let g(θ) =∫_0 ^1 ln( 1−e^(iθ) x^2 )dx find a simple form of g(θ) .θ from R.


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Question Number 36737 by abdo mathsup 649 cc last updated on 04/Jun/18

$l e t g (θ) = \int_{0}^{1} l n (1 - e^{i θ} x^{2}) d x$ $f i n d a s i m p l e f o r m o f g (θ) . θ f r o m R .$
Commented by math khazana by abdo last updated on 05/Aug/18
$we have proved that for ∣z∣=1 ∫_0 ^1 ln(1−zx)dx =(1−(1/z))ln(1−z)−1 but g(θ) =∫_0 ^1 ln(1−(e^(i(θ/2)) x)^2 )dx =∫_0 ^1 ln(1−e^((iθ)/2) x)dx +∫_0 ^1 ln(1 +e^((iθ)/2) x)dx but ∫_0 ^1 ln(1−e^((iθ)/2) x)dx=(1−e^(−((iθ)/2)) )ln(1−e^((iθ)/2) )−1 let find f(z) =∫_0 ^1 ln(1+zx)dx we have for ∣u∣<1 ln^′ (1+u) =(1/(1+u))=Σ_(n=0) ^∞ (−1)^n u^n ⇒ ln(1+u) =Σ_(n=0) ^∞ (((−1)^n )/(n+1))u^(n+1) =Σ_(n=1) ^∞ (−1)^(n−1) (u^n /n) ⇒ln(1+zx) =Σ_(n=1) ^∞ (−1)^(n−1) ((z^n x^n )/n) ⇒ f(z) =Σ_(n=1) ^∞ (−1)^(n−1) (z^n /n) ∫_0 ^1 x^n dx =Σ_(n=1) ^∞ (−1)^(n−1) (z^n /(n(n+1))) =Σ_(n=1) ^∞ (−1)^(n−1) ((1/n)−(1/(n+1)))z^n =Σ_(n=1) ^∞ (−1)^(n−1) (z^n /n) −Σ_(n=1) ^∞ (−1)^(n−1) (z^n /(n+1)) =ln(1+z) −Σ_(n=2) ^∞ (−1)^(n−2) (z^(n−1) /n) =ln(1+z) +(1/z)Σ_(n=2) ^∞ (−1)^(n−1) (z^n /n) =ln(1+z) +(1/z){Σ_(n=1) ^∞ (−1)^(n−1) (z^n /n) −z} =ln(1+z) +(1/z)ln(1+z)−1 =(1+(1/z))ln(1+z)−1⇒ ∫_0 ^1 ln(1+e^((iθ)/2) x)dx =(1+e^(−((iθ)/2)) )ln(1+e^((iθ)/2) )−1⇒ g(θ) =(1−e^((iπ)/2) )ln(1−e^((iθ)/2) ) +(1+e^(−((iθ)/2)) )ln(1+e^(−((iθ)/2)) )−2$
$w e h a v e p r o v e d t h a t f o r ∣ z ∣= 1$ $\int_{0}^{1} l n (1 - z x) d x = (1 - \frac{1}{z}) l n (1 - z) - 1 b u t$ $g (θ) = \int_{0}^{1} l n (1 - {(e^{i \frac{θ}{2}} x)}^{2}) d x$ $= \int_{0}^{1} l n (1 - e^{\frac{i θ}{2}} x) d x + \int_{0}^{1} l n (1 + e^{\frac{i θ}{2}} x) d x b u t$ $\int_{0}^{1} l n (1 - e^{\frac{i θ}{2}} x) d x = (1 - e^{- \frac{i θ}{2}}) l n (1 - e^{\frac{i θ}{2}}) - 1 l e t$ $f i n d f (z) = \int_{0}^{1} l n (1 + z x) d x w e h a v e f o r ∣ u ∣< 1$ ${l n}^{^{'}} (1 + u) = \frac{1}{1 + u} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n} \Rightarrow$ $l n (1 + u) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} u^{n + 1} = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{u^{n}}{n}$ $\Rightarrow l n (1 + z x) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{z^{n} x^{n}}{n} \Rightarrow$ $f (z) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{z^{n}}{n} \int_{0}^{1} x^{n} d x$ $= \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{z^{n}}{n (n + 1)}$ $= \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} (\frac{1}{n} - \frac{1}{n + 1}) z^{n}$ $= \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{z^{n}}{n} - \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{z^{n}}{n + 1}$ $= l n (1 + z) - \sum_{n = 2}^{\infty} {(- 1)}^{n - 2} \frac{z^{n - 1}}{n}$ $= l n (1 + z) + \frac{1}{z} \sum_{n = 2}^{\infty} {(- 1)}^{n - 1} \frac{z^{n}}{n}$ $= l n (1 + z) + \frac{1}{z} {\sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{z^{n}}{n} - z}$ $= l n (1 + z) + \frac{1}{z} l n (1 + z) - 1$ $= (1 + \frac{1}{z}) l n (1 + z) - 1 \Rightarrow$ $\int_{0}^{1} l n (1 + e^{\frac{i θ}{2}} x) d x = (1 + e^{- \frac{i θ}{2}}) l n (1 + e^{\frac{i θ}{2}}) - 1 \Rightarrow$ $g (θ) = (1 - e^{\frac{i π}{2}}) l n (1 - e^{\frac{i θ}{2}}) + (1 + e^{- \frac{i θ}{2}}) l n (1 + e^{- \frac{i θ}{2}}) - 2$
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