prove that # ∫_0 ^( ∞) (( sin^( 3) ( x ).ln( x ))/x) dx =^? (π/8) (−2γ +ln(3)) .....■ m.n


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Question Number 154478 by mnjuly1970 last updated on 18/Sep/21

$You can't use 'macro parameter character #' in math mode$ $\int_{0}^{\infty} \frac{{s i n}^{3} (x) . l n (x)}{x} d x \overset{?}{=} \frac{π}{8} (- 2 γ + l n (3)) . . . . . ◼ m . n$
	Answered by ARUNG_Brandon_MBU last updated on 18/Sep/21

	$\int_{0}^{\infty} \frac{\sin^{3} x \cdot \ln x}{x} d x$ $2 i \sin x = (e^{i x} - e^{- i x})$ $- 8 i \sin^{3} x = e^{3 i x} - 3 e^{i x} + 3 e^{- i x} - e^{- 3 i x}$ $\sin^{3} x = \frac{3}{4} \sin x - \frac{1}{4} \sin 3 x$ $Ω (α) = - \int_{0}^{\infty} \frac{\sin^{3} x}{x^{α}} = \frac{1}{4} \int_{0}^{\infty} (\frac{\sin 3 x}{x^{α}} - \frac{3 \sin x}{x^{α}}) d x = \frac{1}{4} [\frac{π 3^{α - 1} - 3 π}{2 Γ (α) \sin (\frac{π}{2} α)}]$ $Ω^{'} (α) = \frac{1}{8} [\frac{π 3^{α - 1} \ln 3 (Γ (α) \sin (\frac{π}{2} α)) - (π 3^{α - 1} - 3 π) (\frac{π}{2} \cos (\frac{π}{2} α) Γ (α) + \sin (\frac{π}{2} α) Γ^{'} (α))}{Γ^{2} (α) \sin^{2} (\frac{π}{2} α)}]$ $Ω^{'} (1) = \int_{0}^{\infty} \frac{\sin^{3} x \cdot \ln x}{x} d x = \frac{1}{8} [\frac{π \ln 3 + 2 π (Γ^{'} (1))}{Γ^{2} (1) \sin^{2} (\frac{π}{2})}] = \frac{1}{8} (π \ln 3 - 2 π γ) = \frac{π}{8} (\ln 3 - 2 γ)$
	Commented by mnjuly1970 last updated on 18/Sep/21

	$t h a n k y o u s o m u c h s i r A r u n g . .$
	Commented by Ar Brandon last updated on 18/Sep/21
	My pleasure, Sir ��
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