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Question Number 91621 by abdomathmax last updated on 02/May/20

$$calculate{\int }_0^{\mathrm{\infty }}sin\left(x^6\right)dx$$

Commented by mathmax by abdo last updated on 02/May/20

$${\int }_0^{\mathrm{\infty }}sin\left(x^6\right)dx=Im\left({\int }_0^{\mathrm{\infty }}{e}^{{ix}^6}dx\right)wehavebychangement{ix}^6=-t$$$$\Rightarrow x^6=it\Rightarrow x={\left(it\right)}^{\frac{1}{6}}=e^{\frac{i\pi }{12}}{t}^{\frac{1}{6}}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}{e}^{{ix}^6}dx=e^{\frac{i\pi }{12}}{\int }_0^{\mathrm{\infty }}{e}^{-t}\times \frac{1}{6}{t}^{\frac{1}{6}-1}dt$$$$=\frac{1}{6}{e}^{\frac{i\pi }{12}}{\int }_0^{\mathrm{\infty }}{t}^{\frac{1}{6}-1}{e}^{-t}dtwehave\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt$$$$=\frac{1}{6}\mathrm{\Gamma }\left(\frac{1}{6}\right)(cos\left(\frac{\pi }{12}\right)+isin\left(\frac{\pi }{12}\right))\Rightarrow {\int }_0^{\mathrm{\infty }}sin\left(x^6\right)dx=\frac{1}{6}\mathrm{\Gamma }\left(\frac{1}{6}\right)sin\left(\frac{\pi }{12}\right)$$

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