find the value of f(x) = ∫_0 ^π ln(x^2 −2x cosθ +1)dθ with x fromR.


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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
$we have f^′ (x)= ∫_0 ^π ((2x−2cosθ)/(x^(2 ) −2xcosθ +1))dθ =_(θ=2t) ∫_0 ^(2π) ((2x−2cos(2t))/(x^2 −2xcos(2t) +1)) 2dt = 4 ∫_0 ^(2π) ((x−cos(2t))/(x^2 −2xcos(2t)+1))dt changement e^(it) =z give f^′ (x)=4 ∫_(∣z∣=1) ((x−((z^2 +z^(−2) )/2))/(x^2 −2x ((z^2 +z^(−2) )/2) +1)) (dz/(iz)) =4 ∫_(∣z∣=1) ((2x−z^2 −z^(−2) )/(iz(x^2 −xz^2 −xz^(−2) +1)))dz =4 ∫_(∣z∣=1) ((2xz^2 −z^4 −1)/(iz(x^2 z^2 −x z^4 −x +z^2 )))dz =−4i ∫_(∣z∣=1) ((−z^4 +2xz^2 −1)/(z(−xz^4 +(1+x^2 )z^2 −x)))dz =−4i ∫_(∣z∣=1) ((z^4 −2xz^2 +1)/(z{xz^4 −(1+x^2 )z^2 +x}))dz let ϕ(z) = ((z^4 −2xz^2 +1)/(z{ xz^4 −(1+x^2 )z^2 +x})) .poles of ϕ roots of xz^4 −(1+x^2 )z^2 +x Δ =(1+x^2 )^2 −4x^2 =1+2x^2 +x^4 −4x^2 =1−2x^2 +x^4 =(1−x^2 )^2 ⇒ z^2 =((1+x^2 +∣1−x^2 ∣)/(2x)) and z^2 =((1+x^2 −∣1−x^2 ∣)/(2x)) (we suppose that x≠0) case 1 ∣x∣<1 ⇒z^2 =((1+x^2 +1−x^2 )/(2x)) = (1/x) or z^2 =((1+x^2 −1+x^2 )/(2x)) =x so if 0<x<1 we get z =+^− (1/(√x)) and z =+^− (√x) ,0 are poles of ϕ ϕ(z) = ((z^4 −2xz^2 +1)/(xz( z−(1/(√x)))(z+(1/(√x)))(z−(√x))(z+(√x)))) ∫_(∣z∣=1) ϕ(z)dz =2iπ { Res(ϕ,0) +Res(ϕ,(√x)) +Res(ϕ,−(√x))} Res(ϕ,0) = (1/x) Res(ϕ,(√x)) = ((x^2 −2x^2 +1)/(x(√x)(x−(1/x))2(√x))) =((1−x^2 )/(2x^2 (((x^2 −1)/x)))) = ((−1)/(2x)) Res(ϕ,−(√x)) =((x^2 −2x^2 +1)/(−x(√x)(x−(1/x))(−2(√x)))) =((1−x^2 )/(2x^2 (((x^2 −1)/x)))) =((−1)/(2x)) ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ{ (1/x) −(1/(2x)) −(1/(2x))}=0 ⇒ f^′ (x) =0 ⇒ f(x)=c = f(0)=0 and we get the same result if −1<x<0 case 2 ∣x∣>1 we have f(x) = ∫_0 ^π ln(x^2 (1 −(2/x) cosθ + (1/x^2 )))dθ =2π ln∣x∣ + ∫_0 ^π ln( X^2 −2X cosθ +1)dθ =2π ln∣x∣ +0 because ∣X∣ = (1/(∣x∣))<1 ⇒ f(x) =2πln∣x∣ if ∣x∣>1 and f(x)=0 if∣x∣<1 .$
$w e h a v e f^{^{'}} (x) = \int_{0}^{π} \frac{2 x - 2 c o s θ}{x^{2} - 2 x c o s θ + 1} d θ$ $=_{θ = 2 t} \int_{0}^{2 π} \frac{2 x - 2 c o s (2 t)}{x^{2} - 2 x c o s (2 t) + 1} 2 d t$ $= 4 \int_{0}^{2 π} \frac{x - c o s (2 t)}{x^{2} - 2 x c o s (2 t) + 1} d t c h a n g e m e n t$ $e^{i t} = z g i v e$ $f^{^{'}} (x) = 4 \int_{∣ z ∣= 1} \frac{x - \frac{z^{2} + z^{- 2}}{2}}{x^{2} - 2 x \frac{z^{2} + z^{- 