1-decompose-at-simple-elements-U-n-n-x-n-1-x-n-1-2-calculste-0-2pi-dt-x-e-it-

Question Number 50406 by Abdo msup. last updated on 16/Dec/18

1) d e c o m p o s e a t s i m p l e e l e m e n t s

U_{n} = \frac{n x^{n - 1}}{x^{n} - 1}

2) c a l c u l s t e \int_{0}^{2 π} \frac{d t}{x - e^{i t}}

Commented by Abdo msup. last updated on 25/Dec/18

l e t p (x) = x^{n} - 1 t h e r o o t s o f p (x) a r e z_{k} = e^{i \frac{2 k π}{n}}

w i t h k \in [[0, n - 1]] b u t U_{n} = \sum_{k = 0}^{n - 1} \frac{λ_{k}}{x - z_{k}}

λ_{k} = \frac{{n z}_{k}^{n - 1}}{p^{^{'}} (z_{k})} = \frac{n}{z_{k}} \frac{1}{n z_{k}^{n - 1}} = 1 \Rightarrow U_{n} = \sum_{k = 0}^{n - 1} \frac{1}{x - z_{k}}

= \sum_{k = 0}^{n - 1} \frac{1}{x - e^{i \frac{2 k π}{n}}}

Commented by Abdo msup. last updated on 25/Dec/18

2) c h a n g e m e n t e^{i t} = z g i v e

\int_{0}^{2 π} \frac{d t}{x - e^{i t}} = \int_{∣ z ∣= 1} \frac{1}{x - z} \frac{d z}{i z}

= \int_{∣ z ∣= 1} \frac{- i d z}{x z - z^{2}} = \int_{∣ z ∣= 1} \frac{i d z}{z^{2} - x z} l e t

φ (z) = \frac{i}{z^{2} - x z} = \frac{i}{z (z - x)} s o t h e p o l e s x o f φ a r e 0 a n d

x i f ∣ x ∣< 1 \int_{∣ z ∣= 1} φ (z) d z = 2 i π (R e s (φ, 0) + R e s (φ, x)}

R e s (φ, 0) = {l i m}_{z \to 0} z φ (z) = - \frac{i}{x}

R e s (φ, x) = {l i m}_{z \to x} (z - x) φ (z) = \frac{i}{x} \Rightarrow

\int_{∣ z ∣= 1} φ (z) d z = 0

i f ∣ x ∣> 1 \int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, 0)

= 2 i π (\frac{- i}{x}) = \frac{2 π}{x} (w e s u p p o s e x \neq 0)

i f x = 0 \int_{0}^{2 π} \frac{d t}{x - e^{i t}} = - \int_{0}^{2 π} e^{- i t} d t

= - {[- \frac{1}{i} e^{- i t}]}_{0}^{2 π} = \frac{1}{i} (1 - 1) = 0