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Question Number 122137 by sdfg last updated on 14/Nov/20

Answered by Bird last updated on 14/Nov/20

$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{xt-e^t}dt{=}_{e^t=u}{\int }_0^{\mathrm{\infty }}{e}^{-u}{u}^x\frac{du}{u}$$$$={\int }_0^{\mathrm{\infty }}{u}^{x-1}{e}^{-u}du=\mathrm{\Gamma }\left(x\right)withx>0$$

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