find-I-0-2pi-ln-x-e-i-d-and-xfromR-and-x-2-1-

Question Number 28889 by abdo imad last updated on 31/Jan/18

f i n d I = \int_{0}^{2 π} l n (x - e^{i θ}) d θ a n d x f r o m R a n d x^{2} \neq 1 .

Commented by abdo imad last updated on 02/Feb/18

l e t p u t f (x) = \int_{0}^{2 π} l n (x - e^{i θ}) d θ w e h a v e

f^{^{'}} (x) = \int_{0}^{2 π} \frac{d θ}{x - e^{i θ}} = \int_{0}^{2 π} \frac{d θ}{x - c o s θ - i s i n θ}

= \int_{0}^{2 π} \frac{x - c o s θ + i s i n θ}{{(x - c o s θ)}^{2} + {s i n}^{2} θ} d θ

= \int_{0}^{2 π} \frac{x - c o s θ}{x^{2} - 2 x c o s θ + 1} d θ + i \int_{0}^{2 π} \frac{s i n θ}{x^{2} - 2 x c o s θ + 1} d θ

= A (x) + i B (x)

A (x) = \frac{1}{2} \int_{0}^{2 π} \frac{2 x - 2 c o s θ}{x^{2} - 2 x c o s θ + 1} d θ = \frac{1}{2} {[l n ∣ x^{2} - 2 x c o s θ + 1 ∣]}_{θ = 0}^{2 π}

= 0 l e t f i n d B (x) t h e c h . e^{i θ} = z g i v e

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

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

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

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

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

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

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

{x z}^{2} - (1 + x^{2}) z + x = 0 \Rightarrow Δ = {(1 + x^{2})}^{2} - 4 x^{2}

= 1 + 2 x^{2} + x^{4} - 4 x^{2} = {(1 - x^{2})}^{2} \Rightarrow z_{1} = \frac{1 + x^{2} + ∣ 1 - x^{2} ∣}{2 x} a n d

z_{2} = \frac{1 + x^{2} - ∣ 1 - x^{2} ∣}{2 x} (x \neq 0)

c a s e 1 i f ∣ x ∣< 1 \Rightarrow z_{1} = \frac{1}{x} a n d z_{2} = x a n d w e h a v e ∣ z_{1} ∣> 1 a n d ∣ z_{2} ∣< 1

ψ (z) = \frac{z^{2} - 1}{2 x z (z - z_{1}) (z - z_{2})} a n d b y r e s i d u s t h e o r e m

\int_{∣ z ∣= 1} ψ (z) d z = 2 i π (R e s (ψ, 0) + R e s (ψ, z_{2}))

R e s (ψ, 0) = {l i m}_{z \to 0} z ψ (z) = \frac{- 1}{2 x z_{1} . z_{2}} = \frac{- 1}{2 x}

R e s (ψ, z_{2}) = {l i m}_{z \to z_{2}} (z - z_{2}) ψ (z) = \frac{z_{2}^{2} - 1}{2 {x z}_{2} (z_{2} - z_{1})}

= \frac{x^{2} - 1}{2 x^{2} (x - \frac{1}{x})} = \frac{x^{2} - 1}{2 x (x^{2} - 1)} = \frac{1}{2 x}

\int_{∣ z ∣= 1} ψ (z) d z = 2 i π (0) = 0

c a s e 2 i f ∣ x ∣> 1 \Rightarrow z_{1} = x a n d z_{2} = \frac{1}{x} a n d ∣ z_{1} ∣> 1 a n d ∣ z_{2} ∣< 1

\int_{∣ z ∣= 1} ψ (z) d z = 2 i π (R e s (ψ, 0) + R e s (ψ, z_{2}))

R e s (ψ, 0) = \frac{- 1}{2 x} a n d R e s (ψ, z_{2}) = \frac{z_{2}^{2} - 1}{2 {x z}_{2} (z_{2} - z_{1})}

= \frac{\frac{1}{x^{2}} - 1}{2 (\frac{1}{x} - x)} = \frac{1 - x^{2}}{2 x^{2} \frac{1 - x^{2}}{x}} = \frac{1}{2 x} a n d \int_{∣ z ∣= 1} ψ (z) d z = 0 s o

A (x) = B (x) = 0 \Rightarrow f^{^{'}} (x) = 0 \Rightarrow f (x) = λ \forall x / x^{2} \neq 1

Commented by abdo imad last updated on 02/Feb/18

λ = f (0) = \int_{0}^{2 π} l n (- e^{i θ}) d θ = \int_{0}^{2 π} l n (e^{i (π + θ)}) d θ

= \int_{0}^{2 π} i (π + θ) d θ = 2 i π^{2} + i {[\frac{θ^{2}}{2}]}_{0}^{2 π} = 2 i π^{2} + 2 i π^{2} = 4 i π^{2} .

Commented by abdo imad last updated on 02/Feb/18

i f z = r e^{i θ} l n (z) = l n (r) + i θ - π < θ < π . a n d r > 0