2}}{2} + 1} \frac{d z}{i z}$ $= 4 \int_{∣ z ∣= 1} \frac{2 x - z^{2} - z^{- 2}}{i z (x^{2} - {x z}^{2} - {x z}^{- 2} + 1)} d z$ $= 4 \int_{∣ z ∣= 1} \frac{2 {x z}^{2} - z^{4} - 1}{i z (x^{2} z^{2} - x z^{4} - x + z^{2})} d z$ $= - 4 i \int_{∣ z ∣= 1} \frac{- z^{4} + 2 {x z}^{2} - 1}{z (- {x z}^{4} + (1 + x^{2}) z^{2} - x)} d z$ $= - 4 i \int_{∣ z ∣= 1} \frac{z^{4} - 2 {x z}^{2} + 1}{z {{x z}^{4} - (1 + x^{2}) z^{2} + x}} d z$ $l e t φ (z) = \frac{z^{4} - 2 {x z}^{2} + 1}{z {{x z}^{4} - (1 + x^{2}) z^{2} + x}} . p o l e s o f φ$ $r o o t s o f {x z}^{4} - (1 + x^{2}) z^{2} + x$ $Δ = {(1 + x^{2})}^{2} - 4 x^{2} = 1 + 2 x^{2} + x^{4} - 4 x^{2}$ $= 1 - 2 x^{2} + x^{4} = {(1 - x^{2})}^{2} \Rightarrow$ $z^{2} = \frac{1 + x^{2} + ∣ 1 - x^{2} ∣}{2 x} a n d z^{2} = \frac{1 + x^{2} - ∣ 1 - x^{2} ∣}{2 x}$ $(w e s u p p o s e t h a t x \neq 0)$ $c a s e 1 ∣ x ∣< 1 \Rightarrow z^{2} = \frac{1 + x^{2} + 1 - x^{2}}{2 x} = \frac{1}{x}$ $o r z^{2} = \frac{1 + x^{2} - 1 + x^{2}}{2 x} = x s o i f 0 < x < 1 w e g e t$ $z = \bar{+} \frac{1}{\sqrt{x}} a n d z = \bar{+} \sqrt{x}, 0 a r e p o l e s o f φ$ $φ (z) = \frac{z^{4} - 2 {x z}^{2} + 1}{x z (z - \frac{1}{\sqrt{x}}) (z + \frac{1}{\sqrt{x}}) (z - \sqrt{x}) (z + \sqrt{x})}$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π {R e s (φ, 0) + R e s (φ, \sqrt{x})$ $+ R e s (φ, - \sqrt{x})}$ $R e s (φ, 0) = \frac{1}{x}$ $R e s (φ, \sqrt{x}) = \frac{x^{2} - 2 x^{2} + 1}{x \sqrt{x} (x - \frac{1}{x}) 2 \sqrt{x}} = \frac{1 - x^{2}}{2 x^{2} (\frac{x^{2} - 1}{x})}$ $= \frac{- 1}{2 x}$ $R e s (φ, - \sqrt{x}) = \frac{x^{2} - 2 x^{2} + 1}{- x \sqrt{x} (x - \frac{1}{x}) (- 2 \sqrt{x})}$ $= \frac{1 - x^{2}}{2 x^{2} (\frac{x^{2} - 1}{x})} = \frac{- 1}{2 x} \Rightarrow$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π {\frac{1}{x} - \frac{1}{2 x} - \frac{1}{2 x}} = 0 \Rightarrow$ $f^{^{'}} (x) = 0 \Rightarrow f (x) = c = f (0) = 0 a n d w e g e t t h e$ $s a m e r e s u l t i f - 1 < x < 0$ $c a s e 2 ∣ x ∣> 1 w e h a v e$ $f (x) = \int_{0}^{π} l n (x^{2} (1 - \frac{2}{x} c o s θ + \frac{1}{x^{2}})) d θ$ $= 2 π l n ∣ x ∣ + \int_{0}^{π} l n (X^{2} - 2 X c o s θ + 1) d θ$ $= 2 π l n ∣ x ∣ + 0 b e c a u s e ∣ X ∣ = \frac{1}{∣ x ∣} < 1 \Rightarrow$ $f (x) = 2 π l n ∣ x ∣ i f ∣ x ∣> 1 a n d f (x) = 0 i f ∣ x ∣< 1 .$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jul/18
