Question Number 110112 by mathdave last updated on 27/Aug/20
$${show}\:{that}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}\mathrm{sin}\left({x}^{\mathrm{3}} \right){dx}=\frac{\mathrm{1}}{\mathrm{3}}\bullet\frac{\pi}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$
Answered by mnjuly1970 last updated on 27/Aug/20
$${x}^{\mathrm{3}} ={t}\Rightarrow\left\{_{{dx}=\frac{\mathrm{1}}{\mathrm{3}}{t}^{\frac{−\mathrm{2}}{\mathrm{3}}} {dt}} ^{{x}={t}^{\frac{\mathrm{1}}{\mathrm{3}}} } \right. \\ $$$${we}\:{know}::\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}\right)}{{x}^{{p}} }{dx}\:\overset{\mathrm{0}<{p}\leqslant\mathrm{1}} {=}\frac{\pi}{\mathrm{2}\Gamma\left({p}\right){sin}\left(\frac{{p}\pi}{\mathrm{2}}\right)}\:\:\:\:\:\:\: \\ $$$$\Omega=\int_{\mathrm{0}} ^{\:\infty} {xsin}\left({x}^{\mathrm{3}} \right){dx}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right)}{{t}^{\frac{\mathrm{1}}{\mathrm{3}}} }{dt} \\ $$$$\:\Omega=\frac{\mathrm{1}}{\mathrm{3}}\ast\:\frac{\pi}{\mathrm{2}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right){sin}\left(\frac{\pi}{\mathrm{6}}\right)}\:=\frac{\pi}{\mathrm{3}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}\:\clubsuit \\ $$$$\mathscr{M}.\mathscr{N}.\mathscr{J}{uly}\:\mathrm{1970} \\ $$$$\:\:\: \\ $$
Commented by mathdave last updated on 27/Aug/20
$${u}\:{did}\:{well}\:{man} \\ $$
Commented by mnjuly1970 last updated on 27/Aug/20
$${grateful}…{sir}…. \\ $$