Lillianna Andersen

2022-07-07

Prove that $\frac{{\int }_{0}^{\mathrm{\infty }}{x}^{n}{e}^{-{x}^{n}}dx}{{\int }_{0}^{\mathrm{\infty }}{e}^{-{x}^{n}}dx}=\frac{1}{n}$

Johnathan Morse

It suffices to prove that
${\int }_{0}^{\mathrm{\infty }}\left({x}^{n}-1/n\right){\mathrm{e}}^{-{x}^{n}}\mathrm{d}x=0$
or
${\int }_{0}^{\mathrm{\infty }}{\left[-\frac{1}{n}x{\mathrm{e}}^{-{x}^{n}}\right]}^{\prime }\mathrm{d}x=0$
or
${\int }_{0}^{\mathrm{\infty }}{\left[-\frac{1}{n}x{\mathrm{e}}^{-{x}^{n}}\right]}^{\prime }\mathrm{d}x=0$
or
$\left[-\frac{1}{n}x{\mathrm{e}}^{-{x}^{n}}\right]{|}_{0}^{\mathrm{\infty }}=0$
which is true.

Savanah Boone

Consider
${\int }_{0}^{\mathrm{\infty }}{x}^{\alpha }{e}^{-{x}^{\beta }}\phantom{\rule{thinmathspace}{0ex}}dx$
with $\alpha \ge 0,\beta >0. Lettingy={x}^{\beta }gives$
$\begin{array}{r}{\int }_{0}^{\mathrm{\infty }}{x}^{\alpha }{e}^{-{x}^{\beta }}\phantom{\rule{thinmathspace}{0ex}}dx=\frac{1}{\beta }{\int }_{0}^{\mathrm{\infty }}{y}^{\frac{\alpha +1}{\beta }-1}{e}^{-y}\phantom{\rule{thinmathspace}{0ex}}dy=\frac{1}{\beta }\mathrm{\Gamma }\left(\frac{\alpha +1}{\beta }\right)\end{array}$
Where $\mathrm{\Gamma }$ is the Euler-Gamma function. Hence,
$\frac{{\int }_{0}^{\mathrm{\infty }}{x}^{n}{e}^{-{x}^{n}}\phantom{\rule{thinmathspace}{0ex}}dx}{{\int }_{0}^{\mathrm{\infty }}{e}^{-{x}^{n}}\phantom{\rule{thinmathspace}{0ex}}dx}=\frac{\mathrm{\Gamma }\left(\frac{1}{n}+1\right)}{\mathrm{\Gamma }\left(\frac{1}{n}\right)}=\frac{1}{n}$
for all $n>0$

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