \to + \infty} I (ξ)$ $I (ξ) = \int_{- ξ}^{ξ} \frac{d x}{x - α - i β} = \int_{- ξ}^{ξ} \frac{x - α + i β}{{(x - α)}^{2} + β^{2}} d x$ $= \frac{1}{2} {[l n {{(x - α)}^{2} + β^{2}}]}_{- ξ}^{ξ} + i β \int_{- ξ}^{ξ} \frac{d x}{{(x - α)}^{2} + β^{2}}$ $= \frac{1}{2} l n (\frac{{(ξ - α)}^{2} + β^{2}}{{(ξ + α)}^{2}) + β^{2}}) + i β \int_{- ξ}^{ξ} \frac{d x}{{(x - α)}^{2} + β^{2}}$ $c h a 7 g e m e n t x - α =∣ β ∣ u g i v e \int_{- ξ}^{ξ} \frac{d x}{{(x - α)}^{2} + β^{2}}$ $= \int_{\frac{- ξ - α}{∣ β ∣}}^{\frac{ξ - α}{∣ β ∣}} \frac{∣ β ∣ d u}{β^{2} (1 + u^{2})} = \frac{1}{∣ β ∣} {a r c t a n (\frac{ξ - α}{∣ β ∣}) + a r c t a n (\frac{ξ + α}{∣ β ∣})}$ $c a s e 1 β > 0 \Rightarrow {l i m}_{ξ \to + \infty} I (ξ) = i β \frac{1}{β} (\frac{π}{2} + \frac{π}{2}) = i π$ $c a s e 2 β < 0 \Rightarrow {l i m}_{ξ \to + \infty} I (ξ) = - i β \frac{1}{β} (\frac{π}{2} + \frac{π}{2}) = - i π .$
Commented by mathmax by abdo last updated on 01/Aug/19

$f i n a l l y \int_{- \infty}^{+ \infty} \frac{d x}{x - a} = i π i f i m (a) > 0 a n d - i π i f i m (a) < 0$
Commented by mathmax by abdo last updated on 01/Aug/19
$2) let decompose F(x)=(1/(x^4 +1)) x^4 +1 =0 ⇒x^4 =−1 ⇒(re^(iθ) )^4 =e^(i(2k+1)π) ⇒r =1 and θ =(2k+1)(π/4) so the roots are z_k =e^(i(2k+1)(π/4)) k∈[[0,3]] z_0 =e^((iπ)/4) , z_1 =e^(i((3π)/4)) ,z_2 =e^(i(((5π)/4))) ,z_3 =e^(i(((7π)/4))) F(x) =Σ_(i=0) ^3 (λ_i /(x−z_i )) and λ_i =(1/(4z_i ^3 )) =−(1/4)z_i ⇒ ∫_0 ^(+∞) (dx/(x^4 +1)) =(1/2)∫_(−∞) ^(+∞) F(x)dx =(1/2)∫_(−∞) ^(+∞) Σ_(i=0) ^3 (−(1/4))(z_i /(x−z_i ))dx =−(1/8) Σ_(i=0) ^3 ∫_(−∞) ^(+∞) (z_i /(x−z_i ))dx =−(1/8){ z_0 ∫_(−∞) ^(+∞) (dx/(x−z_0 )) +z_1 ∫_(−∞) ^(+∞) (dx/(x−z_1 )) +z_2 ∫_(−∞) ^(+∞) (dx/(x−z_2 )) +z_3 ∫_(−∞) ^(+∞) (dx/(x−z_3 ))} =−(1/8){ iπ(z_0 )+iπ z_1 −iπz_2 −iπz_3 } =−((iπ)/8){ z_0 +z_1 −(z_2 +z_3 )} but z_0 +z_1 =e^((iπ)/4) +e^(i(π−(π/4))) =e^((iπ)/4) −e^(−((iπ)/4)) =2i sin((π/4)) =2i((√2)/2) =i(√2) z_2 +z_3 =e^(i((5π)/4)) +e^((i7π)/4) =e^(i(π+(π/4))) +e^(i(2π−(π/4))) =−e^((iπ)/4) +e^(−((iπ)/4)) =−(e^((iπ)/4) −e^(−((iπ)/4)) ) =−2isin((π/4)) =−2i((√2)/2) =−i(√2) ⇒ ∫_0 ^∞ (dx/(1+x^4 )) =−((iπ)/8){i(√2)+i(√2)} =(π/8)(2(√2)) =((π(√2))/4)$
