Ω= ∫_0 ^( ∞) (( sin^( 3) (x)ln(x))/x) dx=^(??) (π/8) (ln(3)−2γ) −−−−− solution.. Ω=∫_0^ ^( ∞) {(((3/4) sin(x)−(1/4) sin(3x))/x)} ln(x)dx = (3/4) (((−πγ)/2))− (1/4){ ∫_0 ^( ∞) ((sin(3x)ln(x))/x)dx=Ψ} ∴ Ψ := ∫_0 ^( ∞) (( sin(x).[ln(x)−ln(3)])/x)dx := −((πγ)/2) − ((ln(3).π)/2) ∴ Ω := ((−3πγ)/8) +((πγ)/8) +((π.ln(3))/8) := ((−2πγ)/8) + ((π.ln(3))/8) = (π/8) ( ln(3)−2γ )


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 159526 by mnjuly1970 last updated on 18/Nov/21
$Ω= ∫_0 ^( ∞) (( sin^( 3) (x)ln(x))/x) dx=^(??) (π/8) (ln(3)−2γ) −−−−− solution.. Ω=∫_0^ ^( ∞) {(((3/4) sin(x)−(1/4) sin(3x))/x)} ln(x)dx = (3/4) (((−πγ)/2))− (1/4){ ∫_0 ^( ∞) ((sin(3x)ln(x))/x)dx=Ψ} ∴ Ψ := ∫_0 ^( ∞) (( sin(x).[ln(x)−ln(3)])/x)dx := −((πγ)/2) − ((ln(3).π)/2) ∴ Ω := ((−3πγ)/8) +((πγ)/8) +((π.ln(3))/8) := ((−2πγ)/8) + ((π.ln(3))/8) = (π/8) ( ln(3)−2γ )$
$Ω = \int_{0}^{\infty} \frac{{s i n}^{3} (x) l n (x)}{x} d x \overset{? ?}{=} \frac{π}{8} (l n (3) - 2 γ)$ $- - - - -$ $s o l u t i o n . .$ $Ω = \int_{0^{}}^{\infty} {\frac{\frac{3}{4} s i n (x) - \frac{1}{4} s i n (3 x)}{x}} l n (x) d x$ $= \frac{3}{4} (\frac{- π γ}{2}) - \frac{1}{4} {\int_{0}^{\infty} \frac{s i n (3 x) l n (x)}{x} d x = Ψ}$ $∴ Ψ := \int_{0}^{\infty} \frac{s i n (x) . [l n (x) - l n (3)]}{x} d x$ $:= - \frac{π γ}{2} - \frac{l n (3) . π}{2}$ $∴ Ω := \frac{- 3 π γ}{8} + \frac{π γ}{8} + \frac{π . l n (3)}{8}$ $:= \frac{- 2 π γ}{8} + \frac{π . l n (3)}{8} = \frac{π}{8} (l n (3) - 2 γ)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum