find I = ∫_0 ^(2π) ln(x−e^(iθ) )dθ and xfromR and x^2 ≠1.


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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

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