$2) l e t d e c o m p o s e F (x) = \frac{1}{x^{4} + 1}$ $x^{4} + 1 = 0 \Rightarrow x^{4} = - 1 \Rightarrow {({r e}^{i θ})}^{4} = e^{i (2 k + 1) π} \Rightarrow r = 1 a n d θ = (2 k + 1) \frac{π}{4}$ $s o t h e r o o t s a r e z_{k} = e^{i (2 k + 1) \frac{π}{4}} k \in [[0, 3]]$ $z_{0} = e^{\frac{i π}{4}}, z_{1} = e^{i \frac{3 π}{4}}, z_{2} = e^{i (\frac{5 π}{4})}, z_{3} = e^{i (\frac{7 π}{4})}$ $F (x) = \sum_{i = 0}^{3} \frac{λ_{i}}{x - z_{i}} a n d λ_{i} = \frac{1}{4 z_{i}^{3}} = - \frac{1}{4} z_{i} \Rightarrow$ $\int_{0}^{+ \infty} \frac{d x}{x^{4} + 1} = \frac{1}{2} \int_{- \infty}^{+ \infty} F (x) d x = \frac{1}{2} \int_{- \infty}^{+ \infty} \sum_{i = 0}^{3} (- \frac{1}{4}) \frac{z_{i}}{x - z_{i}} d x$ $= - \frac{1}{8} \sum_{i = 0}^{3} \int_{- \infty}^{+ \infty} \frac{z_{i}}{x - z_{i}} d x$ $= - \frac{1}{8} {z_{0} \int_{- \infty}^{+ \infty} \frac{d x}{x - z_{0}} + z_{1} \int_{- \infty}^{+ \infty} \frac{d x}{x - z_{1}} + z_{2} \int_{- \infty}^{+ \infty} \frac{d x}{x - z_{2}} + z_{3} \int_{- \infty}^{+ \infty} \frac{d x}{x - z_{3}}}$ $= - \frac{1}{8} {i π (z_{0}) + i π z_{1} - i π z_{2} - i π z_{3}}$ $= - \frac{i π}{8} {z_{0} + z_{1} - (z_{2} + z_{3})} b u t z_{0} + z_{1} = e^{\frac{i π}{4}} + e^{i (π - \frac{π}{4})}$ $= e^{\frac{i π}{4}} - e^{- \frac{i π}{4}} = 2 i s i n (\frac{π}{4}) = 2 i \frac{\sqrt{2}}{2} = i \sqrt{2}$ $z_{2} + z_{3} = e^{i \frac{5 π}{4}} + e^{\frac{i 7 π}{4}} = e^{i (π + \frac{π}{4})} + e^{i (2 π - \frac{π}{4})} = - e^{\frac{i π}{4}} + e^{- \frac{i π}{4}}$ $= - (e^{\frac{i π}{4}} - e^{- \frac{i π}{4}}) = - 2 i s i n (\frac{π}{4}) = - 2 i \frac{\sqrt{2}}{2} = - i \sqrt{2} \Rightarrow$ $\int_{0}^{\infty} \frac{d x}{1 + x^{4}} = - \frac{i π}{8} {i \sqrt{2} + i \sqrt{2}} = \frac{π}{8} (2 \sqrt{2}) = \frac{π \sqrt{2}}{4}$
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