find-the-value-of-f-x-0-pi-ln-x-2-2x-cos-1-d-with-x-fromR-

Question Number 39633 by abdo mathsup 649 cc last updated on 09/Jul/18

f i n d t h e v a l u e o f

f (x) = \int_{0}^{π} l n (x^{2} - 2 x c o s θ + 1) d θ w i t h x f r o m R .

Commented by math khazana by abdo last updated on 09/Jul/18

w e h a v e f^{^{'}} (x) = \int_{0}^{π} \frac{2 x - 2 c o s θ}{x^{2} - 2 x c o s θ + 1} d θ

=_{θ = 2 t} \int_{0}^{2 π} \frac{2 x - 2 c o s (2 t)}{x^{2} - 2 x c o s (2 t) + 1} 2 d t

= 4 \int_{0}^{2 π} \frac{x - c o s (2 t)}{x^{2} - 2 x c o s (2 t) + 1} d t c h a n g e m e n t

e^{i t} = z g i v e

f^{^{'}} (x) = 4 \int_{∣ z ∣= 1} \frac{x - \frac{z^{2} + z^{- 2}}{2}}{x^{2} - 2 x \frac{z^{2} + z^{- 2}}{2} + 1} \frac{d z}{i z}

= 4 \int_{∣ z ∣= 1} \frac{2 x - z^{2} - z^{- 2}}{i z (x^{2} - {x z}^{2} - {x z}^{- 2} + 1)} d z

= 4 \int_{∣ z ∣= 1} \frac{2 {x z}^{2} - z^{4} - 1}{i z (x^{2} z^{2} - x z^{4} - x + z^{2})} d z

= - 4 i \int_{∣ z ∣= 1} \frac{- z^{4} + 2 {x z}^{2} - 1}{z (- {x z}^{4} + (1 + x^{2}) z^{2} - x)} d z

= - 4 i \int_{∣ z ∣= 1} \frac{z^{4} - 2 {x z}^{2} + 1}{z {{x z}^{4} - (1 + x^{2}) z^{2} + x}} d z

l e t φ (z) = \frac{z^{4} - 2 {x z}^{2} + 1}{z {{x z}^{4} - (1 + x^{2}) z^{2} + x}} . p o l e s o f φ

r o o t s o f {x z}^{4} - (1 + x^{2}) z^{2} + x

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

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

z^{2} = \frac{1 + x^{2} + ∣ 1 - x^{2} ∣}{2 x} a n d z^{2} = \frac{1 + x^{2} - ∣ 1 - x^{2} ∣}{2 x}

(w e s u p p o s e t h a t x \neq 0)

c a s e 1 ∣ x ∣< 1 \Rightarrow z^{2} = \frac{1 + x^{2} + 1 - x^{2}}{2 x} = \frac{1}{x}

o r z^{2} = \frac{1 + x^{2} - 1 + x^{2}}{2 x} = x s o i f 0 < x < 1 w e g e t

z = \bar{+} \frac{1}{\sqrt{x}} a n d z = \bar{+} \sqrt{x}, 0 a r e p o l e s o f φ

φ (z) = \frac{z^{4} - 2 {x z}^{2} + 1}{x z (z - \frac{1}{\sqrt{x}}) (z + \frac{1}{\sqrt{x}}) (z - \sqrt{x}) (z + \sqrt{x})}

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

+ R e s (φ, - \sqrt{x})}

R e s (φ, 0) = \frac{1}{x}

R e s (φ, \sqrt{x}) = \frac{x^{2} - 2 x^{2} + 1}{x \sqrt{x} (x - \frac{1}{x}) 2 \sqrt{x}} = \frac{1 - x^{2}}{2 x^{2} (\frac{x^{2} - 1}{x})}

= \frac{- 1}{2 x}

R e s (φ, - \sqrt{x}) = \frac{x^{2} - 2 x^{2} + 1}{- x \sqrt{x} (x - \frac{1}{x}) (- 2 \sqrt{x})}

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

\int_{∣ z ∣= 1} φ (z) d z = 2 i π {\frac{1}{x} - \frac{1}{2 x} - \frac{1}{2 x}} = 0 \Rightarrow

f^{^{'}} (x) = 0 \Rightarrow f (x) = c = f (0) = 0 a n d w e g e t t h e

s a m e r e s u l t i f - 1 < x < 0

c a s e 2 ∣ x ∣> 1 w e h a v e

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

= 2 π l n ∣ x ∣ + \int_{0}^{π} l n (X^{2} - 2 X c o s θ + 1) d θ

= 2 π l n ∣ x ∣ + 0 b e c a u s e ∣ X ∣ = \frac{1}{∣ x ∣} < 1 \Rightarrow

f (x) = 2 π l n ∣ x ∣ i f ∣ x ∣> 1 a n d f (x) = 0 i f ∣ x ∣< 1 .

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jul/18

x^{2} - 2 x c o s θ + 1

{(x - c o s θ)}^{2} + {s i n}^{2} θ

(x - c o s θ + i s i n θ) (x - c o s θ - i s i n θ)

(x - e^{- i θ}) (x - e^{i θ})

f (x) = \int_{0}^{Π} {l n (x - e^{- i θ}) + l n (x - e^{i θ})} d θ

\frac{d I}{d x} = \int_{0}^{Π} \frac{1}{x - e^{- i θ}} + \frac{1}{x - e^{i θ}} d θ

= \int_{0}^{Π} \frac{e^{i θ}}{{x e}^{i θ} - 1} d θ + \int_{0}^{Π} \frac{e^{- i θ}}{{x e}^{- i θ} - 1} d θ

= \frac{1}{x} \int_{0}^{Π} \frac{e^{i θ}}{e^{i θ} - \frac{1}{x}} d θ + \frac{1}{x} \int_{0}^{Π} \frac{e^{- i θ}}{e^{- i θ} - \frac{1}{x}} d θ

=∣ \frac{1}{i x} l n (e^{i θ} - \frac{1}{x}) + \frac{1}{- i x} l n (e^{- i θ} - \frac{1}{x}) ∣_{0}^{Π}

= \frac{1}{i x} ∣ l n (\frac{e^{i θ} - \frac{1}{x}}{e^{- i θ} - \frac{1}{x}}) ∣_{0}^{Π} e^{i Π} = c o s Π + i s i n Π = - 1

\frac{1}{i x} {l n (\frac{- 1 - \frac{1}{x}}{- 1 - \frac{1}{x}}) - l n (\frac{1 - \frac{1}{x}}{1 - \frac{1}{x}})} = 0

\frac{d I}{d x} = 0

I = c o n s t a n t

p l s c h e c k m y s t e p s \dots w h e t h e r a n y m i s t a k e \dots

Commented by maxmathsup by imad last updated on 09/Jul/18

s i r T a n m a y t h e f u n c t i o n l n a t C i s n o t l i k e l n a t R l o o k l n (- 1) i n R

d o n t e x i s t b u t i n C l n (- 1) = l n (e^{i π}) = i π s o y o u h a v e c o m m i t e d a e r r o r

i n t h e f i n a l l i n e s y o u m u s t e x t r a c t R e (\int) a n d I m (\int) t o h a v e a c o r r e c t

a n s w e r \dots