Advanced-Calculus-sin-1-x-3-x-dx-solution-1-x-t-2-0-sin-t-3-1-t-dt-t

Question Number 136803 by mnjuly1970 last updated on 26/Mar/21

.....Advanced ◂.............▶ Calculus..... 𝛗=∫_(−∞) ^( ∞) ((sin((1/x^3 )))/x)dx=......??? solution:: 𝛗=^((1/x) =t) 2∫_0 ^( ∞) ((sin(t^3 ))/(1/t)).(dt/t^2 ) ............. =2∫_0 ^( ∞) ((sin(t^3 ))/t)dt ......... =^(t^3 =y) (2/3)∫_0 ^( ∞) ((sin(y))/y^(1/3) ).(dy/y^(2/3) )=(2/3)∫_0 ^( ∞) ((sin(y))/y) =^(⟨∫_0 ^( ∞) ((sin(r))/r)dr =(π/2)⟩) (π/3) ........... ........𝛗=∫_(−∞) ^( ∞) ((sin((1/x))^3 )/x)dx=(π/3) ........✓✓✓

$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:…..\mathscr{A}{dvanced}\:\:\:\:\blacktriangleleft………….\blacktriangleright\:\:\:\mathscr{C}{alculus}….. \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{−\infty} ^{\:\infty} \frac{{sin}\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)}{{x}}{dx}=……??? \\ $$$$\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\overset{\frac{\mathrm{1}}{{x}}\:={t}} {=}\:\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}^{\mathrm{3}} \right)}{\frac{\mathrm{1}}{{t}}}.\frac{{dt}}{{t}^{\mathrm{2}} }\:…………. \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}^{\mathrm{3}} \right)}{{t}}{dt}\:……… \\ $$$$\:\:\:\:\:\:\:\:\:\:\overset{{t}^{\mathrm{3}} ={y}} {=}\:\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({y}\right)}{{y}^{\frac{\mathrm{1}}{\mathrm{3}}} }.\frac{{dy}}{{y}^{\frac{\mathrm{2}}{\mathrm{3}}} }=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({y}\right)}{{y}} \\ $$$$\:\:\:\:\overset{\langle\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({r}\right)}{{r}}{dr}\:=\frac{\pi}{\mathrm{2}}\rangle} {=}\:\frac{\pi}{\mathrm{3}}\:……….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……..\boldsymbol{\phi}=\int_{−\infty} ^{\:\infty} \frac{{sin}\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} }{{x}}{dx}=\frac{\pi}{\mathrm{3}}\:\:……..\checkmark\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

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)

$$\phi=\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)}{\mathrm{x}}\mathrm{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \mathrm{x}^{−\mathrm{1}} \mathrm{sin}\left(\mathrm{x}^{−\mathrm{3}} \right)\mathrm{dx} \\ $$$$\:\:\:=\mathrm{2}\centerdot\frac{\mathrm{1}}{\mid−\mathrm{3}\mid}\centerdot\frac{\pi}{\mathrm{2}\Gamma\left(\mathrm{1}−\frac{\mathrm{1}−\mathrm{1}}{−\mathrm{3}}\right)\mathrm{sin}\left[\left(\mathrm{1}−\frac{\mathrm{1}−\mathrm{1}}{−\mathrm{3}}\right)\frac{\pi}{\mathrm{2}}\right]}=\frac{\pi}{\mathrm{3}} \\ $$

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