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I'm trying to solve a improper integral exercise. It follows:

Using the comparison test determine whether the following integral converges or diverges

$$\displaystyle \int_{1}^{\infty} x^3 \cdot e^{-x} \ dx$$

Given that the answer is convergent, I assume I need to compare it to a function that is greater than the given one (that will allow me to conclude that, if the new one is convergent, the given one will also be convergent). I am having a hard time coming up with a function to compare the given one with. Using for example $g(x) = x^3$ would definitely not work (as it is divergent).

Any help is highly appreciated.

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Hint: Show that $x^3\leq e^{x/2}$ for sufficiently large $x$ (you can explicitly determine which) and conclude by taking the “flatten” out function.

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  • $\begingroup$ The point here is that if $f$ and $g$ are locally integrable (it suffices to check that they are continuous), then $\int_0^\infty f$ exists if $\int_0^\infty g$ exists as long as $f(x) \leq g(x)$ for all $x \geq M$. The upshot is that you can be pretty loose with your choice of comparison function as long as it's eventually bigger. $\endgroup$ Sep 10 '20 at 15:34
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we see that $\displaystyle \lim_{x\to+\infty} \; x^2. x^3e^{-x}= 0$, so $\exists \,A \geqslant0 $ s.t : $x^5 e^{-x} \leqslant 1$ for all $x> A.$ then $x^3e^{-x} \leqslant x^3e^{-x} \;\chi_{[0,A]} \;+\dfrac{1}{x^2}\chi_{]A,+\infty[} $

$\displaystyle \int_1^{+\infty} x^3e^{-x}\; \chi_{[0,A]}\;\, dx =\int_1^Ax^3e^{-x}dx $ and $\displaystyle \int_1^{+\infty} \dfrac{1}{x^2}dx $ converges, then $\displaystyle \int_1^{+\infty} x^3e^{-x}\;dx$ converges.

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