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

Question Number 125146 by mathmax by abdo last updated on 08/Dec/20

1) calculate \int_{0}^{2 π} \frac{d θ}{x^{2} - 2 x \cos θ + 1}

2) calculate \int_{0}^{2 π} \frac{\cos θ}{{(x^{2} - 2 xcos θ + 1)}^{2}} d θ

Answered by Dwaipayan Shikari last updated on 08/Dec/20

x^{2} - 2 x c o s θ + 1 = 0

x = c o s θ \pm i s i n θ = e^{\pm i θ}

\int_{0}^{2 π} \frac{1}{(x - e^{i θ}) (x - e^{- i θ})} d x

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

= \frac{1}{2 c o s θ} {[l o g (\frac{x - e^{i θ}}{x - e^{- i θ}})]}_{0}^{2 π} = \frac{1}{2 c o s θ} l o g (\frac{2 π - e^{i θ}}{2 π - e^{- i θ}}) - \frac{θ i}{c o s θ}

Answered by mathmax by abdo last updated on 08/Dec/20

2) we have I (x) = \int_{0}^{\infty} \frac{d θ}{x^{2} - 2 xcos θ + 1} \Rightarrow

I^{^{'}} (x) = \int_{0}^{\infty} \frac{\partial}{\partial x} (\frac{1}{x^{2} - 2 xcos θ + 1}) d θ = - \int_{0}^{\infty} \frac{2 x - 2 \cos θ}{{(x^{2} - 2 xcos θ + 1)}^{2}} d θ

= - 2 x \int_{0}^{\infty} \frac{d θ}{{(x^{2} - 2 xcos θ + 1)}^{2}} + 2 \int_{0}^{\infty} \frac{\cos θ}{{(x^{2} - 2 xcos θ + 1)}^{2}} d θ \Rightarrow

\int_{0}^{\infty} \frac{\cos θ d θ}{{(x^{2} - 2 xcos θ + 1)}^{2}} = \frac{1}{2} {I^{^{'}} (x) + 2 x \int_{0}^{\infty} \frac{d θ}{{(x^{2} - 2 xcos θ + 1)}^{2}}}

I (x) is known rest to calculate I^{^{'}} (x) and \int_{0}^{\infty} \frac{d θ}{{(x^{2} - 2 xcos θ + 1)}^{2}}

\dots be continued . .

Answered by mathmax by abdo last updated on 08/Dec/20

1) let I (x) = \int_{0}^{2 π} \frac{d θ}{x^{2} - 2 xcos θ + 1} changement e^{i θ} = z give

I (x) = \int_{∣ z ∣= 1} \frac{dz}{iz (x^{2} - 2 x \times \frac{z + z^{- 1}}{2} + 1)} = \int_{∣ z ∣= 1} \frac{- idz}{z (x^{2} - xz - {xz}^{- 1} + 1)}

= \int_{∣ z ∣= 1} \frac{- idz}{x^{2} z - {xz}^{2} - x + z} = \int_{∣ z ∣= 1} \frac{- idz}{- {xz}^{2} + (1 + x^{2}) z - x}

= \int_{∣ z ∣= 1} \frac{idz}{{xz}^{2} - (1 + x^{2}) z + x} let w (z) = \frac{i}{{xz}^{2} - (1 + x^{2}) z + x} poles of w ?

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

\Rightarrow z_{1} = \frac{1 + x^{2} + ∣ x^{2} - 1 ∣}{2 x} and z_{2} = \frac{1 + x^{2} - ∣ x^{2} - 1 ∣}{2 x}

case 1 ∣ x ∣> 1 \Rightarrow z_{1} = \frac{1 + x^{2} + x^{2} - 1}{2 x} = x and z_{2} = \frac{1 + x^{2} - x^{2} + 1}{2 x} = \frac{1}{x}

∣ z_{1} ∣ - 1 =∣ x ∣ - 1 > 0 and ∣ z_{2} ∣ - 1 = \frac{1}{∣ x ∣} - 1 < 0 \Rightarrow

\int_{∣ z ∣= 1} w (z) dz = 2 i π Res (w, \frac{1}{x}) but s (x) = \frac{i}{x (z - x) (z - \frac{1}{x})} \Rightarrow

Res (w, \frac{1}{x}) = \frac{i}{x (\frac{1}{x} - x)} = \frac{i}{1 - x^{2}} \Rightarrow \int_{∣ z ∣= 1} w (z) dz = 2 i π \times \frac{i}{1 - x^{2}}

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

case 2 ∣ x ∣< 1 \Rightarrow z_{1} = \frac{1}{x} and z_{2} = x

∣ z_{1} ∣ - 1 = \frac{1}{∣ x ∣} - 1 > 0 (out of circle) and ∣ z_{2} ∣ - 1 =∣ x ∣ - 1 < 0 \Rightarrow

\int_{∣ z ∣= 1} w (z) dz = 2 i π Res (w, x) = 2 i π \times \frac{i}{x (x - \frac{1}{x})} = \frac{- 2 π}{x^{2} - 1} = \frac{2 π}{1 - x^{2}} \Rightarrow

I (x) = \frac{2 π}{1 - x^{2}} [finally I (x) = {\begin{cases} \frac{2 π}{x^{2} - 1} if ∣ x ∣> 1 \\ \frac{2 π}{1 - x^{2}} if ∣ x ∣< 1 \end{cases}