find thevalue of ∫_0 ^∞ (dx/(1 +x^6 )) by using the value of ∫_(−∞) ^(+∞) (dx/(x−z)) with z ∈ C and Im(z)≠0


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Question Number 40007 by math khazana by abdo last updated on 15/Jul/18

$f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{d x}{1 + x^{6}} b y u s i n g t h e$ $v a l u e o f \int_{- \infty}^{+ \infty} \frac{d x}{x - z} w i t h z \in C a n d I m (z) \neq 0$
Commented by prof Abdo imad last updated on 16/Jul/18
$let I = ∫_0 ^∞ (dx/(1+x^6 )) 2I = ∫_(−∞) ^(+∞) (dx/(1+x^6 )) let drcompose inside C(x) F(x)= (1/(x^6 +1)) poles of F? x^6 =−1 ⇒ x_k = e^(i((2k+1)π)/6)) with k∈[[0,5]] x_0 = e^((iπ)/6) , x_1 = e^(i(π/2)) , x_2 = e^((i5π)/6) = −e^(−((iπ)/6)) x_3 = e^(i((7π)/6)) =−e^((iπ)/6) , x_4 =e^((i9π)/6) =e^((i3π)/2) , x_5 = e^((i11π)/6) =e^(−((iπ)/6)) we have λ_i = (1/(6x_i ^5 )) =−(1/6) x_i ⇒ F(x)=−(1/6){ (x_0 /(x−x_0 )) +(x_1 /(x−x_1 )) +(x_2 /(x−x_2 )) +(x_3 /(x−x_3 )) + (x_4 /(x−x_4 )) +(x_5 /(x−x_5 ))} ⇒ ∫_(−∞) ^(+∞) F(x)dx =−(1/6){ x_0 ∫_(−∞) ^(+∞) (dx/(x−x_0 )) +x_1 ∫_(−∞) ^(+∞) (dx/(x−x_1 )) +x_2 ∫_(−∞) ^(+∞) (dx/(x−x_2 )) + x_3 ∫_(−∞) ^(+∞) (dx/(x−x_3 )) + x_4 ∫_(−∞) ^(+∞) (dx/(x−x_4 )) +x_5 ∫_(−∞) ^(+∞) (dx/(x−x_5 ))} but we have proved that ∫_(−∞) ^(+∞) (dx/(x−z)) =iπ if Im(z)>0 and −iπ if Im(z)<0 ∫_(−∞) ^(+∞) F(x)dx =−(1/6){x_0 (iπ) +x_1 (iπ) +x_2 (iπ) +x_3 (−iπ)+x_4 (−iπ) +x_5 (−iπ)} =−((iπ)/6){ x_0 +x_1 +x_2 −x_3 −x_4 −x_5 } =−((iπ)/6){ e^((iπ)/6) + i −e^(−((iπ)/6)) +e^((iπ)/6) −e^((i3π)/2) −e^(−((iπ)/6)) } =−((iπ)/6){ 2i +2(e^((iπ)/6) −e^(−((iπ)/6)) )} =(π/3) −((iπ)/3) 2i (1/2) =(π/3) +(π/3) =((2π)/3) 2I =((2π)/3) ⇒ I =(π/3) .$
$l e t I = \int_{0}^{\infty} \frac{d x}{1 + x^{6}}$ $2 I = \int_{- \infty}^{+ \infty} \frac{d x}{1 + x^{6}} l e t d r c o m p o s e i n s i d e C (x)$ $F (x) = \frac{1}{x^{6} + 1} p o l e s o f F ?$ $x^{6} = - 1 \Rightarrow x_{k} = e^{i \frac{2 k + 1) π}{6}} w i t h k \in [[0, 5]]$ $x_{0} = e^{\frac{i π}{6}}, x_{1} = e^{i \frac{π}{2}}, x_{2} = e^{\frac{i 5 π}{6}} = - e^{- \frac{i π}{6}}$ $x_{3} = e^{i \frac{7 π}{6}} = - e^{\frac{i π}{6}}, x_{4} = e^{\frac{i 9 π}{6}} = e^{\frac{i 3 π}{2}}, x_{5} = e^{\frac{i 11 π}{6}} = e^{- \frac{i π}{6}}$ $w e h a v e λ_{i} = \frac{1}{6 x_{i}^{5}} = - \frac{1}{6} x_{i} \Rightarrow$ $F (x) = - \frac{1}{6} {\frac{x_{0}}{x - x_{0}} + \frac{x_{1}}{x - x_{1}} + \frac{x_{2}}{x - x_{2}} + \frac{x_{3}}{x - x_{3}}$ $+ \frac{x_{4}}{x - x_{4}} + \frac{x_{5}}{x - x_{5}}} \Rightarrow$ $\int_{- \infty}^{+ \infty} F (x) d x = - \frac{1}{6} {x_{0} \int_{- \infty}^{+ \infty} \frac{d x}{x - x_{0}} + x_{1} \int_{- \infty}^{+ \infty} \frac{d x}{x - x_{1}}$ $+ x_{2} \int_{- \infty}^{+ \infty} \frac{d x}{x - x_{2}} + x_{3} \int_{- \infty}^{+ \infty} \frac{d x}{x - x_{3}} + x_{4} \int_{- \infty}^{+ \infty} \frac{d x}{x - x_{4}}$ $+ x_{5} \int_{- \infty}^{+ \infty} \frac{d x}{x - x_{5}}} b u t w e h a v e p r o v e d t h a t$ $\int_{- \infty}^{+ \infty} \frac{d x}{x - z} = i π i f I m (z) > 0 a n d - i π i f I m (z) < 0$ $\int_{- \infty}^{+ \infty} F (x) d x = - \frac{1}{6} {x_{0} (i π) + x_{1} (i π) + x_{2} (i π)$ $+ x_{3} (- i π) + x_{4} (- i π) + x_{5} (- i π)}$ $= - \frac{i π}{6} {x_{0} + x_{1} + x_{2} - x_{3} - x_{4} - x_{5}}$ $= - \frac{i π}{6} {e^{\frac{i π}{6}} + i - e^{- \frac{i π}{6}} + e^{\frac{i π}{6}} - e^{\frac{i 3 π}{2}} - e^{- \frac{i π}{6}}}$ $= - \frac{i π}{6} {2 i + 2 (e^{\frac{i π}{6}} - e^{- \frac{i π}{6}})}$ $= \frac{π}{3} - \frac{i π}{3} 2 i \frac{1}{2} = \frac{π}{3} + \frac{π}{3} = \frac{2 π}{3}$ $2 I = \frac{2 π}{3} \Rightarrow I = \frac{π}{3} .$
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