Comparison tests for improper integrals

Question

The question is: using the comparison test for improper integrals state if

$\int _0^{\infty \:}x^4e^{-x}dx$ converges or not.

Hint: prove that $e^{-x}$ < $x^{-6}$ if x is large enough.

I'm not sure how to prove the hint and I don't particularly see how this would help... any hints would be great!

Answer 1

There is some $M$ large enough so that $$ \left|\frac{x^6}{e^x}\right|<1 $$ for $x>M$, since by L'Hopital's the $\lim_{x\to 0}\frac{x^6}{e^x}=0$.

Then, by monotonicity of integration, we have $$ \int_{M}^\infty e^{-x}x^4\;\mathrm dx\leq \int_{M}^\infty e^{-x}x^{-2}\;\mathrm dx<\infty $$

Of course, $$ \int_0^M e^{-x}x^4\;\mathrm dx $$ 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.

Answer 2

\begin{align*} \int_{0}^{\infty}x^{4}e^{-x}dx&=\int_{0}^{M}x^{4}e^{-x}dx+\int_{M}^{\infty}x^{4}e^{-x}dx\\ &\leq\int_{0}^{M}x^{4}e^{-x}dx+\int_{M}^{\infty}x^{4}x^{-6}dx\\ &<\infty, \end{align*} where $e^{-x}<x^{6}$ for all $x\geq M$.

Answer 3

In explicit terms, the convexity inequality $e^x\geq 1+x$ implies that for any $x\geq 0$ we have $e^{x/6}\geq 1+\frac{x}{6}$ and $e^x\geq \left(1+\frac{x}{6}\right)^6$. In particular

$$ 0\leq \color{red}{\int_{0}^{+\infty}x^4 e^{-x}\,dx\leq} \int_{0}^{+\infty}\frac{x^4}{\left(1+\frac{x}{6}\right)^6}\,dx = 6^5\int_{0}^{+\infty}\frac{z^4\,dz}{(1+z)^6}=\color{red}{\frac{6^5}{5}}.$$ This argument proves $n!\leq\frac{(n+2)^{n+1}}{n+1}$. One may prove much better, like $$ \int_{0}^{+\infty} x^n e^{-x}\,dx \leq \frac{(n+1)^{n+1}}{e^n}$$ by exploiting the log-convexity of the LHS with respect to the $n$-variable.