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Question Number 34910 by abdo imad last updated on 12/May/18

$$find{J}_{n,p}={\int }_0^{\mathrm{\infty }}{x}^n{e}^{-\frac{x^2}{p}}dxwithp>0andnintegr$$

Commented bycandre last updated on 13/May/18

$$u=x^{n-1}\Rightarrow du=(n-1)x^{n-2}dx$$ $$dv={xe}^{-\frac{x^2}{p}}dx\Rightarrow v=\int {xe}^{-\frac{x^2}{p}}dx$$ $$w=-\frac{x^2}{p}\Rightarrow dw=-\frac{2x}{p}dx$$ $$v=-\frac{p}{2}\int e^{w}dw=-\frac{p}{2}{e}^{-\frac{x^2}{p}}$$ $$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{x}^n{e}^{-\frac{x^2}{p}}dx=-{\left[\frac{{px}^{n-1}{e}^{-\frac{x^2}{p}}}{2}\right]}_0^{\mathrm{\infty }}+\frac{(n-1)p}{2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{x}^{n-2}{e}^{-\frac{x^2}{p}}dx$$ $$J_{n,p}=\frac{(n-1){pJ}_{n-2,p}}{2}$$

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