0-pi-4-x-ln-sin-x-dx-

Question Number 159854 by mnjuly1970 last updated on 21/Nov/21

Ω := \int_{0}^{\frac{π}{4}} x . l n (s i n (x)) d x = ?

Answered by mindispower last updated on 22/Nov/21

l n (s i n (x)) = - l n (2) - \sum_{n ⩾ 1} \frac{c o s (2 n x)}{n}

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

= - \frac{l n (2)}{32} π^{2} - \sum_{n ⩾ 1} \frac{1}{n} \int_{0}^{\frac{π}{4}} c o s (2 n x) x d x

= - \frac{l n (2)}{32} π^{2} - \sum_{n ⩾ 1} \frac{1}{n} [\frac{s i n (n \frac{π}{2})}{2 n} . \frac{π}{4} - \frac{1}{2 n} \int_{0}^{\frac{π}{4}} s i n (2 n x) d x

= - \frac{l n (2) π^{2}}{32} + \sum_{n ⩾ 1} (- \frac{π}{8} . \frac{s i n (\frac{n π}{2})}{n^{2}} - \frac{1}{4 n^{3}} [c o s (\frac{n π}{2}) - 1])

= - \frac{l n (2) π^{2}}{32} - \frac{π}{8} . \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} - \frac{1}{4} \sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{8 n^{3}} + \frac{ζ (3)}{4}

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

= - \frac{l n (2)}{32} π^{2} - \frac{π}{8} G + \frac{35}{128} ζ (3)

Commented by mnjuly1970 last updated on 23/Nov/21

t h a n k s a l o t s i r Power \dots

Commented by mindispower last updated on 23/Nov/21

w i t h e p l e a s u r s i r h a v e a n i c e d a y