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

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 $$\int_0^a x^re^{-x} dx + \int_a^\infty e^{-\frac{1}{2}x}$$ 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.

• This is actually one of the more interesting integrals in mathematics. try plugging in $r = 1, 2, 3, 4...$ etc. What do you notice? – Aniruddh Venkatesan Jan 18 at 3:41
• Hint:$$\int\limits_0^{\infty}\mathrm dz\, x^n e^{-x}=n!=\Gamma(n+1)$$ – Frank W. Jan 18 at 3:42
• Did anybody even read the OP's question or do they just want to use the post to mention the gamma function? – Dionel Jaime Jan 18 at 3:49
• @AniruddhVenkatesan I know it equals $n!$ for natural numbers, but for this question I'm mainly interested in for which real values of $r$ the interval converges. – Chase K Jan 18 at 3:53
• @DionelJaime As I recall, the original question didn't mention not using the gamma function, with this being added afterwards. – John Omielan Jan 18 at 3:58

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.