∫_0 ^(π/2) xln sin x dx=?


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Question Number 208692 by Ghisom last updated on 21/Jun/24

$\underset{0}{\int^{π / 2}} x \ln \sin x d x = ?$
	Answered by Berbere last updated on 21/Jun/24

	$l n (s i n (x)) = - l n (2) - \sum_{k ⩾ 1} \frac{c o s (2 k x)}{k}$ $p r o o f l n (s i n (x)) = R e (l n (s i n (x))$ $l n (s i n (x) = l n (\frac{e^{i x} - e^{- i x}}{2 i})$ $l n (s i n (x)) = l n (e^{i x} (1 - e^{- 2 i x})) - l n (2) == i x - l n (2) - Σ \frac{e^{- 2 k i x}}{k}$ $R e l n (s i n (x)) = - l n (2) - Σ \frac{c o s (2 k x)}{k}$ $\int_{0}^{\frac{π}{2}} x l n (s i n (x)) d x = \int_{0}^{\frac{π}{2}} x (- l n (2) - Σ \frac{c o s (2 k x)}{k})$ $= {[- \frac{x^{2}}{2} l n (2)]}_{0}^{\frac{π}{2}} - \sum_{k ⩾ 1} \frac{1}{k} \int_{0}^{\frac{π}{2}} x c o s (2 k x) d \bar{x}$ $\int x c o s (2 k x) d x = x . \frac{s i n (2 k x)}{2 k} + \frac{c o s (2 k x)}{4 k^{2}}$ $. - \frac{π^{2}}{8} l n (2) - \sum_{k} \frac{1}{k} {[\frac{s i n (2 k x)}{2 k} + \frac{c o s (2 k x)}{4 k^{2}}]}_{0}^{\frac{π}{2}}$ $= - \frac{π^{2}}{8} l n (2) - Σ \frac{{(- 1)}^{k} - 1}{4 k^{3}}$ $= - \frac{π^{3}}{8} l n (2) + \sum_{k ⩾ 0} \frac{1}{2 {(2 k + 1)}^{3}} = - \frac{π^{3}}{8} l n (2) + \frac{7}{16} ζ (3)$
	Commented by Ghisom last updated on 22/Jun/24

	$thank you$
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