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Question Number 124920 by mathmax by abdo last updated on 07/Dec/20

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{x}}^{\mathrm{n}}}\mathrm{dx}$$

Answered by Dwaipayan Shikari last updated on 07/Dec/20

$${\int }_0^{\mathrm{\infty }}{e}^{-x^n}dx{x}^n=u\Rightarrow {nx}^{n-1}=\frac{du}{dx}$$$$\Rightarrow \frac{1}{n}{\int }_0^{\mathrm{\infty }}{x}^{1-n}{e}^{-u}du=\frac{1}{n}{\int }_0^{\mathrm{\infty }}{u}^{\frac{1-n}{n}}{e}^{-u}du=\frac{1}{n}\mathrm{\Gamma }\left(\frac{1}{n}\right)=\mathrm{\Gamma }\left(\frac{n+1}{n}\right)$$

Answered by mathmax by abdo last updated on 07/Dec/20

$${\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{x}}^{\mathrm{n}}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}={\mathrm{t}}^{\frac{1}{\mathrm{n}}}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\times \frac{1}{\mathrm{n}}{\mathrm{t}}^{\frac{1}{\mathrm{n}}-1}\mathrm{dt}=\frac{1}{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\frac{1}{\mathrm{n}}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$=\frac{1}{\mathrm{n}}\times \mathrm{\Gamma }\left(\frac{1}{\mathrm{n}}\right)$$

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