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Question Number 144926 by ArielVyny last updated on 30/Jun/21

$${\int }_0^{\mathrm{\infty }}{x}^n{\left(e^{ix}\right)}^{z}dx=???(z\in \mathbb{C})$$

Answered by mathmax by abdo last updated on 30/Jun/21

$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}{\left({\mathrm{e}}^{\mathrm{ix}}\right)}^{\mathrm{z}}\mathrm{dx}\mathrm{let}\mathrm{z}=\mathrm{a}+\mathrm{ib}\Rightarrow$$$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}\left({\mathrm{e}}^{\mathrm{ix}(\mathrm{a}+\mathrm{ib})}\right)\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{iax}-\mathrm{bx}}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}{\mathrm{e}}^{-(\mathrm{b}-\mathrm{ia})\mathrm{x}}\mathrm{dx}{=}_{(\mathrm{b}-\mathrm{ia})\mathrm{x}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{n}}}{{(\mathrm{b}-\mathrm{ia})}^{\mathrm{n}}}{\mathrm{e}}^{-\mathrm{t}}\frac{\mathrm{dt}}{\mathrm{b}-\mathrm{ia}}$$$$=\frac{1}{{(\mathrm{b}-\mathrm{ia})}^{\mathrm{n}+1}}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}=\frac{\mathrm{\Gamma }(\mathrm{n}+1)}{{(\mathrm{b}-\mathrm{ia})}^{\mathrm{n}+1}}$$$$\mathrm{we}\mathrm{have}-\mathrm{iz}=-\mathrm{ia}+\mathrm{b}\Rightarrow \mathrm{\Psi }=\frac{\mathrm{\Gamma }(\mathrm{n}+1)}{{(-\mathrm{iz})}^{\mathrm{n}+1}}=\frac{\mathrm{\Gamma }(\mathrm{n}+1)}{{\mathrm{e}}^{-\mathrm{i}\frac{(\mathrm{n}+1)\pi }{2}}{\mathrm{z}}^{\mathrm{n}+1}}$$$$={\mathrm{e}}^{\frac{\mathrm{i}(\mathrm{n}+1)\pi }{2}}\times \frac{\mathrm{\Gamma }(\mathrm{n}+1)}{{\mathrm{z}}^{\mathrm{n}+1}}$$

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