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Question Number 66170 by mathmax by abdo last updated on 10/Aug/19
calculate ∫_0 ^∞ sin(x^3 )dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:{sin}\left({x}^{\mathrm{3}} \right){dx} \\ $$
Commented by mathmax by abdo last updated on 10/Aug/19
let I =∫_0 ^∞ sin(x^3 )dx ⇒I =−Im(∫_0 ^∞ e^(−ix^3 ) dx) changement ix^3 =t give x^3 =−it ⇒x=(−it)^(1/3) =(−i)^(1/3) t^(1/3) ⇒dx =(1/3)(−i)^(1/3) t^((1/3)−1) ⇒∫_0 ^∞ e^(−ix^3 ) dx=(1/3)(−i)^(1/3) ∫_0 ^∞ e^(−t) t^((1/3)−1) dt =(1/3)(e^(−((iπ)/2)) )^(1/3) Γ((1/3)) =(1/3)e^(−((iπ)/6)) .Γ((1/3)) =(1/3)Γ((1/3))(((√3)/2)−(i/2)) ⇒ Im(∫_0 ^∞ e^(−ix^3 ) dx) =−(1/6)Γ((1/3)) ⇒ ∫_0 ^∞ sin(x^3 )dx =(1/6)Γ((1/3))
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{3}} \right){dx}\:\Rightarrow{I}\:=−{Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ix}^{\mathrm{3}} } {dx}\right) \\ $$$${changement}\:{ix}^{\mathrm{3}} \:={t}\:{give}\:{x}^{\mathrm{3}} =−{it}\:\Rightarrow{x}=\left(−{it}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\left(−{i}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:{t}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\Rightarrow{dx}\:=\frac{\mathrm{1}}{\mathrm{3}}\left(−{i}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{ix}^{\mathrm{3}} } {dx}=\frac{\mathrm{1}}{\mathrm{3}}\left(−{i}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \int_{\mathrm{0}} ^{\infty} {e}^{−{t}} \:\:{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left({e}^{−\frac{{i}\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\frac{\mathrm{1}}{\mathrm{3}}{e}^{−\frac{{i}\pi}{\mathrm{6}}} .\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\frac{\mathrm{1}}{\mathrm{3}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{{i}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ix}^{\mathrm{3}} } {dx}\right)\:=−\frac{\mathrm{1}}{\mathrm{6}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{3}} \right){dx}\:=\frac{\mathrm{1}}{\mathrm{6}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$

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