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Question Number 110175 by mathdave last updated on 27/Aug/20
solve the integral ∫_(−∞) ^(+∞) ((sinx)/( (x)^(1/3) ))dx
$${solve}\:{the}\:{integral} \\ $$$$\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}{x}}{\:\sqrt[{\mathrm{3}}]{{x}}}{dx} \\ $$
Answered by mathmax by abdo last updated on 28/Aug/20
I =∫_(−∞) ^(+∞) ((sinx)/x^(1/3) )dx =2 ∫_0 ^∞ ((sinx)/x^(1/3) )dx =−2 Im(∫_0 ^∞ x^(−(1/3)) e^(−ix) dx) changement ix =t give −x =it ⇒x =−it ⇒ ∫_0 ^∞ x^(−(1/3)) e^(−ix) dx =∫_0 ^∞ (−it)^(−(1/3)) e^(−t) (−i)dt =(−i)^(−(1/3)+1) ∫_0 ^∞ t^(−(1/(3 ))) e^(−t) dt =(e^(−((iπ)/2)) )^(2/3) ∫_0 ^∞ t^((2/3)−1) e^(−t) dt =e^(−((iπ)/3)) Γ((2/3)) =Γ((2/3)).(cos((π/3))−isin((π/3))) ⇒ I =2(Γ((2/3)) sin((π/3)) =(√3)×Γ((2/3))
$$\mathrm{I}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{sinx}}{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }\mathrm{dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sinx}}{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }\mathrm{dx}\:=−\mathrm{2}\:\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{x}^{−\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{e}^{−\mathrm{ix}} \mathrm{dx}\right) \\ $$$$\mathrm{changement}\:\:\mathrm{ix}\:=\mathrm{t}\:\mathrm{give}\:−\mathrm{x}\:=\mathrm{it}\:\Rightarrow\mathrm{x}\:=−\mathrm{it}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{−\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{e}^{−\mathrm{ix}} \mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \left(−\mathrm{it}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{e}^{−\mathrm{t}} \:\left(−\mathrm{i}\right)\mathrm{dt} \\ $$$$=\left(−\mathrm{i}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{−\frac{\mathrm{1}}{\mathrm{3}\:}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:\:=\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} \mathrm{e}^{−\mathrm{t}} \:\mathrm{dt} \\ $$$$=\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:=\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right).\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}\right)−\mathrm{isin}\left(\frac{\pi}{\mathrm{3}}\right)\right)\:\Rightarrow \\ $$$$\mathrm{I}\:=\mathrm{2}\left(\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:\mathrm{sin}\left(\frac{\pi}{\mathrm{3}}\right)\:=\sqrt{\mathrm{3}}×\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right. \\ $$
Answered by mathdave last updated on 28/Aug/20
solution let I=∫_(−∞) ^(+∞) ((sinx)/( (x)^(1/3) ))dx=2∫_0 ^∞ ((sinx)/x^(1/3) )dx from maz identity integral ∫_0 ^∞ ((sin(ax))/x^n )dx=((πa^(n−1) )/(2Γ(n)sin(((nπ)/2)))) taking a=1 and n=(1/3) ∵I=2∫_0 ^∞ ((sin)/x^(1/3) )dx=2•((π×1^((1/3)−1) )/(2Γ((1/3))sin((π/6))))=(π/((1/2)Γ((1/3)))) ∵∫_(−∞) ^(+∞) ((sinx)/( (x)^(1/3) ))dx=((2π)/(Γ((1/3))))=2.3454 by mathdave(28/08/2020)
$${solution} \\ $$$${let}\:{I}=\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}{x}}{\:\sqrt[{\mathrm{3}}]{{x}}}{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{x}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }{dx} \\ $$$${from}\:{maz}\:{identity}\:{integral} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left({ax}\right)}{{x}^{{n}} }{dx}=\frac{\pi{a}^{{n}−\mathrm{1}} }{\mathrm{2}\Gamma\left({n}\right)\mathrm{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)} \\ $$$${taking}\:{a}=\mathrm{1}\:\:{and}\:\:{n}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\because{I}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }{dx}=\mathrm{2}\bullet\frac{\pi×\mathrm{1}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} }{\mathrm{2}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}\right)}=\frac{\pi}{\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$$\because\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}{x}}{\:\sqrt[{\mathrm{3}}]{{x}}}{dx}=\frac{\mathrm{2}\pi}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}=\mathrm{2}.\mathrm{3454} \\ $$$${by}\:{mathdave}\left(\mathrm{28}/\mathrm{08}/\mathrm{2020}\right) \\ $$

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