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Question Number 148816 by ArielVyny last updated on 31/Jul/21

$${\int }_0^{\mathrm{\infty }}{x}^m{e}^{{ix}^n}dx=??$$

Answered by mathmax by abdo last updated on 01/Aug/21

$${\mathrm{A}}_{\mathrm{m}}={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{m}}{\mathrm{e}}^{{\mathrm{ix}}^{\mathrm{n}}}\mathrm{dx}\mathrm{changement}{\mathrm{ix}}^{\mathrm{n}}=-\mathrm{z}\mathrm{give}-{\mathrm{x}}^{\mathrm{n}}=-\mathrm{iz}\Rightarrow$$$${\mathrm{x}}^{\mathrm{n}}=\mathrm{iz}\Rightarrow \mathrm{x}={\mathrm{i}}^{\frac{1}{\mathrm{n}}}{\mathrm{z}}^{\frac{1}{\mathrm{n}}}\Rightarrow {\mathrm{A}}_{\mathrm{m}}={\int }_0^{\mathrm{\infty }}{\mathrm{i}}^{\frac{\mathrm{m}}{\mathrm{n}}}{\mathrm{z}}^{\frac{\mathrm{m}}{\mathrm{n}}}{\mathrm{e}}^{-\mathrm{z}}\frac{1}{\mathrm{n}}{\mathrm{i}}^{\frac{1}{\mathrm{n}}}{\mathrm{z}}^{\frac{1}{\mathrm{n}}-1}\mathrm{dz}$$$$={\mathrm{i}}^{\frac{\mathrm{m}+1}{\mathrm{n}}}{\int }_0^{\mathrm{\infty }}{\mathrm{z}}^{\frac{\mathrm{m}+1}{\mathrm{n}}-1}{\mathrm{e}}^{-\mathrm{z}}\mathrm{dz}\mathrm{but}\mathrm{\Gamma }\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{x}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}(\mathrm{x}>0)\Rightarrow$$$${\mathrm{A}}_{\mathrm{m}}={\left({\mathrm{e}}^{\frac{\mathrm{i}\pi }{2}}\right)}^{\frac{\mathrm{m}+1}{\mathrm{n}}}\mathrm{\Gamma }\left(\frac{\mathrm{m}+1}{\mathrm{n}}\right)={\mathrm{e}}^{\frac{\mathrm{i}\pi (\mathrm{m}+1)}{2\mathrm{n}}}\times \mathrm{\Gamma }\left(\frac{\mathrm{m}+1}{\mathrm{n}}\right)$$

Commented by ArielVyny last updated on 01/Aug/21

$$nicesirbutyouhaveforgot\frac{1}{n}ifyoulook$$$$secondlinethentheansweris$$$${\int }_0^{\mathrm{\infty }}{x}^m{e}^{{ix}^n}dx=\frac{1}{n}\mathrm{\Gamma }\left(\frac{m+1}{n}\right)e^{\frac{i\pi (m+1)}{2n}}$$

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