calculate ∫_0 ^π ln(x^2 −2xsinθ +1)dθ


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Question Number 62145 by maxmathsup by imad last updated on 16/Jun/19

$c a l c u l a t e \int_{0}^{π} l n (x^{2} - 2 x s i n θ + 1) d θ$
Commented by maxmathsup by imad last updated on 17/Jun/19
$let f(x) =∫_0 ^π ln(x^2 −2xsinθ +1)dθ ⇒f^′ (x) =∫_0 ^π ((2x−2sinθ)/(x^2 −2xsinθ+1)) dθ ⇒ ((f^′ (x))/2) =∫_0 ^π ((x−sinθ)/(x^2 −2xsinθ +1))dθ =_(θ=(t/2)) ∫_0 ^(2π) ((x−sin((t/2)))/(x^2 −2x sin((t/2))+1)) (dt/2) ⇒ f^′ (x) =_(z =e^((it)/2) ) ∫_(∣z∣=1) ((x−((z−z^(−1) )/(2i)))/(x^2 −2x((z−z^(−1) )/(2i)))) ((2dz)/(iz)) = ∫_(∣z∣=1) ((2ix−z+z^(−1) )/(−z{ 2ix^2 −2xz +2xz^(−1) })) (2i)dz =−∫_(∣z∣=1) ((−4x−2iz +2iz^(−1) )/(2ix^2 z−2xz^2 +2x))dz =∫ ((4x+2iz−2iz^(−1) )/(−2xz^2 +2ix^2 z +2x))dz =(1/x)∫_(∣z∣=1) ((4xz +2iz^2 −2i)/(z(−2xz^2 +2ix^2 z +2x}))dz (x≠0) =−(1/x)∫_(∣z∣=1) ((iz^2 +2xz −i)/(z{z^2 −ixz −1}))dz let w(z)=((iz^2 +2xz −i)/(z{z^2 −ixz −1})) poles of w? z^2 −ixz −1 =0→Δ^′ =(−ix)^2 +1 =1−x^2 case 1 x>1 ⇒ Δ^′ =(i(√(x^2 −1)))^2 ⇒z_1 =ix+i(√(x^2 −1)) z_2 =ix−i(√(x^2 −1)) and w(z)=((iz^2 +2xz−i)/(z(z−z_1 )(z−z_2 ))) ∣z_1 ∣ −1=∣x+(√(x^2 −1))∣ =x+(√(x^2 −1))−1 =x−1 +(√(x^2 −1))>0 (z_1 is out of circle) ∣z_2 ∣−1 =∣x−(√(x^2 −1))∣−1 =x−1−(√(x^2 −1)) (x−1)^2 −(x^2 −1) =x^2 −2x +1−x^2 +1 =2−2x =2(1−x)<0 ⇒∣z_2 ∣<1 ∫_(∣z∣=1) w(z)dz =2iπ { Res(w,o) +Res(w,z_2 )} Res(w,0) =lim_(z→0) zw(z) =((−i)/(z_1 z_2 )) =((−i)/(−1)) =i Res(w,z_2 ) =lim_(z→z_2 ) (z−z_2 )w(z) =((iz_2 ^2 +2xz_2 −i)/(z_2 (z_2 −z_1 ))) =((iz_2 ^2 +2xz_2 −i)/((ix−i(√(x^2 −1)))^2 +1)) =((i(ix−i(√(x^2 −1)))^2 +2x(ix−(√(x^2 −1))) −i)/(1−(x+(√(x^2 −1)))^2 )) =((−i(x+(√(x^2 −1)))^2 +2ix^2 −2x(√(x^2 −1))−i)/(1−(x^2 +2x(√(x^2 −1))+x^2 −1))) =.... be continued... m$
$l e t f (x) = \int_{0}^{π} l n (x^{2} - 2 x s i n θ + 1) d θ \Rightarrow f^{^{'}} (x) = \int_{0}^{π} \frac{2 x - 2 s i n θ}{x^{2} - 2 x s i n θ + 1} d θ \Rightarrow$ $\frac{f^{^{'}} (x)}{2} = \int_{0}^{π} \frac{x - s i n θ}{x^{2} - 2 x s i n θ + 1} d θ =_{θ = \frac{t}{2}} \int_{0}^{2 π} \frac{x - s i n (\frac{t}{2})}{x^{2} - 2 x s i n (\frac{t}{2}) + 1} \frac{d t}{2}$ $\Rightarrow f^{^{'}} (x) =_{z = e^{\frac{i t}{2}}} \int_{∣ z ∣= 1} \frac{x - \frac{z - z^{- 1}}{2 i}}{x^{2} - 2 x \frac{z - z^{- 1}}{2 i}} \frac{2 d z}{i z}$ $= \int_{∣ z ∣= 1} \frac{2 i x - z + z^{- 1}}{- z {2 {i x}^{2} - 2 x z + 2 {x z}^{- 1}}} (2 i) d z$ $= - \int_{∣ z ∣= 1} \frac{- 4 x - 2 i z + 2 {i z}^{- 1}}{2 {i x}^{2} z - 2 {x z}^{2} + 2 