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 )](Q159526.png)
$$ \\ $$$$\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{sin}^{\:\mathrm{3}} \left({x}\right){ln}\left({x}\right)}{{x}}\:{dx}\overset{??} {=}\frac{\pi}{\mathrm{8}}\:\left({ln}\left(\mathrm{3}\right)−\mathrm{2}\gamma\right) \\ $$$$−−−−− \\ $$$$\:\:\:\:\:\:{solution}.. \\ $$$$\:\:\:\:\Omega=\int_{\mathrm{0}^{\:} } ^{\:\infty} \left\{\frac{\frac{\mathrm{3}}{\mathrm{4}}\:{sin}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{4}}\:{sin}\left(\mathrm{3}{x}\right)}{{x}}\right\}\:{ln}\left({x}\right){dx} \\ $$$$\:\:=\:\:\frac{\mathrm{3}}{\mathrm{4}}\:\left(\frac{−\pi\gamma}{\mathrm{2}}\right)−\:\frac{\mathrm{1}}{\mathrm{4}}\left\{\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\mathrm{3}{x}\right){ln}\left({x}\right)}{{x}}{dx}=\Psi\right\} \\ $$$$\:\:\therefore\:\:\Psi\::=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{sin}\left({x}\right).\left[{ln}\left({x}\right)−{ln}\left(\mathrm{3}\right)\right]}{{x}}{dx} \\ $$$$\:\:\:\:\:\::=\:−\frac{\pi\gamma}{\mathrm{2}}\:\:−\:\frac{{ln}\left(\mathrm{3}\right).\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\:\:\Omega\::=\:\:\frac{−\mathrm{3}\pi\gamma}{\mathrm{8}}\:+\frac{\pi\gamma}{\mathrm{8}}\:\:+\frac{\pi.{ln}\left(\mathrm{3}\right)}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\::=\:\frac{−\mathrm{2}\pi\gamma}{\mathrm{8}}\:+\:\frac{\pi.{ln}\left(\mathrm{3}\right)}{\mathrm{8}}\:=\:\frac{\pi}{\mathrm{8}}\:\left(\:{ln}\left(\mathrm{3}\right)−\mathrm{2}\gamma\:\right) \\ $$$$ \\ $$