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

Question Number 32343 by abdo imad last updated on 23/Mar/18

c a l c u l a t e \int_{0}^{2 π} \frac{d t}{x - e^{i t}} .

Commented by abdo imad last updated on 28/Mar/18

l e t p u t I (x) = \int_{0}^{2 π} \frac{d t}{x - e^{i t}} l e t s u p p o s e x \neq 0

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

c a s e 1 ∣ x^{- 1} e^{i t} ∣ < 1 \Leftrightarrow ∣ x ∣> 1 \Rightarrow

x I (x) = \int_{0}^{2 π} (\sum_{n = 0}^{\infty} x^{- n} e^{i n t}) d t

= \sum_{n = 0}^{\infty} x^{- n} \int_{0}^{2 π} e^{i n t} d t = 2 π + \sum_{n = 1}^{\infty} x^{- n} {[\frac{1}{i n} e^{i n t}]}_{0}^{2 π} = 2 π + 0

= 2 π \Rightarrow I (x) = \frac{2 π}{x} .

c a s e 2 ∣ x^{- 1} e^{i t} ∣> 1 \Leftrightarrow ∣ x ∣< 1 t h e c h . x = \frac{1}{α} \Rightarrow∣ α ∣> 1

I (x) = I (\frac{1}{α}) = \int_{0}^{2 π} \frac{d t}{\frac{1}{α} - e^{i t}} = α \int_{0}^{2 π} \frac{d t}{1 - α e^{i t}}

I (x) = \int_{0}^{2 π} \frac{d t}{x - c o s t - i s i n t} d 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

=_{t = π + u} \int_{- π}^{π} \frac{x + c o s u}{{(x + c o s u)}^{2} + {s i n}^{2} u} d u - i \int_{- π}^{π} \frac{s i n u}{{(x + c o s u)}^{2} + {s i n}^{2} u} d u

= 2 \int_{0}^{π} \frac{x + c o s u}{x^{2} + 2 x c o s u + 1} d u + 0 . c h t a n (\frac{u}{2}) = t g i v e

I (x) = 2 \int_{0}^{+ \infty} \frac{x + \frac{1 - t^{2}}{1 + t^{2}}}{x^{2} + 2 x \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

= 4 \int_{0}^{+ \infty} \frac{x (1 + t^{2}) + 1 - t^{2}}{x^{2} (1 + t^{2}) + 2 x (1 - t^{2})} d t

= 4 \int_{0}^{+ \infty} \frac{x + 1 + (x - 1) t^{2}}{(x^{2} + 2 x + (x^{2} - 2 x) t^{2}) (1 + t^{2})} d t

= \frac{4}{x} \int_{0}^{+ \infty} \frac{(x - 1) t^{2} + x + 1}{(x + 2 + (x - 2) t^{2}) (1 + t^{2})} d t \dots . b e c o n t i n u e d \dots .

Answered by sma3l2996 last updated on 25/Mar/18

l e t z = e^{i t} \Rightarrow d t = \frac{d z}{i z}

\int_{0}^{2 π} \frac{d t}{x - e^{i t}} = \frac{1}{i} \int_{∣ z ∣⩽ 1} \frac{d z}{z (x - z)}

i f x > 1 :

\int_{0}^{2 π} \frac{d t}{x - e^{i t}} = 2 π R e s (f; 0) = \frac{2 π}{x}

i f x ⩽ 1 :

\int_{0}^{2 π} \frac{d t}{x - e^{i t}} = 2 π (R e s (f; 0) + R e s (f; x)) = 2 π (\frac{1}{x} - \frac{1}{x}) = 0