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


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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
	$2) we have I(x)=∫_0 ^∞ (dθ/(x^2 −2xcosθ +1)) ⇒ I^′ (x)=∫_0 ^∞ (∂/∂x)((1/(x^2 −2xcosθ +1)))dθ =−∫_0 ^∞ ((2x−2cosθ)/((x^2 −2xcosθ +1)^2 ))dθ =−2x ∫_0 ^∞ (dθ/((x^2 −2xcosθ +1)^2 )) +2 ∫_0 ^∞ ((cosθ)/((x^2 −2xcosθ+1)^2 ))dθ ⇒ ∫_0 ^∞ ((cosθ dθ)/((x^2 −2xcosθ +1)^2 ))=(1/2){ I^′ (x)+2x ∫_0 ^∞ (dθ/((x^2 −2xcosθ+1)^2 ))} I(x) is known rest to calculate I^′ (x) and ∫_0 ^∞ (dθ/((x^2 −2xcosθ +1)^2 )) ...be continued..$
	$2) we have I (x) = \int_{0}^{\infty} \frac{d θ}{x^{2} - 2 xcos θ + 1} \Rightarrow$ $I^{^{'}} (x) = \int_{0}^{\infty} \frac{\partial}{\partial x} (\frac{1}{x^{2} - 2 xcos θ + 1}) d θ = - \int_{0}^{\infty} \frac{2 x - 2 \cos θ}{{(x^{2} - 2 xcos θ + 1)}^{2}} d θ$ $= - 2 x \int_{0}^{\infty} \frac{d θ}{{(x^{2} - 2 xcos θ + 1)}^{2}} + 2 \int_{0}^{\infty} \frac{\cos θ}{{(x^{2} - 2 xcos θ + 1)}^{2}} d θ \Rightarrow$ $\int_{0}^{\infty} \frac{\cos θ d θ}{{(x^{2} - 2 xcos θ + 1)}^{2}} = \frac{1}{2} {I^{^{'}} (x) + 2 x \int_{0}^{\infty} \frac{d θ}{{(x^{2} - 2 xcos θ + 1)}^{2}}}$ $I (x) is known rest to calculate I^{^{'}} (x) and \int_{0}^{\infty} \frac{d θ}{{(x^{2} - 2 xcos θ + 1)}^{2}}$ $. . . be continued . .$
	Answered by mathmax by abdo last updated on 08/Dec/20
	$1) let I(x)=∫_0 ^(2π) (dθ/(x^2 −2xcosθ +1)) changement e^(iθ) =z give I(x)=∫_(∣z∣=1) (dz/(iz(x^2 −2x×((z+z^(−1) )/2)+1))) =∫_(∣z∣=1) ((−idz)/(z(x^2 −xz−xz^(−1) +1))) =∫_(∣z∣=1) ((−idz)/(x^2 z−xz^2 −x+z)) =∫_(∣z∣=1) ((−idz)/(−xz^2 +(1+x^2 )z−x)) =∫_(∣z∣=1) ((idz)/(xz^2 −(1+x^2 )z+x)) let w(z)=(i/(xz^2 −(1+x^2 )z+x)) poles of w? Δ=(1+x^2 )^2 −4x^2 =x^4 +2x^2 +1−4x^2 =x^4 −2x^2 +1=(x^2 −1)^2 ⇒z_1 =((1+x^2 +∣x^2 −1∣)/(2x)) and z_2 =((1+x^2 −∣x^2 −1∣)/(2x)) case1 ∣x∣>1 ⇒z_1 =((1+x^2 +x^2 −1)/(2x))=x and z_2 =((1+x^2 −x^2 +1)/(2x))=(1/x) ∣z_1 ∣−1=∣x∣−1>0 and ∣z_2 ∣−1 =(1/(∣x∣))−1<0 ⇒ ∫_(∣z∣=1) w(z)dz =2iπ Res(w,(1/x)) but s(x)=(i/(x(z−x)(z−(1/x)))) ⇒ Res(w,(1/x))=(i/(x((1/x)−x)))=(i/(1−x^2 )) ⇒∫_(∣z∣=1) w(z)dz =2iπ×(i/(1−x^2 )) =((−2π)/(1−x^2 ))=((2π)/(x^2 −1))=I(x) case2 ∣x∣<1 ⇒z_1 =(1/x) and z_2 =x ∣z_1 ∣−1 =(1/(∣x∣))−1>0 (out of circle) and ∣z_2 ∣−1=∣x∣−1<0 ⇒ ∫_(∣z∣=1) w(z)dz =2iπ Res(w,x)=2iπ×(i/(x(x−(1/x))))=((−2π)/(x^2 −1))=((2π)/(1−x^2 )) ⇒ I(x)=((2π)/(1−x^2 )) [finally I(x)= { ((((2π)/(x^2 −1)) if ∣x∣>1)),((((2π)/(1−x^2 )) if ∣x∣<1)) :}$
