calculate ∫_0 ^(2π) (dt/(x −e^(it) ))


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Question Number 36931 by maxmathsup by imad last updated on 07/Jun/18

$c a l c u l a t e \int_{0}^{2 π} \frac{d t}{x - e^{i t}}$
Commented by math khazana by abdo last updated on 09/Jun/18
$let f(x) = ∫_0 ^(2π) (dt/(x −e^(it) )) f(x)= ∫_0 ^(2π) (dt/(x−cost −isint)) =∫_0 ^(2π) ((x−cost +i sint)/((x−cost)^2 +sin^2 t))dt =∫_0 ^(2π) ((x−cost)/((x−cost)^2 +sin^2 t))dt +i ∫_0 ^(2π) ((sint)/((x−cost)^2 +sin^2 t))dt =I +iJ I = ∫_0 ^π ((x−cost)/(x^2 −2x cost +1))dt +∫_π ^(2π) ((x−cost)/(x^2 −2xcost +1)) chang.tan((t/2)) =u give ∫_0 ^π ((x−cost)/(x^2 −2x cost +1))dt = ∫_0 ^∞ ((x −((1−u^2 )/(1+u^2 )))/(x^2 −2x ((1−u^2 )/(1+u^2 )) +1)) ((2du)/(1+u^2 )) = ∫_0 ^∞ ((x(1+u^2 ) −1+u^2 )/((1+u^2 )^2 (((x^2 (1+u^2 )−2x+2xu^2 )/(1+u^2 )))))du =∫_0 ^∞ ((x +xu^2 −1+u^2 )/((1+u^2 ){x^2 +x^2 u^2 −2x +2xu^2 )))du =∫_0 ^∞ (((x+1)u^2 +x−1)/((1+u^2 ){ (x^2 +2x)u^2 +x^2 −2x}))du =(1/x)∫_0 ^∞ (((x+1)u^2 +x−1)/((1+u^2 )( (x+2)u^2 +x−2))) du = (1/(x(x+2)))∫_0 ^∞ (((x+1)u^2 +x−1)/((1+u^2 )(u^2 +((x−2)/(x+2)))))du case1 ((x−2)/(x+2))>0 andx(x+2)≠0 let ϕ(z)= (((x+1)z^2 +x−1)/((1+z^2 )(z^(2 ) +((x−2)/(x+2))))) the poles of ϕ are i ,−i, i(√((x−2)/(x+2))),−i(√((x−2)/(x+2))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,i)+Res(ϕ,i(√((x−2)/(x+2))))} Res(ϕ,i) = (((x+1)(−1) +x−1)/(2i(−1 +((x−2)/(x+2))))) =((−2)/(2i((−4)/(x+2)))) =((−2(x+2))/(−8i)) = ((x+2)/(4i)) Res(ϕ,i(√((x−2)/(x+2)))) = ((−(x+1)((x−2)/(x+2)) +x−1)/(2i(√((x−2)/(x+2)))(1−((x−2)/(x+2))))) = (((x−1)(x+2)−(x+1)(x−2))/(2i(x+2)(√((x−2)/(x+2)))(x+2 −x+2))) = ((x^2 +x −2 −(x^2 −x−2))/(8i(x+2)(√((x−2)/(x+2))))) = ((2x)/(8i(√(x^2 −4)))) = (x/(4i(√(x^2 −4)))) ∫_(−∞) ^+ ϕ(z)dz =2iπ{ ((x+2)/(4i)) + (x/(4i(√(x^2 −4))))} =(π/2)(x+2) +((πx)/(2(√(x^2 −4)))) ⇒ ∫_0 ^π ((x−cost)/(x^2 −2x cost +1)) dt =(1/(x(x+2))){(π/4)(x+2) +((πx)/(4(√(x^2 −4)))) .} = (π/(4x)) +(π/(4(x+2)(√(x^2 −4)))) . case 2 ((x−2)/(x+2))<0 ⇒ ϕ(z) = (((x+1)z^2 +x−1)/((1+z^2 )(z^2 − ((2−x)/(2+x))))) and the roots of ϕ are i,−i,(√((2−x)/(2+x))) ,−(√( ((2−x)/(2+x)))) and we follow the same method...$
$l e t f (x) = \int_{0}^{2 π} \frac{d t}{x - e^{i t}}$ $f (x) = \int_{0}^{2 π} \frac{d t}{x - c o s t - i s i n t}$ $= \int_{0}^{2 π} \frac{x - c o s t + i s i n t}{{(x - c o s t)}^{2} + {s i n}^{2} t} d t$ $= \int_{0}^{2 π} \frac{x - c o s t}{{(x - c o s t)}^{2} + {s i n}^{2} t} d t + i \int_{0}^{2 π} \frac{s i n t}{{(x - c o s t)}^{2} + {s i n}^{2} t} d t$ $= I + i J$ $I = \int_{0}^{π} \frac{x - c o s t}{x^{2} - 2 x c o s t + 1} d t + \int_{π}^{2 π} \frac{x - c o s t}{x^{2} - 2 x c o s t + 1}$ $c h a n g . t a n (\frac{t}{2}) = u g i v e$ $\int_{0}^{π} \frac{x - c o s t}{x^{2} - 2 x c o s t + 1} d t = \int_{0}^{\infty} \frac{x - \frac{1 - u^{2}}{1 + u^{2}}}{x^{2} - 2 x \frac{1 - u^{2}}{1 + u^{2}} + 1} \frac{2 d u}{1 + u^{2}}$ $= \int_{0}^{\infty} \frac{x (1 + u^{2}) - 1 + u^{2}}{{(1 + u^{2})}^{2} (\frac{x^{2} (1 + u^{2}) - 2 x + 2 {x u}^{2}}{1 + u^{2}})} d u$ $= \int_{0}^{\infty} \frac{x + {x u}^{2} - 1 + u^{2}}{(1 + u^{2}) {x^{2} + x^{2} u^{2} - 2 x + 2 {x u}^{2})} d u$ $= \int_{0}^{\infty} \frac{(x + 1) u^{2} + x - 1}{(1 + u^{2}) {(x^{2} + 2 x) u^{2} + x^{2} - 2 x}} d u$ $= \frac{1}{x} \int_{0}^{\infty} \frac{(x + 1) u^{2} + x - 1}{(1 + u^{2}) ((x + 2) u^{2} + x - 2)} d u$ $= \frac{1}{x (x + 2)} \int_{0}^{\infty} \frac{(x + 1) u^{2} + x - 1}{(1 + u^{2}) (u^{2} + \frac{x - 2}{x + 2})} d u$ $c a s e 1 \frac{x - 2}{x + 2} > 0 a n d x (x + 2) \neq 0 l e t$ $φ (z) = \frac{(x + 1) z^{2} + x - 1}{(1 + z^{2}) (z^{2} + \frac{x - 2}{x + 2})} t h e p o l e s o f φ a r e$ $i, - i, i \sqrt{\frac{x - 2}{x + 2}}, - i \sqrt{\frac{x - 2}{x + 2}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, i) + R e s (φ, i \sqrt{\frac{x - 2}{x + 2}})}$ $R e s (φ, i) = \frac{(x + 1) (- 1) + x - 1}{2 i (- 1 + \frac{x - 2}{x + 2})} = \frac{- 2}{2 i \frac{- 4}{x + 2}}$ $= \frac{- 2 (x + 2)}{- 8 i} = \frac{x + 2}{4 i}$ $R e s (φ, i \sqrt{\frac{x - 2}{x + 2}}) = \frac{- (x + 1) \frac{x - 2}{x + 2} + x - 1}{2 i \sqrt{\frac{x - 2}{x + 2}} (1 - \frac{x - 2}{x + 2})}$ $= \frac{(x - 1) (x + 2) - (x + 1) (x - 2)}{2 i (x + 2) \sqrt{\frac{x - 2}{x + 2}} (x + 2 - x + 2)}$ $= \frac{x^{2} + x - 2 - (x^{2} - x - 2)}{8 i (x + 2) \sqrt{\frac{x - 2}{x + 2}}}$ $= \frac{2 x}{8 i \sqrt{x^{2} - 4}} = \frac{x}{4 i \sqrt{x^{2} - 4}}$ $\int_{- \infty}^{+} φ (z) d z = 2 i π {\frac{x + 2}{4 i} + \frac{x}{4 i \sqrt{x^{2} - 4}}}$ $= \frac{π}{2} (x + 2) + \frac{π x}{2 \sqrt{x^{2} - 4}} \Rightarrow$ $\int_{0}^{π} \frac{x - c o s t}{x^{2} - 2 x c o s t + 1} d t = \frac{1}{x (x + 2)} {\frac{π}{4} (x + 2) + \frac{π x}{4 \sqrt{x^{2} - 4}} .}$ $= \frac{π}{4 x} + \frac{π}{4 (x + 2) \sqrt{x^{2} - 4}} .$ $c a s e 2 \frac{x - 2}{x + 2} < 0 \Rightarrow$ $φ (z) = \frac{(x + 1) z^{2} + x - 1}{(1 + z^{2}) (z^{2} - \frac{2 - x}{2 + x})} a n d t h e r o o t s o f φ a r e$ $i, - i, \sqrt{\frac{2 - x}{2 + x}}, - \sqrt{\frac{2 - x}{2 + x}} a n d w e f o l l o w t h e s a m e$ $m e t h o d . . .$
Commented by math khazana by abdo last updated on 09/Jun/18

$\int_{π}^{2 π} \frac{x - c o s t}{x^{2} - 2 x c o s t + 1} d t =_{t = π + u} \int_{0}^{π} \frac{x + c o s u}{x^{2} + 2 x c o s u + 1} d u$ $. . .$
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