For what values of $a$ does the function $f(x) = x^a e^{-x}$ converge as $x \rightarrow 0$? It turns out that the answer is $a<-1$ but I don't understand how to arrive at the answer. Any help is appreciated.
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Well, as $e^{-x}$ tends to $1$, you have to have convergence of $x^a$, meaning simply $a\geq0$. Or you must have made a mistake somewhere.
Bcpicao
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Does that mean $\int_0^1x^a e^{-x}dx$ also converge for all $a \geq 0$? – ForumWhiner Mar 27 '20 at 20:07
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Ah, now I get what the question is about! The answer is yes, although think carefully about what you're saying. If the function converges, then $0$ is not an improper bound and you're just integrating a continuous function over a segment. The converse is not true, the fact that that integral converges does not imply that $a\geq0$. Think of this: what is $\int_{0}^{1} \frac{1}{x} dx$ equal to, and why? – Bcpicao Mar 27 '20 at 20:22