give the decomposition of F(x) = (1/(x^(2n) +1)) inside C[x] then find the value of ∫


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Question Number 26757 by abdo imad last updated on 28/Dec/17

$g i v e t h e d e c o m p o s i t i o n o f F (x) = \frac{1}{x^{2 n} + 1} i n s i d e C [x]$ $t h e n f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{d x}{1 + x^{2 n}} n \in N a n d n \neq o$
Commented by abdo imad last updated on 03/Jan/18

$l e t f i n d t h e p o l e s o f F$ $z^{2 n} + 1 = 0 \Leftrightarrow z^{2 n} = e^{i (2 k + 1) π} s o t h e p o l e s o f F a r e$ $z_{k} = e^{i \frac{(2 k + 1) π}{2 n}} k \in [[0, 2 n - 1]]$ $F (x) = \sum_{k = 0}^{2 n - 1} \frac{λ_{k}}{x - z_{k}} a n d λ_{k} = \frac{1}{2 n z_{k}^{2 n - 1}} = - \frac{1}{2 n} z_{k}$ $F (x) = - \frac{1}{2 n} \sum_{k = 0}^{2 n - 1} \frac{z_{k}}{x - z_{k}} b u t z_{0} = e^{i \frac{π}{2 n}} z_{0}^{-} = e^{- i \frac{π}{2 n}} = e^{i \frac{(2 π - \frac{π}{2 n})}{}}$ $= z_{2 n - 1}, z_{1}^{-} = z_{2 n - 2} . . . . .$ $\Rightarrow F (x) = \sum_{k = 0}^{n - 1} (\frac{z_{k}}{x - z_{k}} + \frac{z_{k}^{-}}{x - z_{k}^{-}}) .$
Commented by abdo imad last updated on 03/Jan/18

$F (x) = - \frac{1}{2 n} \sum_{k = 0}^{n - 1} (\frac{z_{k}}{x - z_{k}} + \frac{z_{k}^{-}}{x - z_{k}^{-}})$
Commented by abdo imad last updated on 23/Jan/18

$\int_{0}^{\infty} \frac{d x}{1 + x^{2 n}} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d x}{1 + x^{2 n}}$ $= \frac{- 1}{4 n} (\sum_{k = 0}^{n - 1} z_{k} \int_{R} \frac{d x}{x - z_{k}} + \sum_{k = 0}^{n - 1} z_{k}^{-} \int_{R} \frac{d x}{x - z_{k}^{-}})$ $= \frac{- 1}{4 n} (i π \sum_{k = 0}^{n - 1} z_{k} - i π \sum_{k =}^{n - 1} z_{k}^{-})$ $= \frac{- 1}{4 n} (- 2 π) \sum_{k = 0}^{n - 1} s i n (\frac{(2 k + 1) π}{2 n})$ $= \frac{π}{2 n} \sum_{k = 0}^{n - 1} s i n (\frac{(2 k + 1) π}{2 n}) l e t f i n d$ $A = \sum_{k = 0}^{n - 1} s i n (\frac{2 k + 1) π}{2 n}) = I m (\sum_{k = 0}^{n - 1} e^{i (\frac{(2 k + 1) π}{2 n})}) b u t$ $= e^{i \frac{π}{2 n}} \sum_{k = 0}^{n - 1} {(e^{i \frac{π}{n}})}^{k} = e^{i \frac{π}{2 n}} \frac{2}{1 - e^{i \frac{π}{n}}}$ $= \frac{2 e^{i \frac{π}{2 n}}}{1 - c o s (\frac{π}{n}) - i s i n (\frac{π}{n})} = \frac{e^{i \frac{π}{2 n}}}{{s i n}^{2} (\frac{π}{2 n}) - 2 i s i n (\frac{π}{2 n}) c o s (\frac{π}{2 n})}$ $= \frac{- 1}{i s i n (\frac{π}{2 n})} = \frac{i}{s i n (\frac{π}{2 n})} \Rightarrow A = \frac{1}{s i n (\frac{π}{2 n})} s o$ $\int_{0}^{\infty} \frac{d x}{1 + x^{2 n}} = \frac{π}{2 n s i n (\frac{π}{2 n})} = \frac{\frac{π}{2 n}}{s i n (\frac{π}{2 n})} .$
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