K-0-4-pi-ln-cosx-dx-

Question Number 208245 by Shrodinger last updated on 08/Jun/24

K = \int_{0}^{\frac{4}{π}} l n (c o s x) d x

Answered by mathzup last updated on 09/Jun/24

K = \int_{0}^{\frac{4}{π}} l n (\frac{e^{i x} + e^{- i x}}{2}) d x l e t λ = \frac{4}{π}

= \int_{0}^{λ} (

l n (e^{i x}) + l n (1 + e^{- 2 i x}) - l n 2) d x

= \int_{0}^{λ} i x d x + \int_{0}^{λ} l n (1 + e^{- 2 i x}) d x - λ l n 2

= i \frac{λ^{2}}{2} - λ l n 2 + \int_{0}^{λ} l n (1 + e^{- 2 i x}) d x

(l n {(1 + z)}^{^{'}} = \frac{1}{1 + z} = \sum_{n = 0}^{\infty} {(- 1)}^{n} z^{n} \Rightarrow

l n (1 + z) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} z^{n + 1} + c (c = 0)

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} z^{n} \Rightarrow

\int_{0}^{λ} l n (1 + e^{- 2 i x}) d x = \int_{0}^{λ} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} e^{- 2 i n x} d x

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

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

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

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

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

+ \frac{i}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} (c o s (2 λ) - i s i n (2 λ))

o r I i s r e a l s o

I = - λ l n 2 + \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} s i n (2 λ)

= - \frac{4}{π} l n 2 + \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} s i n (\frac{8}{π})

r e s t t o f i n d t h e v a l u e o f t h i s s e r i e

\dots b e c o n t i n u e d \dots

Commented by Shrodinger last updated on 10/Jun/24

t h a n k s s i r