Question Number 136803 by mnjuly1970 last updated on 26/Mar/21
$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:…..\mathscr{A}{dvanced}\:\:\:\:\blacktriangleleft………….\blacktriangleright\:\:\:\mathscr{C}{alculus}….. \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{−\infty} ^{\:\infty} \frac{{sin}\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)}{{x}}{dx}=……??? \\ $$$$\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\overset{\frac{\mathrm{1}}{{x}}\:={t}} {=}\:\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}^{\mathrm{3}} \right)}{\frac{\mathrm{1}}{{t}}}.\frac{{dt}}{{t}^{\mathrm{2}} }\:…………. \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}^{\mathrm{3}} \right)}{{t}}{dt}\:……… \\ $$$$\:\:\:\:\:\:\:\:\:\:\overset{{t}^{\mathrm{3}} ={y}} {=}\:\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({y}\right)}{{y}^{\frac{\mathrm{1}}{\mathrm{3}}} }.\frac{{dy}}{{y}^{\frac{\mathrm{2}}{\mathrm{3}}} }=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({y}\right)}{{y}} \\ $$$$\:\:\:\:\overset{\langle\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({r}\right)}{{r}}{dr}\:=\frac{\pi}{\mathrm{2}}\rangle} {=}\:\frac{\pi}{\mathrm{3}}\:……….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……..\boldsymbol{\phi}=\int_{−\infty} ^{\:\infty} \frac{{sin}\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} }{{x}}{dx}=\frac{\pi}{\mathrm{3}}\:\:……..\checkmark\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by Ar Brandon last updated on 26/Mar/21
![φ=∫_(−∞) ^∞ ((sin((1/x^3 )))/x)dx=2∫_0 ^∞ x^(−1) sin(x^(−3) )dx =2∙(1/(∣−3∣))∙(π/(2Γ(1−((1−1)/(−3)))sin[(1−((1−1)/(−3)))(π/2)]))=(π/3)](https://www.tinkutara.com/question/Q136814.png)
$$\phi=\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)}{\mathrm{x}}\mathrm{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \mathrm{x}^{−\mathrm{1}} \mathrm{sin}\left(\mathrm{x}^{−\mathrm{3}} \right)\mathrm{dx} \\ $$$$\:\:\:=\mathrm{2}\centerdot\frac{\mathrm{1}}{\mid−\mathrm{3}\mid}\centerdot\frac{\pi}{\mathrm{2}\Gamma\left(\mathrm{1}−\frac{\mathrm{1}−\mathrm{1}}{−\mathrm{3}}\right)\mathrm{sin}\left[\left(\mathrm{1}−\frac{\mathrm{1}−\mathrm{1}}{−\mathrm{3}}\right)\frac{\pi}{\mathrm{2}}\right]}=\frac{\pi}{\mathrm{3}} \\ $$