x} d z$ $= \int \frac{4 x + 2 i z - 2 {i z}^{- 1}}{- 2 {x z}^{2} + 2 {i x}^{2} z + 2 x} d z = \frac{1}{x} \int_{∣ z ∣= 1} \frac{4 x z + 2 {i z}^{2} - 2 i}{z (- 2 {x z}^{2} + 2 {i x}^{2} z + 2 x}} d z (x \neq 0)$ $= - \frac{1}{x} \int_{∣ z ∣= 1} \frac{{i z}^{2} + 2 x z - i}{z {z^{2} - i x z - 1}} d z l e t w (z) = \frac{{i z}^{2} + 2 x z - i}{z {z^{2} - i x z - 1}} p o l e s o f w ?$ $z^{2} - i x z - 1 = 0 \to Δ^{^{'}} = {(- i x)}^{2} + 1 = 1 - x^{2}$ $c a s e 1 x > 1 \Rightarrow Δ^{^{'}} = {(i \sqrt{x^{2} - 1})}^{2} \Rightarrow z_{1} = i x + i \sqrt{x^{2} - 1}$ $z_{2} = i x - i \sqrt{x^{2} - 1} a n d w (z) = \frac{{i z}^{2} + 2 x z - i}{z (z - z_{1}) (z - z_{2})}$ $∣ z_{1} ∣ - 1 =∣ x + \sqrt{x^{2} - 1} ∣ = x + \sqrt{x^{2} - 1} - 1 = x - 1 + \sqrt{x^{2} - 1} > 0 (z_{1} i s o u t o f c i r c l e)$ $∣ z_{2} ∣ - 1 =∣ x - \sqrt{x^{2} - 1} ∣ - 1 = x - 1 - \sqrt{x^{2} - 1}$ ${(x - 1)}^{2} - (x^{2} - 1) = x^{2} - 2 x + 1 - x^{2} + 1 = 2 - 2 x = 2 (1 - x) < 0 \Rightarrow∣ z_{2} ∣< 1$ $\int_{∣ z ∣= 1} w (z) d z = 2 i π {R e s (w, o) + R e s (w, z_{2})}$ $R e s (w, 0) = {l i m}_{z \to 0} z w (z) = \frac{- i}{z_{1} z_{2}} = \frac{- i}{- 1} = i$ $R e s (w, z_{2}) = {l i m}_{z \to z_{2}} (z - z_{2}) w (z) = \frac{{i z}_{2}^{2} + 2 {x z}_{2} - i}{z_{2} (z_{2} - z_{1})} = \frac{{i z}_{2}^{2} + 2 {x z}_{2} - i}{{(i x - i \sqrt{x^{2} - 1})}^{2} + 1}$ $= \frac{i {(i x - i \sqrt{x^{2} - 1})}^{2} + 2 x (i x - \sqrt{x^{2} - 1}) - i}{1 - {(x + \sqrt{x^{2} - 1})}^{2}}$ $= \frac{- i {(x + \sqrt{x^{2} - 1})}^{2} + 2 {i x}^{2} - 2 x \sqrt{x^{2} - 1} - i}{1 - (x^{2} + 2 x \sqrt{x^{2} - 1} + x^{2} - 1)} = . . . . b e c o n t i n u e d . . .$ $m$
	Answered by perlman last updated on 16/Jun/19
	$let A(θ)=∫_0 ^π ln(x^2 −2xcos(θ)+1)dθ θ∈IR\{2kπ,k∈IZ} A(θ)=[xln(x^2 −2cos(θ)x+1)]_0 ^π −∫((2x^2 −2xcos(θ))/(x^2 −2cos(θ)x+1))dx =πln(π^2 −2cos(θ)π+1)−2∫_0 ^π ((x^2 −xcos(θ))/(x^2 −2cos(θ)x+1))dx ((x^2 −xcos(θ))/(x^2 −2cos(θ)x+1))=1+((xcos(θ)−1)/(x^2 −2cos(θ)x+1)) A(θ)=πln(π^2 −2cos(θ)π+1)−2∫_0 ^π 1+((xcos(θ)−1)/(x^2 −2cos(θ)x+1))dx the 2 nd integral =−2π −2∫_0 ^π ((xcos(θ)−1)/(x^2 −2cos(θ)x+1))dx=−2π−∫_0 ^π ((cos(θ)[2x−2cos(θ)]−2+2cos^2 (θ))/(x^2 −2cos(θ)x+1))dx =−2π−cos(θ)∫_0 ^π ((2x−2cos(θ))/(x^2 −2cos(θ)x+1))dx+2∫((sin^2 (θ))/(x^2 −2cos(θ)x+1))dx =−2π−cos(θ)[ln[x^2 −2cos(θ)x+1⌉_0 ^π +2sin^2 θ∫_0 ^π (1/((x−cos(θ))^2 +sin^2 (θ)))dx =2π−cos(θ)ln(π2−2cos(θ)π+1)+2∫_0 ^π (1/(((x/(sin(θ)))−cot (θ))^2 +1))dx if sin≠0 =−2π−cos(θ)ln(π^2 −2cos(θ)π+1)+2sin(θ){tan^(−1) {(x/(sinθ))−cot (θ)}_0 ^π } =−2π−cos(θ)ln{π^2 −2cos(θ)π+1}+2sin(θ){tan^(−1) {(π/(sin(θ)))−cot (θ)}−tan^(−1) {cot (−θ)} =−2π−cos(θ)ln{π^2 −2cos(θ)π+1}+2sin(θ){arctan{(π/(sin(θ)))−cot(θ)}−arctan(tg((π/2)+θ)) A(θ)=−2π+(π−cos(θ))ln(π^2 −2cos(θ)x+1)+2sin(θ)arctan(((π−cos(θ))/(sin(θ))))−2sin(θ)arctan(tan((π/2)+θ)) arctan(tg(a))=a if a∈]((−π)/2).