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Question Number 140194 by mnjuly1970 last updated on 05/May/21

p l e a s e i n t e g r a t e ::

f (x) = \int_{0}^{1} {\frac{1}{z} l o g (\frac{z^{2} + 2 z c o s (x) + 1}{{(z + 1)}^{2}})} d z

Answered by Dwaipayan Shikari last updated on 05/May/21

\int_{0}^{1} \frac{l o g (z^{2} + 2 z c o s x + 1)}{z} - 2 \frac{l o g (z + 1)}{z} d z

= \int_{0}^{1} \frac{l o g (1 + {z e}^{i x}) + l o g (1 + {z e}^{- i x})}{z} - 2 Σ \frac{{(- 1)}^{n + 1}}{n^{2}}

= Σ \int_{0}^{1} {(- 1)}^{n + 1} \frac{1}{n} (\frac{z^{n} e^{i n x}}{z} + \frac{z^{n} e^{- i n x}}{z}) d z - \frac{π^{2}}{6}

= 2 Σ {(- 1)}^{n + 1} \frac{c o s (n x)}{n^{2}} - \frac{π^{2}}{6} = 2 ξ - \frac{π^{2}}{6} = - \frac{x^{2}}{2}

\int_{0}^{x} Σ {(- 1)}^{n + 1} \frac{s i n (n x)}{n^{2}} d x = \frac{π^{2}}{12} - \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1} c o s (n x)}{n^{2}}

\Rightarrow \int_{0}^{x} \frac{x}{2} d x = \frac{π^{2}}{12} - ξ \Rightarrow ξ = \frac{π^{2}}{12} - \frac{x^{2}}{4}

Commented by mnjuly1970 last updated on 05/May/21

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