calculate-0-2pi-ln-x-2-2xcos-1-d-

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) = \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 ∣