let-g-0-1-ln-1-e-i-x-2-dx-find-a-simple-form-of-g-from-R-

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

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