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Question Number 143628 by ArielVyny last updated on 16/Jun/21

$${\int }_x^{\propto }{t}^{\alpha -1}{e}^{it}dt=??$$

Answered by mathmax by abdo last updated on 16/Jun/21

$${\int }_{\mathrm{x}}^{\mathrm{\infty }}{\mathrm{t}}^{\alpha -1}{\mathrm{e}}^{\mathrm{it}}\mathrm{dt}{=}_{\mathrm{it}=-\mathrm{z}}{\int }_{-\mathrm{ix}}^{-\mathrm{i}\mathrm{\infty }}{\left(\mathrm{iz}\right)}^{\alpha -1}{\mathrm{e}}^{-\mathrm{z}}\frac{-\mathrm{dz}}{\mathrm{i}}$$$$=-{\mathrm{i}}^{\alpha }{\int }_{-\mathrm{ix}}^{-\mathrm{i}\mathrm{\infty }}{\mathrm{z}}^{\alpha -1}{\mathrm{e}}^{-\mathrm{z}}\mathrm{dz}=-{\mathrm{e}}^{\frac{\mathrm{i}\pi \alpha }{2}}{\int }_{-\mathrm{ix}}^{-\mathrm{i}\mathrm{\infty }}{\mathrm{z}}^{\alpha -1}{\mathrm{e}}^{-\mathrm{z}}\mathrm{dz}$$

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