$x^2 −2xcosθ+1 (x−cosθ)^2 +sin^2 θ (x−cosθ+isinθ)(x−cosθ−isinθ) (x−e^(−iθ) )(x−e^(iθ) ) f(x)=∫_0 ^Π {ln(x−e^(−iθ) )+ln(x−e^(iθ) )}dθ (dI/dx)=∫_0 ^Π (1/(x−e^(−iθ) ))+(1/(x−e^(iθ) )) dθ =∫_0 ^Π (e^(iθ) /(xe^(iθ) −1))dθ+∫_0 ^Π (e^(−iθ) /(xe^(−iθ) −1))dθ =(1/x)∫_0 ^Π (e^(iθ) /(e^(iθ) −(1/x)))dθ+(1/x)∫_0 ^Π (e^(−iθ) /(e^(−iθ) −(1/x)))dθ =∣(1/(ix))ln(e^(iθ) −(1/x))+(1/(−ix))ln(e^(−iθ) −(1/x))∣_0 ^Π =(1/(ix))∣ln(((e^(iθ) −(1/x))/(e^(−iθ) −(1/x))))∣_0 ^Π e^(iΠ) =cosΠ+isinΠ=−1 (1/(ix)){ln(((−1−(1/x))/(−1−(1/x))))−ln(((1−(1/x))/(1−(1/x))))}=0 (dI/dx)=0 I=constant pls check my steps...whether any mistake...$
$x^{2} - 2 x c o s θ + 1$ ${(x - c o s θ)}^{2} + {s i n}^{2} θ$ $(x - c o s θ + i s i n θ) (x - c o s θ - i s i n θ)$ $(x - e^{- i θ}) (x - e^{i θ})$ $f (x) = \int_{0}^{Π} {l n (x - e^{- i θ}) + l n (x - e^{i θ})} d θ$ $\frac{d I}{d x} = \int_{0}^{Π} \frac{1}{x - e^{- i θ}} + \frac{1}{x - e^{i θ}} d θ$ $= \int_{0}^{Π} \frac{e^{i θ}}{{x e}^{i θ} - 1} d θ + \int_{0}^{Π} \frac{e^{- i θ}}{{x e}^{- i θ} - 1} d θ$ $= \frac{1}{x} \int_{0}^{Π} \frac{e^{i θ}}{e^{i θ} - \frac{1}{x}} d θ + \frac{1}{x} \int_{0}^{Π} \frac{e^{- i θ}}{e^{- i θ} - \frac{1}{x}} d θ$ $=∣ \frac{1}{i x} l n (e^{i θ} - \frac{1}{x}) + \frac{1}{- i x} l n (e^{- i θ} - \frac{1}{x}) ∣_{0}^{Π}$ $= \frac{1}{i x} ∣ l n (\frac{e^{i θ} - \frac{1}{x}}{e^{- i θ} - \frac{1}{x}}) ∣_{0}^{Π} e^{i Π} = c o s Π + i s i n Π = - 1$ $\frac{1}{i x} {l n (\frac{- 1 - \frac{1}{x}}{- 1 - \frac{1}{x}}) - l n (\frac{1 - \frac{1}{x}}{1 - \frac{1}{x}})} = 0$ $\frac{d I}{d x} = 0$ $I = c o n s t a n t$ $p l s c h e c k m y s t e p s . . . w h e t h e r a n y m i s t a k e . . .$
Commented by maxmathsup by imad last updated on 09/Jul/18

$s i r T a n m a y t h e f u n c t i o n l n a t C i s n o t l i k e l n a t R l o o k l n (- 1) i n R$ $d o n t e x i s t b u t i n C l n (- 1) = l n (e^{i π}) = i π s o y o u h a v e c o m m i t e d a e r r o r$ $i n t h e f i n a l l i n e s y o u m u s t e x t r a c t R e (\int) a n d I m (\int) t o h a v e a c o r r e c t$ $a n s w e r . . .$
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