kloseyq

2022-01-02

The question is: using the comparison test for improper integrals state if
${\int }_{0}^{\mathrm{\infty }}{x}^{4}{e}^{-x}dx$ converges or not.

Neil Dismukes

There is some M large enough so that
$|\frac{{x}^{6}}{{e}^{x}}|<1$
for $x>M$, since by LHopitals the $\underset{x\to 0}{lim}\frac{{x}^{6}}{{e}^{x}}=0$
Then, by monotonicity of integration, we have
${\int }_{M}^{\mathrm{\infty }}{e}^{-x}{x}^{4}dx\le {\int }_{M}^{\mathrm{\infty }}{e}^{-x}{x}^{-2}dx<\mathrm{\infty }$
Of course,
${\int }_{0}^{M}{e}^{-x}{x}^{4}dx$
poses no problem as the integrand is bounded and the interval is finite. Note also you could integrate this by parts without too much trouble and evaluate the integral directly.

Philip Williams

${\int }_{0}^{\mathrm{\infty }}{x}^{4}{e}^{-x}dx={\int }_{0}^{M}{x}^{4}{e}^{-x}dx+{\int }_{M}^{\mathrm{\infty }}{x}^{4}{e}^{-x}dx$
$\le {\int }_{0}^{M}{x}^{4}{e}^{-x}dx+{\int }_{M}^{\mathrm{\infty }}{x}^{4}{x}^{-6}dx$
where ${e}^{-x}<{x}^{6}$ for all $x\ge M$

karton

In explicit terms, the convexity inequality ${e}^{x}\ge 1+x$ implies that for any $\ge 0$ we have ${e}^{x/6}\ge 1+\frac{x}{6}$ and ${e}^{x}\ge \left(1+\frac{x}{6}{\right)}^{6}$. In particular
$0\le {\int }_{0}^{+\mathrm{\infty }}{x}^{4}{e}^{-x}dx\le {\int }_{0}^{+\mathrm{\infty }}\frac{{x}^{4}}{\left(1+\frac{x}{6}{\right)}^{6}}dx={6}^{5}{\int }_{0}^{+\mathrm{\infty }}\frac{{z}^{4}dz}{\left(1+z{\right)}^{6}}=\frac{{6}^{5}}{5}$
This argument proves $n!\le \frac{\left(n+2{\right)}^{n+1}}{n+1}$. One may prove much better, like
${\int }_{0}^{+\mathrm{\infty }}{x}^{n}{e}^{-x}dx\le \frac{\left(n+1{\right)}^{n+1}}{{e}^{n}}$
by exploiting the log-convexity of the LHS with respect to the n-variable

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