calculate ∫_0 ^(2π) ln(x^2 −2xcosθ +1)dθ


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Question Number 130536 by mathmax by abdo last updated on 26/Jan/21

$calculate \int_{0}^{2 π} \ln (x^{2} - 2 xcos θ + 1) d θ$
	Answered by Ar Brandon last updated on 26/Jan/21

	$1 . 1 - 2 acos (x) + a^{2}$ $= \cos^{2} - 2 acos (x) + a^{2} + \sin^{2} (x)$ $= {(\cos (x) - a)}^{2} + \sin^{2} (x) > 0$ $2 . 1 - 2 a \cos \frac{2 k π}{n} + a^{2}$ $= {(\cos \frac{2 k π}{n} - a)}^{2} + \sin^{2} \frac{2 k π}{n}$ $= {(\cos \frac{2 k π}{n} - a)}^{2} - {(isin \frac{2 k π}{n})}^{2}$ $= (\cos \frac{2 k π}{n} - isin \frac{2 k π}{n} - a) (\cos \frac{2 k π}{n} + isin \frac{2 k π}{n} - a)$ $= (a - e^{2 ik π / n}) (a - e^{- 2 ik π / n})$ $3 . S e e M r 1549442205 P V T^{'} s e x p l a n a t i o n o n Q 107498$ $4 . I = \int_{0}^{2 π} \ln (1 - 2 a \cos (x) + a^{2}) dx$ $= \lim_{n \to \infty} \frac{2 π}{n} \underset{k = 1}{\sum^{n}} \ln (1 - 2 a \cos (\frac{2 π k}{n}) + a^{2})$ $= \lim_{n \to \infty} \frac{2 π}{n} \ln \underset{k = 1}{\prod^{n}} (1 - 2 a \cos (\frac{2 π k}{n}) + a^{2})$ $= \lim_{n \to \infty} \frac{2 π}{n} \ln {(a^{n} - 1)}^{2} = \lim_{n \to \infty} \frac{4 π}{n} \ln (a^{n} - 1)$ $= 4 π \lim_{n \to \infty} [\frac{\ln (a^{n} - 1)}{n}] = 4 π \lim_{n \to \infty} [\frac{a^{n} lna}{a^{n} - 1}]$ $= 4 π \ln (a)$
	Answered by mathmax by abdo last updated on 26/Jan/21
	$let f(x)=∫_0 ^(2π) ln(x^2 −2xcosθ+1)dθ ⇒f^′ (x)=∫_0 ^(2π) ((2x−2cosθ)/(x^2 −2xcosθ +1))dθ =_(e^(iθ) =z) ∫_(∣z∣=1) ((2x−2((z+z^(−1) )/2))/(x^2 −2x((z+z^(−1) )/2)+1))(dz/(iz)) =∫_(∣z∣=1) ((2x−z−z^(−1) )/(x^2 +1−xz−xz^(−1) )iz))dz =(1/i)∫_(∣z∣=1) ((2x−z−z^(−1) )/((x^2 +1)z−xz^2 −x))dz =(1/i)∫_(∣z∣=1) ((2xz−z^2 −1)/(−xz^2 +(x^2 +1)z−x))dz =−i∫_(∣z∣=1) ((z^2 −2xz+1)/(z(xz^2 −(x^2 +1)z+x)))dz ϕ(z)=((z^2 −2xz+1)/(z(xz^2 −(x^2 +1)z+x))) poles of ϕ! Δ=(x^2 +1)^2 −4x^2 =x^4 +2x^2 +1−4x^2 =(x^2 −1)^2 ⇒z_1 =((x^2 +1+∣x^2 −1∣)/(2x)) and z_2 =((x^2 +1−∣x^2 −1∣)/(2x)) case1 ∣x∣>1 ⇒z_1 =((2x^2 )/(2x))=x and z_2 =(2/(2x))=(1/x) ⇒∣z_2 ∣=(1/(∣x∣))<1 ⇒ ∫_R ϕ(z)dz =2iπ{Res(ϕ,z_2 )} ϕ(z)=((z^2 −2xz+1)/(zx(z−z_1 )(z−z_2 ))) ⇒Res(ϕ,o) =(1/(z_1 z_2 ))=1 Res(ϕ,z_2 )=((z_2 ^2 −2xz_2 +1)/(z_2 x((1/x)−x)))=(((1/x^2 )−2+1)/(1−x^2 ))×x=((1−x^2 )/(x^2 (1−x^2 ))).x=(1/x) ∫_R ϕ(z)dz =2iπ{(1/x)} =((2iπ)/x) ⇒f^′ (x)=−i(((2iπ)/x))=((2π)/x) ⇒ f(x)=2πln∣x∣ +c_0 c_0 =f(1) =∫_0 ^(2π) ln(2−2cosθ)dθ =2πln(2)+∫_0 ^(2π) ln(2sin^2 ((θ/2)))dθ =4πln(2)+2∫_0 ^(2π) ln(sin((θ/2)))dθ((θ/2)=t) =4πln(2)+4∫_0 ^π ln(sin(t))dt but ∫_0 ^π ln(sint)dt =∫_0 ^(π/2) ln(sint)dt +∫_(π/2) ^π ln(sint)dt = =−(π/2)ln2−(π/2)ln2 =−πln2 ⇒c_0 =4πln2−4πln2 =0 ⇒ f(x)=2πln∣x∣ case 2 ∣x∣<1 we get z_1 =(1/x) and z_2 =x due to symetrie we get the same value ⇒f(x)=2πln∣x∣$
