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Question Number 82867 by mind is power last updated on 24/Feb/20

$$helloprovethat{\int }_0^{+\mathrm{\infty }}sin\left(x^4\right)dx=sin\left(\frac{\pi }{8}\right){\int }_0^{+\mathrm{\infty }}{e}^{-x^4}dx?$$$$verrynicedayGoodBlessYou$$

Commented by abdomathmax last updated on 25/Feb/20

$${\int }_0^{\mathrm{\infty }}sin\left(x^4\right)dx=Im\left({\int }_0^{\mathrm{\infty }}{e}^{{ix}^4}dx\right)butchsngement$$$${ix}^4=-t^{4}give{x}^4={it}^4\Rightarrow x=i^{\frac{1}{4}}t\Rightarrow$$$${\int }_0^{\mathrm{\infty }}{e}^{{ix}^4}dx={\int }_0^{\mathrm{\infty }}{e}^{-t^4}{i}^{\frac{1}{4}}dt={\left(e^{\frac{i\pi }{2}}\right)}^{\frac{1}{4}}{\int }_0^{\mathrm{\infty }}{e}^{-t^4}dt$$$$=e^{\frac{i\pi }{8}}{\int }_0^{\mathrm{\infty }}{e}^{-t^4}dt=(cos\left(\frac{\pi }{8}\right)+isin\left(\frac{\pi }{8}\right)){\int }_0^{\mathrm{\infty }}{e}^{-t^4}dt\Rightarrow$$$${\int }_0^{\mathrm{\infty }}sin\left(x^4\right)dx=sin\left(\frac{\pi }{8}\right){\int }_0^{\mathrm{\infty }}{e}^{-t^4}dt$$$$alsochang.t^4=ugivet=u^{\frac{1}{4}}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}{e}^{-t^4}dt={\int }_0^{\mathrm{\infty }}{e}^{-u}\frac{1}{4}{u}^{\frac{1}{4}-1}du$$$$=\frac{1}{4}\mathrm{\Gamma }\left(\frac{1}{4}\right)\Rightarrow {\int }_0^{\mathrm{\infty }}sin\left(x^4\right)dx=\frac{1}{4}sin\left(\frac{\pi }{8}\right)\mathrm{\Gamma }\left(\frac{1}{4}\right)$$$$=\frac{\sqrt{2-\sqrt{2}}}{8}\times \mathrm{\Gamma }\left(\frac{1}{4}\right)$$

Commented by mind is power last updated on 25/Feb/20

$$nicesolutionSirthankyou$$

Commented by msup trace by abdo last updated on 25/Feb/20

$$youarewelcome$$

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