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.
Best Answer
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.