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Question Number 122109 by sdfg last updated on 14/Nov/20

Answered by Dwaipayan Shikari last updated on 14/Nov/20

$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt$$$$\mathrm{\Gamma }\left(x\right)={\int }_1^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt+{\int }_0^1{t}^{x-1}{e}^{-t}dt$$$$={\int }_1^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt+\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\int }_0^1\frac{t^{n+x-1}}{n!}dt$$$$={\int }_1^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n!(n+x)}$$

Answered by Dwaipayan Shikari last updated on 14/Nov/20

$$s^x{\int }_0^{\mathrm{\infty }}{e}^{-st}{t}^{x-1}dx$$$$={\mathit{s}}^{\mathit{x}}{\int }_0^{\mathrm{\infty }}\frac{1}{\mathit{s}}{\mathit{e}}^{-\mathit{u}}{\left(\frac{\mathit{u}}{\mathit{s}}\right)}^{\mathit{x}-1}\mathit{d}\mathit{u}\mathit{s}\mathit{t}=\mathit{u}\Rightarrow \mathit{s}=\frac{\mathit{d}\mathit{u}}{\mathit{d}\mathit{x}}$$$$={\int }_0^{\mathrm{\infty }}{\mathit{e}}^{-\mathit{u}}{\mathit{u}}^{\mathit{x}-1}\mathit{d}\mathit{u}\equiv \mathbf{\Gamma }\left(\mathit{x}\right)$$

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