0-4-x-log-1-tanx-dx-

Question Number 156923 by MathSh last updated on 17/Oct/21

Ω = \underset{0}{\int^{\frac{π}{4}}} x \log (1 + \tan x) dx = ?

Answered by mindispower last updated on 18/Oct/21

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

l n (c o s (x)) = - \sum_{k ⩾ 1} \frac{{(- 1)}^{k} c o s (2 k x)}{k} - l n (2) . . f o u r i e s e r e

a = - \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k} \int_{0}^{\frac{π}{4}} x c o s (2 k x) d x - l n (2) \int_{0}^{\frac{π}{4}} x d x

= - \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k} {{[\frac{x s i n (2 k x)}{2 k}]}_{0}^{\frac{π}{4}} + \frac{1}{4 k^{2}} {[c o s (2 k x)]}_{0}^{\frac{π}{4}} - \frac{π^{2}}{32} l n (2)

= - (\sum_{k ⩾ 1} \frac{{(- 1)}^{k} s i n (\frac{k π}{2})}{2 k^{2}} + \frac{{(- 1)}^{k} c o s (\frac{k π}{2})}{4 k^{3}})

s p i t k = 2 m, k = 2 m + 1 w e g e t w i t h e s i n (π m) = 0

s i n (π m + \frac{π}{2}) = {(- 1)}^{m}, c o s (m π) = {(- 1)}^{m}, c o s (m π + \frac{π}{2}) = {(- 1)}^{m}

= \sum_{m ⩾ 0} \frac{{(- 1)}^{m}}{2 {(2 m + 1)}^{2}} - \sum_{m ⩾ 1} \frac{{(- 1)}^{m}}{32 m^{3}}

= \frac{1}{2} β (- 1) - \frac{1}{32} {L i}_{3} (- 1) = \frac{G}{2} + \frac{3}{4.32} ζ (3) - \frac{π^{2} l n (2)}{32}

\int_{0}^{\frac{π}{4}} l n (c o s (x)) d x, \int_{0}^{\frac{π}{4}} l n (t g (x)) = G

\int_{0}^{\frac{π}{2}} l n (s i n x) d x = - \frac{π}{2} l n (2)

c o s (x) = \frac{1}{2} l n (\frac{s i n (x) c o s (x)}{s i n (x)} c o s (x)), x \in [0, \frac{π}{2}]

= \frac{1}{2} l n (s i n (2 x)) - \frac{l n (t g (x))}{2}

\Rightarrow \int_{0}^{\frac{π}{4}} l n (c o s (x)) = \frac{π}{8} l n (2) - \frac{G}{2}

Ω = \int_{0}^{\frac{π}{4}} x l n (\frac{\sqrt{2} s i n (x + \frac{π}{4})}{c o s (x)}) d x

= \frac{l n (2)}{2} \int_{0}^{\frac{π}{4}} x d x + \int_{0}^{\frac{π}{4}} x l n (s i n (x + \frac{π}{4})) d x - \int_{0}^{\frac{π}{4}} x l n (c o s (x)) d x

= \frac{l n (2)}{32} π^{2} + \int_{0}^{\frac{π}{4}} \frac{π}{4} l n (c o s (x)) d x - 2 \int_{0}^{\frac{π}{4}} x l n (c o s (x)) d x

= \frac{l n (2)}{32} π^{2} + \frac{π}{4} (\frac{π l n 2}{8} - \frac{G}{2}) - 2 (\frac{G}{2} + \frac{3 ζ (1)}{128} - \frac{π^{2} l n (2)}{32})

= - (\frac{π}{8} + 1) G + \frac{π^{2} l n (2)}{8} - \frac{3}{64} ζ (3)

Commented by MathSh last updated on 18/Oct/21

Perfect dear Ser, thank you so much

Commented by mindispower last updated on 19/Oct/21

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