decompose-inside-C-x-F-x-n-x-m-1-with-m-n-2-then-find-0-x-n-x-m-1-dx-

Question Number 30580 by abdo imad last updated on 23/Feb/18

d e c o m p o s e i n s i d e C [x] F = \frac{x^{n}}{x^{m} + 1} w i t h m ⩾ n + 2

t h e n f i n d \int_{0}^{\infty} \frac{x^{n}}{x^{m} + 1} d x .

Commented by prof Abdo imad last updated on 24/Feb/18

z^{m} + 1 = 0 \Leftrightarrow z^{m} = e^{i (2 k + 1) π} s o t h e r o o t s o f t h i s

e q u a t i o n a r e z_{k} = e^{i (2 k + 1) \frac{π}{m}} / k \in [[0, m - 1]] \Rightarrow

F (x) = \sum_{k = 0}^{m - 1} \frac{λ_{k}}{x - z_{k}} w e h a v e λ_{k} = \frac{z_{k}^{n}}{m z_{k}^{m - 1}} \Rightarrow

λ_{k} = - \frac{1}{m} z_{k}^{n + 1} \Rightarrow F (x) = - \frac{1}{m} \sum_{k = 0}^{m - 1} \frac{z_{k}^{n + 1}}{x - z_{k}} .

2) t h e c h . x^{m} = t g i v e

\int_{0}^{\infty} \frac{x^{n}}{x^{m} + 1} d x = \int_{0}^{\infty} \frac{t^{\frac{n}{m}}}{1 + t} \frac{1}{m} t^{\frac{1}{m} - 1} d t

= \int_{0}^{\infty} \frac{1}{m} \frac{t^{\frac{n + 1}{m} - 1}}{1 + t} d t = \frac{1}{m} \frac{π}{s i n (π \frac{n + 1}{m})} w e h a v e

u s e d t h e r e s u l t \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)} i f 0 < a < 1 .