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

Beginner2022-01-03Added 37 answers

There is some M large enough so that

$\left|\frac{{x}^{6}}{{e}^{x}}\right|<1$

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.

for

Then, by monotonicity of integration, we have

Of course,

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

Beginner2022-01-04Added 39 answers

where

karton

Expert2022-01-09Added 613 answers

In explicit terms, the convexity inequality

This argument proves

by exploiting the log-convexity of the LHS with respect to the n-variable

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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