(π/2)[ if a∈]−(π/2)+kπ.(π/2)+kπ[ arctantg(a)=a−kπ$
	$l e t A (θ) = \int_{0}^{π} l n (x^{2} - 2 x c o s (θ) + 1) d θ$ $θ \in I R ∖ {2 k π, k \in I Z}$ $A (θ) = {[x l n (x^{2} - 2 c o s (θ) x + 1)]}_{0}^{π} - \int \frac{2 x^{2} - 2 x c o s (θ)}{x^{2} - 2 c o s (θ) x + 1} d x$ $= π l n (π^{2} - 2 c o s (θ) π + 1) - 2 \int_{0}^{π} \frac{x^{2} - x c o s (θ)}{x^{2} - 2 c o s (θ) x + 1} d x$ $\frac{x^{2} - x c o s (θ)}{x^{2} - 2 c o s (θ) x + 1} = 1 + \frac{x c o s (θ) - 1}{x^{2} - 2 c o s (θ) x + 1}$ $A (θ) = π l n (π^{2} - 2 c o s (θ) π + 1) - 2 \int_{0}^{π} 1 + \frac{x c o s (θ) - 1}{x^{2} - 2 c o s (θ) x + 1} d x$ $t h e 2 n d i n t e g r a l$ $= - 2 π - 2 \int_{0}^{π} \frac{x c o s (θ) - 1}{x^{2} - 2 c o s (θ) x + 1} d x = - 2 π - \int_{0}^{π} \frac{c o s (θ) [2 x - 2 c o s (θ)] - 2 + 2 {c o s}^{2} (θ)}{x^{2} - 2 c o s (θ) x + 1} d x$ $= - 2 π - c o s (θ) \int_{0}^{π} \frac{2 x - 2 c o s (θ)}{x^{2} - 2 c o s (θ) x + 1} d x + 2 \int \frac{{s i n}^{2} (θ)}{x^{2} - 2 c o s (θ) x + 1} d x$ $= - 2 π - c o s (θ) [l n [x^{2} - 2 c o s (θ) x + 1 ⌉_{0}^{π} + 2 {s i n}^{2} θ \int_{0}^{π} \frac{1}{{(x - c o s (θ))}^{2} + \sin^{2} (θ)} d x$ $= 2 π - c o s (θ) l n (π 2 - 2 c o s (θ) π + 1) + 2 \int_{0}^{π} \frac{1}{{(\frac{x}{\sin (θ)} - \cot (θ))}^{2} + 1} d x i f s i n \neq 0$ $= - 2 π - c o s (θ) l n (π^{2} - 2 c o s (θ) π + 1) + 2 s i n (θ) {\tan^{- 1} {\frac{x}{s i n θ} - \cot (θ)}_{0}^{π}}$ $= - 2 π - c o s (θ) l n {π^{2} - 2 c o s (θ) π + 1} + 2 s i n (θ) {\tan^{- 1} {\frac{π}{s i n (θ)} - \cot (θ)} - \tan^{- 1} {\cot (- θ)}$ $= - 2 π - c o s (θ) l n {π^{2} - 2 c o s (θ) π + 1} + 2 s i n (θ) {a r c t a n {\frac{π}{s i n (θ)} - c o t (θ)} - a r c t a n (t g (\frac{π}{2} + θ))$ $A (θ) = - 2 π + (π - c o s (θ)) l n (π^{2} - 2 c o s (θ) x + 1) + 2 s i n (θ) a r c t a n (\frac{π - c o s (θ)}{s i n (θ)}) - 2 s i n (θ) a r c t a n (t a n (\frac{π}{2} + θ))$ $a r c t a n (t g (a)) = a i f a \in] \frac{- π}{2} . \frac{π}{2} [$ $i f a \in] - \frac{π}{2} + k π . \frac{π}{2} + k π [$ $a r c t a n t g (a) = a - k π$
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