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Question Number 137656 by rs4089 last updated on 05/Apr/21

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

Answered by Dwaipayan Shikari last updated on 05/Apr/21

$${\int }_0^{\mathrm{\infty }}sin\left(x^p\right)dx{x}^p=u$$$$=\frac{1}{p}{\int }_0^{\mathrm{\infty }}{u}^{\frac{1}{p}-1}sin\left(u\right)du=\frac{1}{2ip}{\int }_0^{\mathrm{\infty }}{u}^{\frac{1}{p}-1}{e}^{iu}-u^{\frac{1}{p}-1}{e}^{-iu}du$$$$=\frac{1}{2ip}{\int }_0^{\mathrm{\infty }}i{\left(ig\right)}^{\frac{1}{p}-1}dg-{(-i)}^{\frac{1}{p}}{j}^{\frac{1}{p}-1}{e}^{-j}dj$$$$=\frac{1}{2ip}\mathrm{\Gamma }\left(\frac{1}{p}\right)\left(2isin\frac{\pi }{2p}\right)=\mathrm{\Gamma }\left(\frac{p+1}{p}\right)sin\left(\frac{\pi }{2p}\right)$$

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