	$1) let I (x) = \int_{0}^{2 π} \frac{d θ}{x^{2} - 2 xcos θ + 1} changement e^{i θ} = z give$ $I (x) = \int_{∣ z ∣= 1} \frac{dz}{iz (x^{2} - 2 x \times \frac{z + z^{- 1}}{2} + 1)} = \int_{∣ z ∣= 1} \frac{- idz}{z (x^{2} - xz - {xz}^{- 1} + 1)}$ $= \int_{∣ z ∣= 1} \frac{- idz}{x^{2} z - {xz}^{2} - x + z} = \int_{∣ z ∣= 1} \frac{- idz}{- {xz}^{2} + (1 + x^{2}) z - x}$ $= \int_{∣ z ∣= 1} \frac{idz}{{xz}^{2} - (1 + x^{2}) z + x} let w (z) = \frac{i}{{xz}^{2} - (1 + x^{2}) z + x} poles of w ?$ $Δ = {(1 + x^{2})}^{2} - {4 x}^{2} = x^{4} + {2 x}^{2} + 1 - {4 x}^{2} = x^{4} - {2 x}^{2} + 1 = {(x^{2} - 1)}^{2}$ $\Rightarrow z_{1} = \frac{1 + x^{2} + ∣ x^{2} - 1 ∣}{2 x} and z_{2} = \frac{1 + x^{2} - ∣ x^{2} - 1 ∣}{2 x}$ $case 1 ∣ x ∣> 1 \Rightarrow z_{1} = \frac{1 + x^{2} + x^{2} - 1}{2 x} = x and z_{2} = \frac{1 + x^{2} - x^{2} + 1}{2 x} = \frac{1}{x}$ $∣ z_{1} ∣ - 1 =∣ x ∣ - 1 > 0 and ∣ z_{2} ∣ - 1 = \frac{1}{∣ x ∣} - 1 < 0 \Rightarrow$ $\int_{∣ z ∣= 1} w (z) dz = 2 i π Res (w, \frac{1}{x}) but s (x) = \frac{i}{x (z - x) (z - \frac{1}{x})} \Rightarrow$ $Res (w, \frac{1}{x}) = \frac{i}{x (\frac{1}{x} - x)} = \frac{i}{1 - x^{2}} \Rightarrow \int_{∣ z ∣= 1} w (z) dz = 2 i π \times \frac{i}{1 - x^{2}}$ $= \frac{- 2 π}{1 - x^{2}} = \frac{2 π}{x^{2} - 1} = I (x)$ $case 2 ∣ x ∣< 1 \Rightarrow z_{1} = \frac{1}{x} and z_{2} = x$ $∣ z_{1} ∣ - 1 = \frac{1}{∣ x ∣} - 1 > 0 (out of circle) and ∣ z_{2} ∣ - 1 =∣ x ∣ - 1 < 0 \Rightarrow$ $\int_{∣ z ∣= 1} w (z) dz = 2 i π Res (w, x) = 2 i π \times \frac{i}{x (x - \frac{1}{x})} = \frac{- 2 π}{x^{2} - 1} = \frac{2 π}{1 - x^{2}} \Rightarrow$ $I (x) = \frac{2 π}{1 - x^{2}} [finally I (x) = {\begin{cases} \frac{2 π}{x^{2} - 1} if ∣ x ∣> 1 \\ \frac{2 π}{1 - x^{2}} if ∣ x ∣< 1 \end{cases}$
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