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Question Number 151599 by mathdanisur last updated on 22/Aug/21

$$\mathrm{\Omega }=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\cos \left({\mathrm{x}}^{\mathbf{n}}\right)\mathrm{dx}=?$$

Answered by Lordose last updated on 22/Aug/21

$$$$$$\mathrm{\Omega }\stackrel{\mathrm{x}={\mathrm{x}}^{\frac{1}{\mathrm{n}}}}{=}\mathfrak{Re}\frac{1}{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\frac{1}{\mathrm{n}}-1}{\mathrm{e}}^{-\mathbf{i}\mathrm{x}}\mathrm{dx}\stackrel{\mathrm{x}=\frac{\mathrm{x}}{\mathrm{i}}}{=}\mathfrak{Re}(\frac{1}{\mathrm{n}}\cdot {\mathbf{i}}^{-\frac{1}{\mathrm{n}}}{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\frac{1}{\mathrm{n}}-1}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx})$$$$\mathrm{\Omega }=\mathfrak{Re}(\frac{1}{\mathrm{n}}\cdot {\mathbf{i}}^{-\frac{1}{\mathrm{n}}}\mathbf{\Gamma }\left(\frac{1}{\mathrm{n}}\right))=\frac{1}{\mathbf{n}}\mathbf{\Gamma }\left(\frac{1}{\mathbf{n}}\right)\mathbf{cos}\left(\frac{\mathit{\pi }}{2\mathbf{n}}\right)$$

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