let f(x)= ∫_0 ^(π/2) (dθ/(1+x e^(iθ) )) with ∣x∣<1 1) developp f(x) at integr serie 2) calculate f(x) 3) find the value of ∫_0 ^(π/2) (e^(iθ) /((1+x e^(iθ) )^2 )) 4) calculate ∫


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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 bymaxmathsup 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 bymaxmathsup by imad last updated on 30/Jun/18
$2) we have f(x) =(π/2) −i Σ_(n=1) ^∞ (−1)^n e^((inπ)/2) x^n +i Σ_(n=1) ^∞ (−1)^n x^n but Σ_(n=1) ^∞ (−1)^n x^n −1= (1/(1+x)) −1 = ((−x)/(1+x)) Σ_(n=1) ^∞ (−1)^n e^((inπ)/2) x^n = Σ_(n=0) ^∞ (−e^((iπ)/2) x)^n −1 = (1/(1+e^((iπ)/2) x)) −1 = (1/(1+ix)) −1 = ((1−ix)/(1+x^2 )) −1 = ((−x^2 )/(1+x^2 )) −i(x/(1+x^2 )) ⇒ f(x) = (π/2) −i( ((−x^2 )/(1+x^2 )) −((ix)/(1+x^2 ))) −((ix)/(1+x)) =(π/2) −(x/(1+x^2 )) +i{ (x^2 /(1+x^2 )) −(x/(1+x))}$
$2) w e h a v e f (x) = \frac{π}{2} - i \sum_{n = 1}^{\infty} {(- 1)}^{n} e^{\frac{i n π}{2}} x^{n} + i \sum_{n = 1}^{\infty} {(- 1)}^{n} x^{n} b u t$ $\sum_{n = 1}^{\infty} {(- 1)}^{n} x^{n} - 1 = \frac{1}{1 + x} - 1 = \frac{- x}{1 + x}$ $\sum_{n = 1}^{\infty} {(- 1)}^{n} e^{\frac{i n π}{2}} x^{n} = \sum_{n = 0}^{\infty} {(- e^{\frac{i π}{2}} x)}^{n} - 1$ $= \frac{1}{1 + e^{\frac{i π}{2}} x} - 1 = \frac{1}{1 + i x} - 1 = \frac{1 - i x}{1 + x^{2}} - 1 = \frac{- x^{2}}{1 + x^{2}} - i \frac{x}{1 + x^{2}} \Rightarrow$ $f (x) = \frac{π}{2} - i (\frac{- x^{2}}{1 + x^{2}} - \frac{i x}{1 + x^{2}}) - \frac{i x}{1 + x}$ $= \frac{π}{2} - \frac{x}{1 + x^{2}} + i {\frac{x^{2}}{1 + x^{2}} - \frac{x}{1 + x}}$
Commented bymaxmathsup 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 bymaxmathsup by imad last updated on 30/Jun/18
$4) let I =∫_0 ^(π/2) (dθ/(2+e^(iθ) )) = (1/2) ∫_0 ^(π/2) (dθ/(1+(1/2)e^(iθ) )) =(1/2)f((1/2)) ⇒2I =(π/2) −(1/(2(1+(1/4)))) +i{ (1/(4(1+(1/4)))) −(1/(2(1+(1/2))))} =(π/2) (2/5) +i{(1/5) −(1/3)} = (π/5) −(2/(15)) i ⇒ I =(π/(10)) −(i/(15))$
$4) l e t I = \int_{0}^{\frac{π}{2}} \frac{d θ}{2 + e^{i θ}} = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{d θ}{1 + \frac{1}{2} e^{i θ}} = \frac{1}{2} f (\frac{1}{2})$ $\Rightarrow 2 I = \frac{π}{2} - \frac{1}{2 (1 + \frac{1}{4})} + i {\frac{1}{4 (1 + \frac{1}{4})} - \frac{1}{2 (1 + \frac{1}{2})}}$ $= \frac{π}{2} \frac{2}{5} + i {\frac{1}{5} - \frac{1}{3}} = \frac{π}{5} - \frac{2}{15} i \Rightarrow I = \frac{π}{10} - \frac{i}{15}$
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