find-thevalue-of-0-dx-1-x-6-by-using-the-value-of-dx-x-z-with-z-C-and-Im-z-0-

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

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} .