calculate-0-2pi-dt-1-2pcost-p-2-if-p-lt-1-

Question Number 37349 by math khazana by abdo last updated on 12/Jun/18

c a l c u l a t e \int_{0}^{2 π} \frac{d t}{1 - 2 p c o s t + p^{2}} i f ∣ p ∣< 1

Commented by math khazana by abdo last updated on 13/Jun/18

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

I = \int_{∣ z ∣= 1} \frac{1}{1 - 2 p \frac{z + z^{- 1}}{2} + p^{2}} \frac{d z}{i z}

= \int_{∣ z ∣= 1} \frac{- i d z}{z (1 + p^{2} - p z - {p z}^{- 1})}

= \int_{∣ z ∣ = 1} \frac{- i d z}{(1 + p^{2}) z - {p z}^{2} - p}

= \int_{∣ z ∣= 1} \frac{i d z}{{p z}^{2} - (1 + p^{2}) z + p} l e t c o n s i d e r t h e

c o m p l e x f u n c t i o n φ (z) = \frac{i}{{p z}^{2} - (1 + p^{2}) z + p}

p o l e s o f φ ?

Δ = {(1 + p^{2})}^{2} - 4 p^{2} = 1 + 2 p^{2} + p^{4} - 4 p^{2}

= p^{4} + 1 - 2 p^{2} = {(p^{2} - 1)}^{2}

z_{1} = \frac{1 + p^{2} + ∣ p^{2} - 1 ∣}{2 p} = \frac{1 + p^{2} + 1 - p^{2}}{2 p} = \frac{1}{p}

z_{2} = \frac{1 + p^{2} - ∣ p^{2} - 1 ∣}{2 p} = \frac{1 + p^{2} - 1 + p^{2}}{2 p} = p

∣ z_{1} ∣= \frac{1}{∣ p ∣} > 1 b e c a u s e ∣ p ∣< 1 a n d ? w e s u p p o s e t h a t

p \neq 0

∣ z_{2} ∣ =∣ p ∣< 1 s o

\int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{2}) b u t w e h a v e

φ (z) = \frac{i}{p (z - z_{1}) (z - z_{2})} \Rightarrow

R e s (φ, z_{2}) = {l i m}_{z \to z_{2}} (z - z_{2}) φ (z)

= \frac{i}{p (z_{2} - z_{1})} = \frac{i}{p (p - \frac{1}{p})} = \frac{i}{(p^{2} - 1)} \Rightarrow

\int_{∣ z ∣= 1} φ (z) d z = 2 i π \frac{i}{p^{2} - 1} = \frac{- 2 π}{p^{2} - 1} = \frac{2 π}{1 - p^{2}}

I = \frac{2 π}{1 - p^{2}}

i f p = 0 I = \int_{0}^{2 π} d t = 2 π .