calculate-0-2pi-dt-x-e-it-

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

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 \dots

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

\dots