let-f-x-0-pi-2-d-1-x-e-i-with-x-lt-1-1-developp-f-x-at-integr-serie-2-calculate-f-x-3-find-the-value-of-0-pi-2-e-i-1-x-e-i-2-4-calculate-0-pi-2

Question Number 38706 by abdo mathsup 649 cc last updated on 28/Jun/18

l e t f (x) = \int_{0}^{\frac{π}{2}} \frac{d θ}{1 + x e^{i θ}} w i t h ∣ x ∣< 1

1) d e v e l o p p f (x) a t i n t e g r s e r i e

2) c a l c u l a t e f (x)

3) f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} \frac{e^{i θ}}{{(1 + x e^{i θ})}^{2}}

4) c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d θ}{2 + e^{i θ}}

Commented by maxmathsup by imad last updated on 30/Jun/18

2) w e h a v e f (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{i n} e^{\frac{i n π}{2}} x^{n} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{i n} x^{n}

1) f (x) = \int_{0}^{\frac{π}{2}} (\sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} e^{i n θ}) d θ = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} \int_{0}^{\frac{π}{2}} e^{i n θ} d θ

= \frac{π}{2} + \sum_{n = 1}^{\infty} {(- 1)}^{n} x^{n} {[\frac{1}{i n} e^{i n θ}]}_{0}^{\frac{π}{2}}

= \frac{π}{2} - i \sum_{n = 1}^{\infty} {(- 1)}^{n} x^{n} (e^{\frac{i n π}{2}} - 1)

f (x) = \frac{π}{2} - i \sum_{n = 1}^{\infty} {(- 1)}^{n} (e^{\frac{i n π}{2}} - 1) x^{n}

Commented by maxmathsup by imad last updated on 30/Jun/18

2) w e h a v e f (x) = \frac{π}{2} - i \sum_{n = 1}^{\infty} {(- 1)}^{n} e^{\frac{i n π}{2}} x^{n} + i \sum_{n = 1}^{\infty} {(- 1)}^{n} x^{n} b u t

\sum_{n = 1}^{\infty} {(- 1)}^{n} x^{n} - 1 = \frac{1}{1 + x} - 1 = \frac{- x}{1 + x}

\sum_{n = 1}^{\infty} {(- 1)}^{n} e^{\frac{i n π}{2}} x^{n} = \sum_{n = 0}^{\infty} {(- e^{\frac{i π}{2}} x)}^{n} - 1

= \frac{1}{1 + e^{\frac{i π}{2}} x} - 1 = \frac{1}{1 + i x} - 1 = \frac{1 - i x}{1 + x^{2}} - 1 = \frac{- x^{2}}{1 + x^{2}} - i \frac{x}{1 + x^{2}} \Rightarrow

f (x) = \frac{π}{2} - i (\frac{- x^{2}}{1 + x^{2}} - \frac{i x}{1 + x^{2}}) - \frac{i x}{1 + x}

= \frac{π}{2} - \frac{x}{1 + x^{2}} + i {\frac{x^{2}}{1 + x^{2}} - \frac{x}{1 + x}}

Commented by maxmathsup by imad last updated on 30/Jun/18

3) w e h a v e f (x) = \int_{0}^{\frac{π}{2}} \frac{d θ}{1 + x c o s θ} \Rightarrow f^{^{'}} (x) = \int_{0}^{\frac{π}{2}} \frac{- s i n θ}{{(1 + c o s θ)}^{2}} d θ \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{s i n θ}{{(1 + c o s θ)}^{2}} d θ = - f^{^{'}} (x) .

Commented by maxmathsup by imad last updated on 30/Jun/18

4) l e t I = \int_{0}^{\frac{π}{2}} \frac{d θ}{2 + e^{i θ}} = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{d θ}{1 + \frac{1}{2} e^{i θ}} = \frac{1}{2} f (\frac{1}{2})

\Rightarrow 2 I = \frac{π}{2} - \frac{1}{2 (1 + \frac{1}{4})} + i {\frac{1}{4 (1 + \frac{1}{4})} - \frac{1}{2 (1 + \frac{1}{2})}}

= \frac{π}{2} \frac{2}{5} + i {\frac{1}{5} - \frac{1}{3}} = \frac{π}{5} - \frac{2}{15} i \Rightarrow I = \frac{π}{10} - \frac{i}{15}