Convergence for improper integral $\int_0^\infty x^re^{-x} \,dx$

Question

I'm trying to find for which values $r$ the following improper integral converges. $$\int_0^\infty x^re^{-x}\, dx$$ What I have so far is that $x^r < e^{\frac{1}{2}x}$ for $x \geq a$, which splits the integral into $$\int_0^a x^re^{-x} \,dx + \int_a^\infty e^{-\frac{1}{2}x} \, dx$$ We know the latter interval converges, but I don't know what to do with the first one. (For reference, graphing the functions reveals the answer to be $x > -1$.)

Edit: I would like a proof without the gamma function. Preferably one that uses the comparison test to compare limits.

Answer 1

Note that for all $r \in \mathbb{R}$,

$$\lim_{x \to \infty} \frac{x^r e^{-x}}{x^{-2}} = \lim_{x \to \infty} \frac{x^{r+2}}{e^x} = 0 $$

since the exponential function tends to infinity faster than any polynomial. Hence, by the limit comparison test the integral over $[1, \infty)$ converges since $\int_1^\infty x^{-2}\, dx = 1$.

See if you can finish by finding the condition on $r$ such that the integral over $[0,1]$ converges.

Answer 2

\begin{align} & \text{For small enough }a>0, \\[8pt] & x^r > x^r e^{-x} > 0.9 x^r \text{ when } 0<x<a. \end{align} Therefore $\displaystyle \int_0^a x^r e^{-x} \, dx$ converges for precisely those values of $r$ for which $\displaystyle\int_0^a x^r\,dx$ converges.