	$let f (x) = \int_{0}^{2 π} \ln (x^{2} - 2 xcos θ + 1) d θ \Rightarrow f^{^{'}} (x) = \int_{0}^{2 π} \frac{2 x - 2 \cos θ}{x^{2} - 2 xcos θ + 1} d θ$ $=_{e^{i θ} = z} \int_{∣ z ∣= 1} \frac{2 x - 2 \frac{z + z^{- 1}}{2}}{x^{2} - 2 x \frac{z + z^{- 1}}{2} + 1} \frac{dz}{iz}$ $= \int_{∣ z ∣= 1} \frac{2 x - z - z^{- 1}}{x^{2} + 1 - xz - {xz}^{- 1}) iz} dz = \frac{1}{i} \int_{∣ z ∣= 1} \frac{2 x - z - z^{- 1}}{(x^{2} + 1) z - {xz}^{2} - x} dz$ $= \frac{1}{i} \int_{∣ z ∣= 1} \frac{2 xz - z^{2} - 1}{- {xz}^{2} + (x^{2} + 1) z - x} dz = - i \int_{∣ z ∣= 1} \frac{z^{2} - 2 xz + 1}{z ({xz}^{2} - (x^{2} + 1) z + x)} dz$ $φ (z) = \frac{z^{2} - 2 xz + 1}{z ({xz}^{2} - (x^{2} + 1) z + x)} poles of φ!$ $Δ = {(x^{2} + 1)}^{2} - {4 x}^{2} = x^{4} + {2 x}^{2} + 1 - {4 x}^{2} = {(x^{2} - 1)}^{2}$ $\Rightarrow z_{1} = \frac{x^{2} + 1 + ∣ x^{2} - 1 ∣}{2 x} and z_{2} = \frac{x^{2} + 1 - ∣ x^{2} - 1 ∣}{2 x}$ $case 1 ∣ x ∣> 1 \Rightarrow z_{1} = \frac{{2 x}^{2}}{2 x} = x and z_{2} = \frac{2}{2 x} = \frac{1}{x} \Rightarrow∣ z_{2} ∣= \frac{1}{∣ x ∣} < 1 \Rightarrow$ $\int_{R} φ (z) dz = 2 i π {Res (φ, z_{2})}$ $φ (z) = \frac{z^{2} - 2 xz + 1}{zx (z - z_{1}) (z - z_{2})} \Rightarrow Res (φ, o) = \frac{1}{z_{1} z_{2}} = 1$ $Res (φ, z_{2}) = \frac{z_{2}^{2} - {2 xz}_{2} + 1}{z_{2} x (\frac{1}{x} - x)} = \frac{\frac{1}{x^{2}} - 2 + 1}{1 - x^{2}} \times x = \frac{1 - x^{2}}{x^{2} (1 - x^{2})} . x = \frac{1}{x}$ $\int_{R} φ (z) dz = 2 i π {\frac{1}{x}} = \frac{2 i π}{x} \Rightarrow f^{^{'}} (x) = - i (\frac{2 i π}{x}) = \frac{2 π}{x} \Rightarrow$ $f (x) = 2 π \ln ∣ x ∣ + c_{0} c_{0} = f (1) = \int_{0}^{2 π} \ln (2 - 2 \cos θ) d θ$ $= 2 π \ln (2) + \int_{0}^{2 π} \ln ({2 \sin}^{2} (\frac{θ}{2})) d θ = 4 π \ln (2) + 2 \int_{0}^{2 π} \ln (\sin (\frac{θ}{2})) d θ (\frac{θ}{2} = t)$ $= 4 π \ln (2) + 4 \int_{0}^{π} \ln (\sin (t)) dt but$ $\int_{0}^{π} \ln (sint) dt = \int_{0}^{\frac{π}{2}} \ln (sint) dt + \int_{\frac{π}{2}}^{π} \ln (sint) dt =$ $= - \frac{π}{2} \ln 2 - \frac{π}{2} \ln 2 = - π \ln 2 \Rightarrow c_{0} = 4 π \ln 2 - 4 π \ln 2 = 0 \Rightarrow$ $f (x) = 2 π \ln ∣ x ∣$ $case 2 ∣ x ∣< 1 we get z_{1} = \frac{1}{x} and z_{2} = x due to symetrie we get the$ $same value \Rightarrow f (x) = 2 π \ln ∣ x ∣$
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