Question-136806

Question Number 136806 by mnjuly1970 last updated on 26/Mar/21

Answered by Dwaipayan Shikari last updated on 26/Mar/21

2∫_0 ^∞ ((sin((1/x^3 )))/( (x)^(1/3) ))dx (1/x^3 )=u⇒−(3/x^4 )=(du/dx) =(2/3)∫_0 ^∞ x^(4−(1/3)) sin(u)du =(2/3)∫_0 ^∞ ((sin(u))/u^((11)/9) )du=(π/(3Γ(((11)/9))sin(((11π)/(18))))) ∫_0 ^∞ ((sin(u))/u^m )du=(1/(Γ(m)))∫_0 ^∞ ∫_0 ^∞ e^(−ut) t^(m−1) sin(u)dtdu =(1/(2iΓ(m)))∫_0 ^∞ (t^(m−1) /(t−i))−(t^(m−1) /(t+i))dt=(1/(Γ(m)))∫_0 ^∞ (t^(m−1) /(t^2 +1))dt =(1/(2Γ(m)))∫_0 ^∞ (u^((m/2)−1) /((u+1)^((m/2)+1−(m/2)) ))du=(π/(2Γ(m)sin(((mπ)/2))))

$$\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)}{\:\sqrt[{\mathrm{3}}]{{x}}}{dx}\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }={u}\Rightarrow−\frac{\mathrm{3}}{{x}^{\mathrm{4}} }=\frac{{du}}{{dx}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}}} {sin}\left({u}\right){du} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({u}\right)}{{u}^{\frac{\mathrm{11}}{\mathrm{9}}} }{du}=\frac{\pi}{\mathrm{3}\Gamma\left(\frac{\mathrm{11}}{\mathrm{9}}\right){sin}\left(\frac{\mathrm{11}\pi}{\mathrm{18}}\right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({u}\right)}{{u}^{\boldsymbol{{m}}} }{du}=\frac{\mathrm{1}}{\Gamma\left({m}\right)}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {e}^{−{ut}} {t}^{{m}−\mathrm{1}} {sin}\left({u}\right){dtdu} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\Gamma\left({m}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{m}−\mathrm{1}} }{{t}−{i}}−\frac{{t}^{{m}−\mathrm{1}} }{{t}+{i}}{dt}=\frac{\mathrm{1}}{\Gamma\left({m}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{m}−\mathrm{1}} }{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\Gamma\left({m}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{{m}}{\mathrm{2}}−\mathrm{1}} }{\left({u}+\mathrm{1}\overset{\frac{{m}}{\mathrm{2}}+\mathrm{1}−\frac{{m}}{\mathrm{2}}} {\right)}}{du}=\frac{\pi}{\mathrm{2}\Gamma\left({m}\right){sin}\left(\frac{{m}\pi}{\mathrm{2}}\right)} \\ $$

Commented by mnjuly1970 last updated on 26/Mar/21

$${thank}\:{you}\:{so}\:{much}… \\ $$

Commented by Dwaipayan Shikari last updated on 26/Mar/21

$${I}\:{had}\:{a}\:{typo}.\:{Kindly}\:{recheck}! \\ $$

Answered by Ar Brandon last updated on 26/Mar/21

∅=∫_(−∞) ^∞ ((sin(x^(−3) ))/( (x)^(1/3) ))dx=2∫_0 ^∞ x^(−(1/3)) sin(x^(−3) )dx =2∙(1/(∣−3∣))∙(π/(2Γ(1−((−(1/3)+1)/(−3)))sin[(1−((−(1/3)+1)/(−3)))(π/2)])) =(1/3)∙(π/(Γ(((11)/9))sin(((11)/(18))π)))

$$\emptyset=\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{x}^{−\mathrm{3}} \right)}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}}}\mathrm{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \mathrm{x}^{−\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{sin}\left(\mathrm{x}^{−\mathrm{3}} \right)\mathrm{dx} \\ $$$$\:\:\:=\mathrm{2}\centerdot\frac{\mathrm{1}}{\mid−\mathrm{3}\mid}\centerdot\frac{\pi}{\mathrm{2}\Gamma\left(\mathrm{1}−\frac{−\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}}{−\mathrm{3}}\right)\mathrm{sin}\left[\left(\mathrm{1}−\frac{−\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}}{−\mathrm{3}}\right)\frac{\pi}{\mathrm{2}}\right]} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}\centerdot\frac{\pi}{\Gamma\left(\frac{\mathrm{11}}{\mathrm{9}}\right)\mathrm{sin}\left(\frac{\mathrm{11}}{\mathrm{18}}\pi\right)} \\ $$

Commented by mnjuly1970 last updated on 26/Mar/21

$$\:\:{grateful}… \\ $$$$ \\ $$

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