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Question Number 124389 by Engr_Jidda last updated on 03/Dec/20

Answered by Dwaipayan Shikari last updated on 03/Dec/20

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

Commented by Engr_Jidda last updated on 03/Dec/20

$$\mathrm{thank}\mathrm{you}?\mathrm{sir}$$

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